A Reform Approach to Business Calculus
A Reform Approach to Business Calculus Marcel B. Finan Arkansas Tech University c °All Rights Reserved January 3, 2003
2
PREFACE This supplement consists of my lectures of a freshmen-level mathematics class offered at Arkansas Tech University. The lectures are designed to accompany the textbook ”Applied Calculus” written by the Harvard Consortium. The lectures cover sections from Chapters 1, 2, 3, 4, 5, 6, and 7. These chapters are basically well suited for a one semester course in Business Calculus. Marcel B. Finan January 2003
Contents 1 Library of Functions 1.1 What is a Function? . . . . . . . . . . . 1.2 Linear Functions . . . . . . . . . . . . . 1.3 Rate of Change . . . . . . . . . . . . . . 1.4 Applications of Functions to Economics 1.5 Exponential Functions . . . . . . . . . . 1.6 Logarithms and Their Properties . . . . 1.7 Exponential Growth and Decay . . . . . 1.8 Building New Functions from Old Ones 1.9 Polynomial and Rational Functions . . . 2 The 2.1 2.2 2.3 2.4 2.5
Derivative of a Function Instantaneous Rate of Change . . . . The derivative Function . . . . . . . Leibniz Notation for The Derivative The Second Derivative . . . . . . . . Marginal Cost and Revenue . . . . .
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5 5 11 15 18 21 28 33 36 49
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55 55 59 62 63 65
3 Rules of Differentiation 69 3.1 Derivative Formulas for Power and Polynomials . . . . . . . . . . 69 3.2 Derivative Formulas for Exponential and Logarithmic Functions . 71 3.3 Derivatives of Composite Functions: The Chain Rule . . . . . . . 73 3.4 The Product and Quotient Rules . . . . . . . . . . . . . . . . . . 74 4 Applications of the Derivative 4.1 Using First and Second Derivatives . . . . . 4.2 Concavity and Points of Inflection . . . . . 4.3 Global Maxima and Minima . . . . . . . . . 4.4 Applications of Optimization to Marginality 4.5 Average Cost . . . . . . . . . . . . . . . . . 4.6 Elasticity of Demand . . . . . . . . . . . . . 3
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77 77 80 82 84 87 89
4 5 The 5.1 5.2 5.3 5.4 5.5 5.6
CONTENTS Definite Integral Measuring The Distance Traveled . . . . The Definite Integral . . . . . . . . . . . The Definite Integral as Area . . . . . . Interpretations of the Definite Integral . The Fundamental Theorem of Calculus . The Definite Integral as an Average . .
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93 . 93 . 97 . 99 . 101 . 102 . 105
6 Antiderivatives and Indefinite Integrals 107 6.1 Finding Antiderivatives Graphically and Numerically . . . . . . . 107 6.2 Analytical Construction of Antiderivatives . . . . . . . . . . . . . 110
Chapter 1
Library of Functions This chapter lays the foundation for the topics that we will discuss in this course. It consists of a brief review of the following family of functions: linear, exponential, logarithmic, trigonometric, polynomial, and rational functions.
1.1
What is a Function?
Functions play a crucial role in mathematics. The goal of this section is to introduce this important concept. Before giving the definition of a function we first consider an application.
Example 1 The sales tax on an item is 6%. So if p denotes the price of the item and C the total cost of buying the item then if the item is sold at $ 1 then the cost is 1 + (0.06)(1) = $1.06. If the item is sold at $2 then the cost of buying the item is 2 + (0.06)(2) = $2.12 and so on. Thus we have a relationship between the quantities C and p such that each value of p determines exactly one value of C. In this case, we say that C is a function of p. The relationship between C and p can be expressed by a table, a graph, or a formula. The chart below gives the total cost of buying an item at price p as a function of p. p C
1 1.06
2 2.12
3 3.18
4 4.24
5 5.30
6 6.36
7 7.42
The graph of the function C is obtained by plotting the data in the above table to obtain 5
6
CHAPTER 1. LIBRARY OF FUNCTIONS 7 6 5 4 3 2 1
1
2
3
4
5
6
7
The formula that describes the relationship between C and p is given by C(p) = 1.06p. By a function we mean a relationship f between two quantities x and y such that each value of x yields a unique value y. We call x the input and y the output. In function notation we write y = f (x) Note that y depends on x so that we call x the independent variable and y the dependent variable. Exercise 1 Using the following table, determine which variable is a function of what variable. x y
17 73
18 77
19 69
20 73
21 75
22 75
23 70
Solution. Since each value of x determines exactly one value of y then y is a function of x. On the contrary, x is not a function of y since the input 73 has two output values 17 and 20. So x is not a function of y. Now, most of the functions that we will encounter in this course have formulas. For example, the area A of a circle is a function of its radius r. In function notation, we write A = πr2 . However, there are functions that can not be represented by a formula. For example, if D is a day in July and T is lowest temperature in that day, then T is a function of D. Note that T can not be expressed as a formula in D. Functions of this nature, are mostly represented by either a graph or a table of numerical data. Exercise 2 The table below shows the daily low temperature for a one-week period in New
1.1. WHAT IS A FUNCTION?
7
York City during July. (a) What was the low temperature on July 19? (b) When was the low temperature 73◦ F ? (c) Is the daily low temperature a function of the date?Explain. (d) Can you express T as a formula? D T
17 73
18 77
19 69
20 73
21 75
22 75
23 70
Solution. (a) The low temperature on July 19 was 69◦ F. (b) On July 17 and July 20 the low temperature was 73◦ F. (c) T is a function of D since each value of D determines exactly one value of T. (d) T can not be expressed as a formula in terms of D otherwise there would be no need for meteorologists. Next, we present a relationship between two quantities x and y that is not a function. Exercise 3 Let x and y be two quantities related by the equation x2 + y 2 = 4. (a) Is x a function of y? Explain. (b) Is y a function of x? Explain. Solution. (a) For y = 0 we have two values of x, namely, x = −2 and x = 2. So x is not a function of y. (b) For x = 0 we have two values of y, namely, y = −2 and y = 2. So y is not a function of x. Suppose that the graph of a relationship between two quantities x and y is given. To say that y is a function of x means that for each value of x there is exactly one value of y. Graphically, this means that each vertical line must intersect the graph at most once. Hence, to determine if a graph is a function one uses the following test: Vertical Line Test: A graph is a function if and only if every vertical line crosses the graph at most once. According to the vertical line test and the definition of a function, if a vertical line cuts the graph for example twice, the graph could not be the graph of a function since we have two y values for the same x − value and this violates the definition of a function.
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CHAPTER 1. LIBRARY OF FUNCTIONS
Exercise 4 Which of the graphs (a) through (i) represent y as a function of x? (Note that an open circle indicates a point that is not included in the graph; a solid dot indicates a point that is included in the graph.)
c
a
b
d
g
e
..
f
.. h
. . .. ..
j
..
Solution. (a), (d), and (g) are functions. The rest are not by the vertical line test. • Domain and Range of a Function If we√ try to find the possible input values that can be used in the function y = x − 2 we see that we must restrict x to the interval [2, ∞), that is x ≥ 2. Similarly, the function y = x12 takes only certain values for the output, namely, y > 0. The above discussion leads to the following definitions: By the domain of a function we mean all possible input values. Graphically, the domain is part of the horizontal axis. The range of a function is the collection of all possible output values. The range is part of the vertical axis. The domain and range of a function can be found either algebraically or graphically. • Finding the Domain and the Range Algebraically When finding the domain of a function, ask yourself what values can’t be used. Your domain is everything else. There are simple basic rules to consider: • The domain of all polynomial functions, i.e. functions of the form f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , is the Real numbers IR.
1.1. WHAT IS A FUNCTION?
9
• Square root functions can not contain a negative underneath the radical. Set the expression under the radical greater than or equal to zero and solve for the variable. This will be your domain. • Rational functions, i.e. ratios of polynomials, can not have zeros in the denominator. Determine which values of the input cause the denominator to equal zero, and set your domain to be everything else. Exercise 5 Find, algebraically, the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx2
(b) y =
√1 x−4
(c) y = 2 + x1 .
Solution. (a) Since the function is a polynomial then its domain is the interval (−∞, ∞). To find the range, note that y ≥ 0 for all x. So the range is the interval [0, ∞). 1 (b) The domain of y = √x−4 consists of all numbers such that x − 4 > 0 or x > 4. That is, the interval (4, ∞). To find the range, notice that y > 0 for all values of x in the domain of y. Thus, the range is the interval (0, ∞). (c) The domain of y = 2 + x1 is the interval (−∞, 0) ∪ (0, ∞). To find the range, 1 write x in terms of y to obtain x = y−2 . The values of y for which this later formula is defined is the range of the given function, that is, (−∞, 2) ∪ (2, ∞). Remark. Note that the domain of the function y = πx2 of the previous problem consists of all real numbers. If this function is used to model a real-world situation, that is, if the x stands for the radius of a circle and y is the corresponding area then the domain of y in this case consists of all numbers x ≥ 0. In general, for a word problem the domain is the set of all x values such that the problem makes sense. • Finding the Domain and the Range Graphically We often use a graphing calculator to find the domain and range of functions. In general, the domain will be the set of all x values that has corresponding points on the graph. We note that if there is an asymptote (shown as a vertical line on the TI series) we do not include that x value in the domain. To find the range, we seek the top and bottom of the graph. The range will be all points from the top to the bottom (minus the breaks in the graph). Exercise 6 Use a graphing calculator to find the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx2
(b) y =
√1 x−4
(c) y = 2 + x1 .
Solution. (a) The graph of y = πx2 is given below
10
CHAPTER 1. LIBRARY OF FUNCTIONS
2 1.8 1.6 1.4 1.2 y 1 0.8 0.6 0.4 0.2 –2
0
–1
1 x
2
The domain is the set (−∞, ∞) and the range is [0, ∞). 1 (b) The graph of y = √x−4 is given below 8 7 6 5 y4 3 2 1 4
5
6 x
7
8
The domain is the set (4, ∞) and the range is (0, ∞). (c) The graph of y = 2 + x1 is given below 20
y 10
–2
–1
0
1 x
2
–10
–20
The domain is the set (−∞, 0) ∪ (0, ∞) and the range is (−∞, 2) ∪ (2, ∞).
1.2. LINEAR FUNCTIONS
11
Recommended Problems (pp. 4 - 6): 1, 3, 5, 10, 13, 15, 17, 24, 25.
1.2
Linear Functions
This section is designed to introduce students to the concept of linear functions. So, the first question that comes to mind is the question of recognizing a linear function. For any function f (x), the average rate of change of f (x) from x = a to x = b is the difference quotient f (b) − f (a) b−a In general, the average rate of change of a function is different on different intervals. A linear function is a function with the property that the average rate of change on any interval is the same. We say that y is changing at a constant rate with respect to x. In this case, the graph is a straight line ( and thus the term linear). It follows, that if f is linear and if [a, b] and [c, d] are two intervals of equal length, i.e. b − a = d − c then f (b) − f (a) f (d) − f (c) = b−a d−c and this implies that f (b) − f (a) = f (d) − f (c). That is, if a function y is such that equal increments in x corresponds to equal increments in y then y is a linear function of x. Exercise 7 Which of the following tables could represent a linear function? x 0 5 10 15
f(x) 10 20 30 40
x 0 10 20 30
g(x) 20 40 50 55
Solution. Since equal increments in x yield equal increments in y then f (x) is a linear 50−40 function. On the contrary, since 40−20 10−0 6= 20−10 then g(x) is not linear. Now, suppose that f (x) is a linear function of x. Then f changes at a constant rate m. That is, if we pick two points (0, f (0)) and (x, f (x)) then m=
f (x) − f (0) . x−0
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CHAPTER 1. LIBRARY OF FUNCTIONS
That is, f (x) = mx + f (0). This is the function notation of the linear function f (x). Another notation is the equation notation, y = mx + f (0). We will denote the number f (0) by b. In this case, the linear function will be written as f (x) = mx + b or y = mx + b. Since b = f (0) then the point (0, b) is the point where the line crosses the vertical line. We call it the y-intercept. So the y-intercept is the output corresponding to the input x = 0, sometimes known as the initial value of y. If we pick any two points (x1 , y1 ) and (x2 , y2 ) on the graph of f (x) = mx + b then we must have y2 − y1 m= . x2 − x1 We call m the slope of the line. Exercise 8 The value of a new computer equipment is $20,000 and the value drops at a constant rate so that it is worth $ 0 after five years. Let V (t) be the value of the computer equipment t years after the equipment is purchased. (a) Find the slope m and the y-intercept b. (b) Find a formula for V (t). Solution. (a) Since V (0) = 20, 000 and V (5) = 0 then the slope of V (t) is m=
0 − 20, 0002 = −4, 000 5−0
and the vertical intercept is V (0) = 20, 000. (b) A formula of V (t) is V (t) = −4, 000t + 20, 000. Exercise 9 The table below shows the cost of selling various amounts of coffee per day from a cart. (a) Using the table, show that the relationship appears to be linear. (b) Plot the data in the table. (c) Find the slope of the line. Explain what this means in the context of the given situation. x(cups per day) C(dollars)
0 50.00
5 51.25
10 52.50
50 62.50
100 75.00
200 100.00
Solution. (a) Since an increment of 5 in x yields an increment of 1.25 in y then the table shows that equal increments in x yield equal increments in y so that y is linear. (b) The graph is given below
1.2. LINEAR FUNCTIONS
13
100 90 80 70 60 50 0
50
100
150
200
(c) The slope of the line is m = 51.25−50.00 = 0.25. This number means that an 5−0 increase in production by one cup will increase cost by 25 cents. • Formulas for Linear Functions Recall that f is linear if and only if f (x) can be written in the form f (x) = mx + b. So the problem of finding the formula of f is equivalent to finding the quantities m and b. Suppose that we know two points on the graph of f (x), say (x1 , f (x1 )) and (x2 , f (x2 )). Since the slope m is just the average rate of change of f (x) on the interval [x1 , x2 ] then f (x2 ) − f (x1 ) m= . x2 − x1 To find b, we use one of the points in the formula of f (x); say we use the first point. Then f (x1 ) = mx1 + b. Solving for b we find b = f (x1 ) − mx1 . Example 2 Let’s find the formula of a linear function given by a table of data values. The table below gives data for a linear function. Find the formula. x f(x)
1.2 0.736
1.3 0.614
1.4 0.492
1.5 0.37
Solution. We use the first two points to find the value of m : m=
f (1.3) − f (1.2) 0.614 − 0.736 = = −1.22 1.3 − 1.2 1.3 − 1.2
To find b we can use the first point to obtain 0.736 = −1.22(1.2) + b. Solving for b we find b = 2.2 Thus, f (x) = −1.22x + 2.2
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CHAPTER 1. LIBRARY OF FUNCTIONS
Example 3 Suppose that the graph of a linear function is given and two points on the graph are known. For example, the graph below is the graph of a linear function going through the points (100, 1) and (160, 1.6). Find the formula.
(160,6)
(100,1)
Solution. The slope m is found as follows: m=
6−1 = 0.083. 160 − 100
To find b we use the first point to obtain 1 = 0.083(100) + b. Solving for b we find b = −7.3. So the formula for f (x) = −7.3 + 0.083x. Example 4 Sometimes a linear function is given by a verbal description as in the following problem: In a college meal plan you pay a membership fee; then all your meals are at a fixed price per meal. If 30 meals cost $152.50 and 60 meals cost $250 then find the formula for the cost C of a meal plan in terms of the number of meals n. Solution. We find m first:
250 − 152.50 = $3.25/meal. 60 − 30 To find b or the membership fee we use the point (30, 152.50) in the formula C = mn + b to obtain 152.50 = 3.25(30) + b. Solving for b we find b = $55. Thus, C = 3.25n + 55. m=
So far we have represented a linear function by the expression y = mx + b. This is known as the slope-intercept form of the equation of a line. Now, if the slope m of a line is known and one point (x0 , y0 ) is given then by taking any point (x, y) on the line and using the definition of m we find y − y0 = m. x − x0 Cross multiply to obtain: y − y0 = m(x − x0 ). This is known as the point-slope form of a line.
1.3. RATE OF CHANGE
15
Example 5 Find the equation of the line passing through the point (100, 1) and with slope m = 0.01. Solution. Using the above formula we have: y − 1 = 0.01(x − 100) or y = 0.01x. Note that the form y = mx + b can be rewritten in the form Ax + By = C
(1.1)
where A = m, B = −1, and C = b. The form (??) is known as the standard form of a linear equation. Example 6 Rewrite in standard form: 3x + 2y + 40 = x − y. Solution. Subtracting x − y + 40 from both sides to obtain 2x + 3y = −40.
Recommended Problems (pp. 11 - 3): 1, 3, 7, 9, 11, 12, 15, 17, 23, 25, 29.
1.3
Rate of Change
In this section we will introduce the concept of the average rate of change of a function. We start with an example. Example 7 (Average Speed) During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car; that is, it shows your speed at a particular instant in time. The instantaneous speed of an object is not to be confused with the average speed. Average speed is a measure of the distance traveled in a given period of time. That is, Average Speed =
Distance traveled Time elapsed
So if the trip to school takes 0.2 hours (i.e. 12 minutes) and the distance traveled is 5 miles then the average speed of your car is 5 miles = 25miles/hour 0.2 hours This says that on the average, your car was moving with a speed of 25 miles per hour. During your trip, there may have been times that you were stopped Ave. Speed =
16
CHAPTER 1. LIBRARY OF FUNCTIONS
and other times that your speedometer was reading 50 miles per hour; yet on the average you were moving with a speed of 25 miles per hour. Looking at the above example, we see that the average speed over the interval [0,0.2] is s(0.2) − s(0) Average Speed = 0.2 − 0 where s(t) is the position function of the car at time t hours. Example 8 A freely falling body experiencing no air resistance falls s(t) = 16t2 feet in t seconds. Complete the following table time interval Average velocity
[1.8,2]
[1.9,2]
[1.99,2]
[1.999,2]
[2,2.0001]
[2,2.001]
[2,2.01]
[1.8,2] 60.8
[1.9,2] 62.4
[1.99,2] 63.84
[1.999,2] 63.98
[2,2.0001] 64.0016
[2,2.001] 64.016
[2,2.01] 64.16
Solution. time interval Average velocity
The average speed is a example of the average rate of a function. In general, the average rate of change of a function f (x) from x = a to x = b is the difference quotient f (b) − f (a) . b−a Geometrically, this quantity represents the slope of the secant line going through the points (a, f (a)) and (b, f (b)) on the graph of f (x). Exercise 10 Calculate the average rate of change of f (x) = x2 between x = −2 and x = 1. Solution. Ave. rate of change =
f(1) − f(−2) −3 = = −1 1 − (−2) 3
• Average Rate of Change and Monotone Functions We say that a function is increasing if its graph climbs as x moves from left to right. That is, the function values increase as x increases. It is said to be decreasing if its graph falls as x moves from left to right. This means that the function values decrease as x increases. As an application of the average rate of change, we can use such quantity to decide whether a function is increasing or decreasing. A function is monotone if it is either increasing or decreasing.
1.3. RATE OF CHANGE
17
If a function f is increasing on an interval I then by taking any two points in the interval I, say a < b, we see that f (a) < f (b) and in this case f (b) − f (a) > 0. b−a Thus, if the average rate of change is positive in an interval then the function is increasing in that interval. Similarly, if the average rate of change is negative in an interval I then the function is decreasing. Exercise 11 The table below gives values of a function w = f (t). Is this function increasing or decreasing? t 0 4 8 12 16 20 24 w 100 58 32 24 20 18 17 Solution. The average of w over the interval [0, 4] is w(4) − w(0) 58 − 100 = = −10.5 4−0 4−0 The average rate of change of the remaining intervals are given in the chart below time interval Average
[0,4] -10.5
[4,8] -6.5
[8,12] -2
[12,16] -1
[16, 20] -0.5
[20,24] -0.25
Since the average rate of change is always negative on [0, 24] then the function is decreasing over there. Exercise 12 Determine the intervals where the function is increasing and decreasing 10 8 6 y 4 2 –4
–2
0 –2 –4 –6 –8 –10
2
x
4
Solution. The function is increasing on (−∞, −1) ∪ (1, ∞) and decreasing on the interval (−1, 1).
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CHAPTER 1. LIBRARY OF FUNCTIONS
• Average Rate of Change and Concavity We say that the graph of a function is concave up or opens up if the average rate of change is increasing. The graph is concave down or opens down if the average rate of change is decreasing. Exercise 13 Find the intervals over which: (a) The function is increasing; decreasing. (b) The graph is concave up; concave down. 4
Fig. 2
3 y 2 1 –1
0
1
2 x
3
4
–1
Solution. The graph is concave up on the interval (1, ∞) and concave down on the interval (−∞, 1). Exercise 14 The table below gives values of a function w = f (t). Is this function concave up or concave down? t 0 4 8 12 16 20 24 w 100 58 32 24 20 18 17 Solution. Completing the following table Interval
[0,4] [4,8] [8,12] [12,16] [16,20] [20,24] -10.5 -6.5 -2 -1 -0.5 -0.25 we see that the average rate of change is increasing so that the graph is concave up. f (b)−f (a) b−a
Recommended Problems (pp. 19 - 21): 1, 2, 3, 4, 5, 7, 10, 12, 13, 16, 17, 25, 27, 30.
1.4
Applications of Functions to Economics
The goal of this section is to exhibit some functions used in business and economics.
1.4. APPLICATIONS OF FUNCTIONS TO ECONOMICS
19
• The Cost Function The cost function C gives the cost C(q) of manufacturing a quantity q of some good. A linear cost function has the form C(q) = mq + b, where the y-intercept b is called the fixed cost, i.e. the costs incurred even if nothing is produced, and the slope m is called the marginal cost or the variable costs per unit. Exercise 15 What is the cost function of manufacturing a product with fixed cost of $400 and variable costs of $40 per item, assuming the function is linear? Solution. The cost function is C(q) = 40q + 400. Exercise 16 Values of a linear cost function are shown below. What are the fixed costs and the marginal cost? Find a formula for the cost function. q C(q)
0 5000
5 5020
10 5040
15 5060
20 5080
Solution. The fixed costs are b = C(0) = $5, 000, the marginal cost is m=
5020 − 5000 =4 5−0
The cost function is C(q) = 4q + 5, 000 • The Revenue Function A revenue function R gives the total revenue R(q) from the sale of a quantity q at a unit price p dollars. Thus, R(q) = pq. Exercise 17 A company that makes a certain brand of chairs has fixed costs of $5, 000 and marginal cost of $30 per chair. The company sells the chairs for $50 each. Find formulas for the cost and revenue functions. Solution. The cost function is C(q) = 30q+5000. The revenue function is R(q) = pq = 30q.
20
CHAPTER 1. LIBRARY OF FUNCTIONS
• The Profit Function Profit is defined to be revenue minus cost. That is P (q) = R(q) − C(q). The break-even point is the point where the profit is zero, i.e. R(q) = C(q). Exercise 18 A company has cost and revenue functions, in dollars, given by C(q) = 6, 000 + 10q and R(q) = 12q. (a) Graph the functions C(q) and R(q) on the same coordinate axes. (b) Find the break-even point and illustrate it graphically. (c) When does the company make a profit? Loses money? Solution. (a) The graph is given below, where the red line is the graph of the cost function and the green one is the graph of the revenue function. 60000 50000 40000 y30000 20000 10000 0
2000 4000 6000 8000 10000 x
(b) The break-even point is the point of intersection of the two lines. To find the point, set 12q = 10q + 6000 and solve for q to find q = 3000. Thus, the break-even point is the point (3000, 36000). (c) The company makes profit for q > 3000 and loses money for q < 3000. • Marginal Cost, Marginal Revenue, and Marginal Profit The marginal cost is the approximate cost of producing an additional item. The marginal revenue is the approximate revenue from selling an additional item. The marginal profit is the approximate profit from selling an additional item. Exercise 19 Consider the cost and revenue functions of the previous exercise. Find the marginal cost, marginal revenue, and marginal profit.
1.5. EXPONENTIAL FUNCTIONS
21
Solution. The marginal cost is the slope of C(q). That is, $10 dollars per item. The marginal cost is $12 dollar per item and since P (q) = R(q) − C(q) = 2q − 6000 then the marginal profit is $2 dollars per item. • Supply and Demand Curves The quantity q manufactured and sold depends on the unit price p. In general, when the price goes up then manufacturers are willing to supply more of the product whereas consumers are going to reduce their buyings. Since consumers and manufacturers react differently to changes in price, there are two curves relating p and q. The supply curve is the quantity that producers are willing to make at a given price. Thus, increasing price will increase quantity. The demand curve is the amount that will be bought by consumers at a given price. Thus, decreasing price will increase quantity. Even though quantity is a function of price, it is the tradition to use the vertical y-axis for the variable p and the horizontal x-axis for the variable q. The supply and demand curves intersect at point (q ∗ , p∗ ) called the point of equilibrium. We call p∗ the equilibrium price and q ∗ the equilibrium quantity. Exercise 20 Find the equilibrium point for the supply function S(p) = 3p−50 and the demand function D(p) = 100 − 2p. Solution. Setting the equation S(p∗ ) = D(p∗ ) to obtain 3p∗ − 50 = 100 − 2p∗ . By adding 2p∗ + 50 to both sides we obtain 5p∗ = 150. Solving for p∗ we find p∗ = 30. Substitute this value in S(p) we find q ∗ = 3(30) − 50 = 40. recommended Problems (pp. 29 - 32): 1, 3, 4, 5, 7, 8, 11, 14, 18, 20, 24.
1.5
Exponential Functions
Linear functions are functions that change at a constant rate. For example, if f (x) = mx + b then f (x + 1) = m(x + 1) + b = f (x) + m. Exponential functions are functions that change at a constant percent rate. That is, if f (x) is an exponential function then f (x + 1) = f (x) + r%f (x). Before we go on to discuss exponential problems, we should pause and discuss some of the terms that are frequently used in these problems. First let’s talk about the terms factor and percent. For example, suppose you invested $100, and after a time, your investment was worth $300. The final value ($300) would be three (3) times the initial value. We would say that your investment had increased by a factor of 3.
22
CHAPTER 1. LIBRARY OF FUNCTIONS
The general equation for percentage changes is Nf inal − Ninitial × 100 = % change Ninitial Thus, a $100 investment that increased to $300 (increased by a factor of 3), had a percentage increase of: 300 − 100 × 100 = 200% 100 Example 9 A quantity increases from 10 to 12. By what percent has it increased? Solution. The increase is
2 10
= .2 = 20%.
Exercise 21 If we multiply a quantity A by 1.15 then A is increased by what percent? Solution. Since 1.15A = A + 0.15A = A + 15%A then multiplying A by 1.15 is equivalent to increasing A by 15%. Exponential functions are used to model increasing quantities such as population problems. Example 10 Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells doubled every minute. That is, the population cell is increasing at the constant rate of 100%. Write an equation to determine the number (population) of cells after one hour. Solution. The table below shows the number of cells for the first 5 minutes. Let P (t) be the number of cells after t minutes. t 0 1 2 3 4 5
P(t) 1 2 4 8 16 32
At time 0, i.e t=0, the number of cells is 1 or 20 = 1. After 1 minute, when t = 1, there are two cells or 21 = 2. After 2 minutes, when t = 2, there are 4 cells or 22 = 4. Therefore, one formula to estimate the number of cells (size of population) after t minutes is the equation (model) f (t) = 2t .
1.5. EXPONENTIAL FUNCTIONS
23
Now, to determine the number of cells after one hour we convert to minutes to obtain t = 60 minutes so that f (60) = 260 = 1.15 × 1018 cells. Exponential functions can also model decreasing quantities. Example 11 If you start a biology experiment with 5,000,000 cells and 45% of the cells are dying every minute, how long will it take to have less than 50,000 cells? Solution. Let P (t) is the number of cells after t minutes. Then P (t + 1) = P (t) − 45%P (t) or P (t + 1) = 0.55P (t). By constructing a table data we find t 0 1 2 3 4 5 6 7 8
P(t) 5,000,000 2,750,000 1,512,500 831,875 457,531.25 251,642.19 138,403.20 76,121.76 41,866.97
So it takes 8 minutes for the population to reduce to less than 50,000 cells. A formula of P (t) is P (t) = 5, 000, 000(0.55)t . From the previous two examples, we see that an exponential function has the general form P (t) = b · at , a > 0 a 6= 1. Since b = P (0) then we call b the initial value. Since P (t + 1) = kat+1 = a(kat ) = aP (t) then we call a the growth factor. Also, since P (t) = P (t) + r%P (t) where r = a−1 then we call r the percent growth rate. Note that for a > 1 (or r < 1) the exponential function represents a growth model whereas for 0 < a < 1 ( or r > 1) the function represents a decay model. Remark. Why a is restricted to a > 0 and a 6= 1? Since t is allowed√ to have any value then a negative a will create meaningless expressions such as a (if t = 12 ). Also, for a = 1 the function P (t) = b is called a constant function and its graph is a horizontal line. Exercise 22 Suppose you are offered a job at a starting salary of $40,000 per year. To strengthen the offer, the company promises annual raises of 6% per year for the first 10 years. Let P (t) be your salary after t years. Find a formula for P (t) and then compute your projected salary after 4 years from now.
24
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. A formula of P (t) is P (t) = 40, 000(1.06)t . After four years, the projected salary is P (3) = 40, 000(1.06)4 ≈ 50, 499.08. Exercise 23 The amount in milligrams of a drug in the body t hours after taking a pill is given by A(t) = 25(0.85)t . (a) What is the initial dose given? (b) What percent of the drug leaves the body each hour? (c) What is the amount of drug left after 10 hours? Solution. (a) Initial dose give is A(0) = 25 mg. (b) r = a − 1 = 0.85 − 1 = −.15 so that 15% of the drug leaves the body each hour. (c) A(10) = 25(0.85)10 ≈ 4.92 mg. • Graphs of Exponential Functions Recall that an exponential function with base a and initial value b is a function of the form f (x) = b · ax . Since b = f (0) then (0, b) is the vertical intercept of f (x). In this section we consider graphs of exponential functions. Let’s see the effect of the parameter b on the graph of f (x) = bax . Example 12 Graph, on the same axes, the exponential functions f1 (x) = 2 · (1.1)x , f2 (x) = (1.1)x , and f3 (x) = 0.75(1.1)x . Solution. Using a graphing calculator, we find 18 16 14 12 10 8 6 4 2 0 0.20.40.60.8 1 1.21.41.61.8 2 x
where f1 is in red, f2 in green and f3 in yellow. Note that these functions have the same growth factor but different b and therefore different vertical intercepts.
1.5. EXPONENTIAL FUNCTIONS
25
We know that the slope of a linear function measures the steepness of the graph. Similarly, the parameter a measures the steepness of the graph of an exponential function.
Example 13 Graph, on the same axes, the exponential functions f( x) = 2 · 2x , f2 (x) = 2 · 3x , and f3 (x) = 2 · 4x . Solution. Using a graphing calculator we find
6 5 4 3 2 0 0.05 0.1 0.15 0.2 0.25 0.3 x
where f1 is in red, f2 in green, and f3 in yellow. • Observations. (i) Note that the greater the value of a, the more rapidly the graph rises. That is, the growth factor a affects the steepness of an exponential function. (ii) Since a > 1 then the graphs represent exponential growths. (iii) Note that as x decreases, the function values approach the x-axis. Symbolically, as x → −∞, y → 0.
Example 14 Graph, on the same axes, the exponential functions f1 (x) = 2·2−x = 2 · 3−x , and f3 (x) = 2 · 4−x = 32x . Solution. Using a graphing calculator we find
2 2x , f2 (x)
=
26
CHAPTER 1. LIBRARY OF FUNCTIONS 2 1.5 1 0.5
0
1
2 x
3
4
where f1 is in red, f2 in green, and f3 in yellow. • Observations. (i) Note that the smaller the value of a, the more rapidly the graph falls. (ii) Since 0 < a < 1 then the graphs represent exponential decays. (iii) Note that as x increases, the function values approach the x-axis. Symbolically, as x → ∞, y → 0. • General Observations (i) For a > 1, as x decreases, the function values get closer and closer to 0. Symbolically, as x → −∞, y → 0. For 0 < a < 1, as x increases, the function values gets closer and closer to the x-axis. That is, as x → ∞, y → 0. We call the x-axis, a horizontal asymptote. (ii) The domain of an exponential function consists of the set of all real numbers whereas the range consists of the set of all positive real numbers. (iii) Since the average rate of change is the slope of a secant line connecting two points on a graph then the average rate of change of the exponential functions discussed above is always increasing. This means that the graph opens up. We also say that the graph is concave up ( If the average rate of change of a function is decreasing then the graph opens down and we say that the graph is concave down). Exponential Functions Versus Linear Functions Next, we consider the question of recognizing whether a function given by a table of values is exponential or linear. We know that for a linear function, equal increments in x correspond to equal increments in y. For an exponential function let us first assume that we have a formula for the function, say f (x) = bax . Then f (x+1) f (x) = a. Thus, if the ratio of consecutive y-values of a function is the same then the function is an exponential function. Example 15 Decide if the function is linear or exponential?Find a formula for each case.
1.5. EXPONENTIAL FUNCTIONS
x 0 1 2 3 4
f(x) 12.5 13.75 15.125 16.638 18.301
x 0 1 2 3 4
27
g(x) 0 2 4 6 8
Solution. 15.125 16.638 18.301 Since 13.75 12.5 ≈ 13.75 ≈ 15.125 ≈ 16.638 ≈ 1.1 then f (x) is an exponential function. To find a formula for f (x) = bax we use the first two points obtaining 12.5 = f (0) = b and 13.75 = f (1) = ba = 12.5a. Hence, a = 13.75 12.5 ≈ 1.1 so that f (x) = 12.5(1.1)x . On the other hand, equal increments in x corresponds to equal increments in the g-values so that g(x) is linear, say g(x) = mx + b. Since g(0) = 0 then b = 0. Also, 2 = g(1) = m so that g(x) = 2x. Finally, we consider the question of finding a formula for an exponential function. The above example shows how to find linear and exponential functions using two data points. The next example illustrates how to find the formula of an exponential function given two points on its graph. Example 16 Find a formula for the exponential function whose graph is given below
(-1,2.5) (1,1.6)
Solution. Write f (x) = bax . Since f (−1) = 2.5 then ba−1 = 2.5. Similarly, ba = 1.6. 2 Taking the ratio we find baba−1 = 1.6 2.5 or a = .64 or a = 0.8. From ba = 1.6 we x find that b = 1.6 = 2 so that f (x) = 2(0.8) . 0.8 Recommended Problems (pp. 37 - 9): 1, 3, 7, 8, 10, 11, 13, 15, 20, 25.
28
1.6
CHAPTER 1. LIBRARY OF FUNCTIONS
Logarithms and Their Properties
In this section we find an algebraic way to solve equations of the form ax = b where a and b are positive constants. For this, we introduce two functions that are found in today’s calculators, namely, the functions log x and ln x. If x > 0 then we define log x to be a number y that satisfies the equality 10y = x. For example, log 100 = 2 since 102 = 100. Similarly, log 0.01 = −2 since 10−2 = 0.01. We call log x the common logarithm of x. Thus, y = log x if and only if 10y = x. Example 17 (a) Rewrite log 30 = 1.477 using exponents instead of logarithms. (b) Rewrite 100.8 = 6.3096 using logarithms instead of exponents. Solution. (a) log30 = 1.477 is equivalent to 101.477 = 30. (b) 100.8 = 6.3096 is equivalent to log 6.3096 = 0.8. Example 18 Without a calculator evaluate the following expressions: (a) log 1
(b) log 100
(c) log ( √110 )
(d) 10log 100
(e) 10log (0.01)
Solution. (a) log 1 = 0 since 100 = 1. (b) log 100 = log 1 = 0 by (a). 1 (c) log ( √110 ) = log 10− 2 = − 21 . (d) 10log 100 = 102 = 100. (e) 10log (0.01) = 10−2 = 0.01. • Properties of Logarithms (i) Since 10x = 10x we can write log 10x = x (ii) Since log x = log x then 10log x = x (iii) log 1 = 0 since 100 = 1. (iv) log 10 = 1 since 101 = 10. (v) Suppose that m = log a and n = log b. Then a = 10m and b = 10n . Thus, a · b = 10m · 10n = 10m+n . Rewriting this using logs instead of exponents, we see that log (a · b) = m + n = log a + log b.
1.6. LOGARITHMS AND THEIR PROPERTIES (vi) If, in (v), instead of multiplying we divide, that is using logs again we find ³a´ log = log a − log b. b
29 a b
=
10m 10n
= 10m−n then
(vii) It follows from (vi) that if a = b then log a − log b = log 1 = 0 that is log a = log b. (viii) Now, if n = log b then b = 10n . Taking both sides to the power k we find bk = (10n )k = 10nk . Using logs instead of exponents we see that log bk = nk = k log b that is log bk = k log b. Example 19 Solve the equation: 4(1.171)x = 7(1.088)x . Solution. ¡ ¢x Rewriting the equation into the form 1.171 = 74 and then using properties 1.088 (vii) and (viii) to obtain µ ¶ 1.171 7 x log = log . 1.088 4 Thus, x=
log 74 ¡ ¢. log 1.171 1.088
Exercise 24 Solve the equation log (2x + 1) + 3 = 0. Solution. Subtract 3 from both sides to obtain log (2x + 1) = −3. Switch to exponential form to get 2x + 1 = 10−3 = 0.001. Subtract 1 and then divide by 2 to obtain x = −0.4995. Remarks. • All of the above arguments are valid for the function ln x for which we replace the number 10 by the number e = 2.718 · · · . That is, y = ln x is equivalent to x = ey , ln (a · b) = ln a + ln b, etc. We call ln x the natural logarithm of x. • Keep in mind the following: log (a + b) 6= log a + log b log (a − b) 6= log a − log b log (ab) ¡ ¢ 6= logaa · log b log ab 6= log ¡ ¢ log b log a1 6= log1 a (Give an example for each case!) • The Logarithmic Function
30
CHAPTER 1. LIBRARY OF FUNCTIONS
We have seen that the notation y = log x is equivalent to 10y = x. Since 10 raised to any power is always positive then the domain of the function log x consists of all positive numbers. That is, log x cannot be used with negative numbers. Now, let us sketch the graph of this function by first constructing the following chart: x 0 0.001 0.01 0.1 1 10 100 1000
log x underfined -3 -2 -1 0 1 2 3
Average Rate of Change 111.11 11.11 1.11 0.11 0.011 0.0011
From the chart we see that the graph is always increasing. Since the average rate of change is decreasing then the graph is always concave down. Now plotting these points and connecting them with a smooth curve to obtain 4 2
0
20
40
x
60
80
100
–2
We observe the following: (a) The range of log x consists of all real numbers. (b) The graph never crosses the y-axis since a positive number raised to any power is always positive. (c) The graph crosses the x-axis at x = 1. (d) As x gets closer and closer to 0 from the right the function log x decreases without bound. That is, as x → 0+ , x → −∞. We call the y-axis a vertical asymptote. In general, if a function increases or decreases without bound as x gets closer to a number a then we say that the line x = a is a vertical asymptote. Next, let’s graph the function y = 10x by using the above process:
1.6. LOGARITHMS AND THEIR PROPERTIES x -3 -2 -1 0 1 2 3
10x 0.001 0.01 0.1 1 10 100 1000
31
Average Rate of Change 0.009 0.09 0.9 9 90 900
Note that this chart can be obtained from the chart of log x discussed above by interchanging the variables x and y. This means, that the graph of y = 10x (in red) is a reflection of the graph of y = log x (in blue) about the line y = x (in green) as seen in the picture below. 3 2 y 1 –2
–1
0
1 x
2
–1 –2 –3
Example 20 Sketch the graphs of the functions y = ln x and y = ex on the same axes.
Solution. The functions y = ln x and y = ex are inverses of each other like the functions y = log x and y = 10x . So their graphs are reflections of one another across the line y = x as shown below
32
CHAPTER 1. LIBRARY OF FUNCTIONS
• Continuous Growth Rate and the Number e For any positive number c, eln c = c. Thus, any positive number a can be written in the form a = ek where k = ln a. If a > 1 then ln a > 0 and therefore k > 0. If 0 < a < 1 then ln a < 0 and thus k < 0. Now, suppose that Q(t) = bat . Then Q(t) = b(ek )t = bekt . Note that for k > 0 ek > 1 so that Q(t) represents an exponential growth and if k < 0 then ek < 1 so that Q(t) is an exponential decay. We call the constant k the continuous growth rate. Example 21 If f (t) = 3(1.072)t is rewritten as f (t) = 3ekt , find k. Solution. By comparison of the two functions we find ek = 1.072. Take the ln of both sides to obtain k = ln (1.072). Example 22 A population increases from its initial level of 7.3 million at the continuous rate of 2.2% per year. Find a formula for the population P (t) as a function of the yeat t. When does the population reach 10 million? Solution. We are given the intial value 7.3 million and the continuous growth rate k = 0.022. Therefore, P (t) = 7.3e0.022t . Next,we want to find the time when P (t) =
1.7. EXPONENTIAL GROWTH AND DECAY
33
10. That is , 7.3e0.022t = 10. Divide both sides by 7.3 and then take the ln of ln 10 both sides to obtain 0.022t = ln 7.3 . Thus, t=
1 ln 10 ≈ 14.3. 0.022 ln 7.3
Next, in order to convert from Q(t) = bekt to Q(t) = bat we let a = ek . For example, to convert the formula Q(t) = 7e0.3t to the form Q(t) = bat we let b = 7 and a = e0.3 ≈ 1.35. Thus, Q(t) = 7(1.35)t . Example 23 Find the annual percent rate and the continuous percent growth rate of Q(t) = 200(0.886)t . Solution. The annual percent of decrease is r = a − 1 = 0.886 − 1 = −0.114 = −11.4%. To find the continuous percent growth rate we let ek = 0.886 and solve for k by taking the ln of both sides and obtain k = ln (0.886) ≈ −0.121 = −12.1%. Recommended Problems (pp. 42 - 3): 3, 5, 9, 11, 15, 17, 19, 22, 23, 26, 27, 31.
1.7
Exponential Growth and Decay
In this section, we consider some applications of exponential functions. • Doubling Time In some exponential models one is interested in finding the time for an exponential growing quantity to double. We call this time the doubling time. To find it, we start with the equation b · at = 2b or at = 2. Solving for t we find log 2 t = log a. Example 24 Find the doubling time of a population growing according to P = P0 e0.2t . Solution. Setting the equation P0 20.2t = 2P0 and dividing both sides by P0 to obtain 2 20.2t = 2. Take ln of both sides to obtain 0.2t = ln 2. Thus, t = ln 0.2 ≈ 3.47. • Half-Life On the other hand, if a quantity is decaying exponentially then the time required for the quantity to reduce into half is called the half-life. To find it, we start with the equation bat = 2b and we divide both sides by b to obtain at = 0.5. Take the log of both sides to obtain t log a = log (0.5). Solving for t we find t = loglog(0.5) a .
34
CHAPTER 1. LIBRARY OF FUNCTIONS
Example 25 The half-life of Iodine-123 is about 13 hours. You begin with 50 grams of this substance. What is a formula for the amount of Iodine-123 remaining after t hours? Solution. Since the problem involves exponential decay then if Q(t) is the quantity remaining after t hours then Q(t) = 50at with 0 < t < 1. But Q(13) = 25. That 1 is, 50a13 = 25 or a13 = 0.5. Thus a = (0.5) 13 ≈ 0.95 and Q(t) = 50(0.95)t . • Compound Interest The term compound interest refers to a procedure for computing interest whereby the interest for a specified interest period is added to the original principal. The resulting sum becomes a new principal for the next interest period. The interest earned in the earlier interest periods earn interest in the future interest periods. The future value is the total amount due on the maturity date of the loan or the investment. The maturity value is calculated by using the compound amount formula which is given by A = P (1 +
r nt ) n
where P is the present value or principal, r is the nominal annual interest rate, n is the number of periods in a year, and t is the time in years. Possible values of n are: 1 (annually), 2 (semiannually), 4 (quarterly), 12 (monthly), and 365 (daily). When interest is compounded more frequently than once a year, the account effectively earns more than the nominal rate. Thus, we distinguish between nominal rate and effective rate. The effective annual rate tells how much interest the investment actually earns. The quantity (1 + nr )n − 1 is known as the effective interest rate. Example 26 Translating a value to the future is referred to as compounding. What will be the maturity value of an investment of $15, 000 invested for four years at 9.5% compounded semi-annually? Solution. Using the formula for compound interest with P = $15, 000, t = 4, n = 2, and r = .095 we obtain µ ¶8 0.95 A = 15, 000 1 + ≈ $21, 743.20 2 Example 27 Translating a value to the present is referred to as discounting. We call (1 + nr )−nt the discount factor. What principal invested today will amount to $8, 000 in 4 years if it is invested at 8% compounded quarterly?
1.7. EXPONENTIAL GROWTH AND DECAY
35
Solution. The present value is found using the formula µ ¶−16 ³ r ´−nt 0.08 P =A 1+ = 8, 000 1 + ≈ $5, 827.57 n 4 Example 28 What is the effective rate of interest corresponding to a nominal interest rate of 5% compounded quarterly? Solution. effective rate =
µ ¶4 0.05 1+ − 1 ≈ 0.051 = 5.1% 4
Continuous Compound Interest When the compound formula is used over smaller time periods the interest becomes slightly larger and larger. That is, frequent compounding earns a higher effective rate, though the increase is small. This suggests compounding more and more, or equivalently, finding the value of A in the long run. The example at previous section shows ¡ ¢n the end of the that as n → ∞ the expression 1 + nr approaches er so that A = P ert . This formula is known as the continuous compound formula. In this case, the annual effective interest rate is found using the formula er − 1. Example 29 Find the effective rate if $1000 is deposited at 5% annual interest rate compounded continuously. Solution. The effective interest rate is e0.05 − 1 ≈ 0.05127 = 5.127% Example 30 Which is better: An account that pays 8% annual interest rate compounded quarterly or an account that pays 7.95% compounded continuously? Solution. The effective rate corresponding to the first option is µ ¶4 0.08 1+ − 1 ≈ 8.24% 4 That of the second option e0.0795 − 1 ≈ 8.27% Thus, we see that 7.95% compounded continuously is better than 8% compounded quarterly. Recommended Problems (pp. 48 - 51): 5, 7, 8, 13, 19, 22, 28, 29, 30, 33.
36
1.8
CHAPTER 1. LIBRARY OF FUNCTIONS
Building New Functions from Old Ones
In this section we discuss various ways for building new functions from old ones. New functions can be obtained by composing functions, using arithmetic combinations, and finally by making changes to either the input or the output of a function. • Composition of Functions In this section we will discuss a procedure for building new functions from old ones known as the composition of functions. We start with an example of a real-life situation where composite functions are applied. Example 31 You have two money machines, both of which increase any money inserted into them. The first machine doubles your money. The second adds five dollars. The money that comes out is described by f (x) = 2x in the first case, and g(x) = x + 5 in the second case, where x is the number of dollars inserted. The machines can be hooked up so that the money coming out of one machine goes into the other. Find formulas for each of the two possible composition machines. Solution. Suppose first that x dollars enters the first machine. Then the amount of money that comes out is f (x) = 2x. This amount enters the second machine. The final amount coming out is given by g(f (x)) = f (x) + 5 = 2x + 5. Now, if x dollars enters the second machine first, then the amount that comes out is g(x) = x + 5. If this amount enters the second machine then the final amount coming out is f (g(x)) = 2(x + 5) = 2x + 10. The function f (g(x)) is called the composition of the functions f and g; the function g(f (x)) is called the composition of the functions g and f. In general, suppose we are given two functions f and g such that the range of g is contained in the domain of f so that the output of g can be used as input for f. We define a new function, called the composition of f and g, by the formula (f ◦ g)(x) = f (g(x)). In a similar way, we can define the composition of g and f to be the function (g ◦ f )(x) = g(f (x)) so that the output of f is the input of g. Using a Venn diagram we have
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
37
• Composition of Functions Defined by Tables Example 32 Complete the following table x f(x) g(x) f(g(x))
0 1 5
1 0 2
2 5 3
3 2 1
4 3 4
5 4 8
0 1 5 4
1 0 2 5
2 5 3 2
3 2 1 0
4 3 4 3
5 4 8 undefined
Solution. x f(x) g(x) f(g(x))
• Composition of Functions Defined by Formulas Example 33 Suppose that f (x) = 2x + 1 and g(x) = x2 − 3. (a) Find f ◦ g and g ◦ f. (b) Calculate (f ◦ g)(5) and (g ◦ f )(−3). (c) Are f ◦ g and g ◦ f equal? Solution. (a) (f ◦ g)(x) = f (g(x)) = f (x2 − 3) = 2(x2 − 3) + 1 = 2x2 − 5. Similarly,
38
CHAPTER 1. LIBRARY OF FUNCTIONS
(g ◦ f )(x) = g(f (x)) = g(2x + 1) = (2x + 1)2 − 3 = 4x2 + 4x − 2. (b) (f ◦ g)(5) = 2(5)2 − 5 = 45 and (g ◦ f )(−3) = 4(−3)2 + 4(−3) − 2 = 22. (c) f ◦ g 6= g ◦ f. With only one function you can build new functions using composition of the function with itself. Also, there is no limit on the number of functions that can be composed. Example 34 Suppose that f (x) = 2x + 1 and g(x) = x2 − 3. (a) Find (f ◦ f )(x). (b) Find [f ◦ (f ◦ g)](x). Solution. (a) (f ◦ f )(x) = f (f (x)) = f (2x + 1) = 2(2x + 1) + 1 = 4x + 3. (b) [f ◦ (f ◦ g)](x) = f (f (g(x))) = f (f (x2 − 3)) = f (2x2 − 5) = 2(2x2 − 5) + 1 = 4x2 − 9. • Decomposition of Functions If a formula for (f ◦ g)(x) is given then the process of finding the formulas for f and g is called decomposition. Example 35 √ Decompose (f ◦ g)(x) = 1 − 4x2 . Solution. √ 2 One possible √ answer is f (x) = x and g(x) = 1 − 4x . Another possible answer 2 is f (x) = 1 − x and g(x) = 2x. • Combinations of Functions In this section we are going to construct new functions from old ones using the operations of addition, subtraction, multiplication, and division. Let f (x) and g(x) be two given functions. Then for all x in the common domain of these two functions we define new functions as follows: • Sum: (f + g)(x) = f (x) + g(x). • Difference: (f − g)(x) = f (x) − g(x). • Product: (f ³ ·´g)(x) = f (x) · g(x). • Division:
f g
(x) =
f (x) g(x)
provided that g(x) 6= 0.
In the following example we see how to construct the four functions discussed above when the individual functions are defined by formulas. Example 36 √ Let f (x) = x + 1 and g(x) = x + 3. Find the common domain and then find a formula for each of the functions f + g, f − g, f · g, fg .
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
39
Solution. The domain of f (x) consists of all real numbers whereas the domain of g(x) consists of all numbers x ≥ 3. Thus, the common domain is the interval [−3, ∞). For any x in this domain we have √ (f + g)(x) = x + 1 + √x + 3 (f − g)(x) = x+1− √ x+3 √ (f x x+3+ x+3 ³ · ´g)(x) = f √x+1 provided x > −3. (x) = g x+3 In the next example, we see how to evaluate the four functions when the individual functions are given in numerical forms. Example 37 Suppose the functions following table: x -1 -1 f(x) 8 2 g(x) -1 -5 (f + g)(x) (f − g)(x) (f · g)(x) ( fg )(x)
f and g are given in numerical forms. Complete the 0 7 -11
1 -1 7
1 -5 8
3 -3 9
Solution. x f(x) g(x) (f + g)(x) (f − g)(x) (f · g)(x) ( fg )(x)
-1 8 -1 7 9 -8 -8
-1 2 -5 -3 7 -10 - 25
0 7 -11 -4 18 -77 7 - 11
1 -1 7 6 -8 -7 - 17
1 -5 8 3 -13 -40 - 85
3 -3 9 6 -12 -27 - 13
• Vertical and Horizontal Translations In this section we want to discuss the effect of change to the input of a function as well as to the output. More precisely, we will discuss graphing new functions of the form f (x + k) and f (x) + k where k is a given constant. First, let’s look at the following example. Example 38 Let f (x) = x2 . (a) Use a calculator y3 = f (x) + 3. What (b) Use a calculator y3 = f (x) − 3. What
to graph the functions y1 = f (x) + 1, y2 = f (x) + 2 and can you say about these graphs? to graph the functions y1 = f (x) − 1, y2 = f (x) − 2 and can you say about these graphs?
40
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. (a) The graph of y1 = f (x) + 1 = x2 + 1 (green color) is obtained by lifting the graph of f (x) (red color) 1 unit upward. Similarly, the graph of y2 = f (x) + 2 = x2 + 2 (yellow color) is the graph of f (x) lifted two units up and the graph of y3 = f (x) + 3 = x2 + 3 (blue color) is the graph of f (x) lifted up three units. 10 8 6 y 4 2
–4
–2
0
2
x
4
(b) The graph of y1 = f (x) − 1 = x2 − 1 (green color) is obtained by shifting the graph of f (x) (red color) 1 unit downward. Similarly, the graph of y2 = f (x) − 2 = x2 − 2 (yellow color) is the graph of f (x) shifted two units down and the graph of y3 = f (x) − 3 = x2 − 3 (blue color) is the graph of f (x) shifted down three units. 5 4 y
3 2 1
–4
–2
0 –1
2
x
4
–2 –3
It follows that the graph of f (x) + k with k > 0 is a vertical translation of the graph of f (x), k units upward, whereas for k < 0 it is a shift by k units downward. Next, we discuss horizontal shifts. Example 39 Let f (x) = x2 .
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES (a) Use a calculator y3 = f (x + 3). What (b) Use a calculator y3 = f (x − 3). What
41
to graph the functions y1 = f (x + 1), y2 = f (x + 2) and can you say about these graphs? to graph the functions y1 = f (x − 1), y2 = f (x − 2) and can you say about these graphs?
Solution. (a) The graph of y1 = f (x + 1) = (x + 1)2 (green color) is obtained by shifting the graph of f (x) (red color) 1 unit to the left. Similarly, the graph of y2 = f (x + 2) = (x + 2)2 (yellow color) is the graph of f (x) shifted two units to the left and the graph of y3 = f (x + 3) = (x + 3)2 (blue color) is the graph of f (x) shifted to the left by three units. 10 8 6 y 4 2
–4
0
–2
2
x
4
(b) The graph of y1 = f (x − 1) = (x − 1)2 (green color) is obtained by shifting the graph of f (x) (red color) 1 unit to the right. Similarly, the graph of y2 = f (x − 2) = (x − 2)2 (yellow color) is the graph of f (x) shifted two units to the right and the graph of y3 = f (x − 3) = (x − 3)2 (blue color) is the graph of f (x) shifted to the right by three units. 10 8 6 y 4 2
–4
–2
0
2
x
4
It follows that the graph of f (x + k) with k > 0 is a horizontal translation
42
CHAPTER 1. LIBRARY OF FUNCTIONS
of the graph of f (x) k units to the left whereas for k < 0 it is a shift by k units to the right. Exercise 25 Suppose S(d) gives the height of high tide in Seattle on a specific day, d, of the year. Use shifts of the function S(d) to find formulas of each of the following functions: (a) T (d), the height of high tide in Tacoma on day d, given that high tide in Tacoma is always one foot higher than high tide in Seattle. (b) P (d), the height of high tide in Portland on day d, given that high tide in Portland is the same height as the previous day’s hight in Seattle. Solution. (a) T (d) = S(d) + 1. (b) P (d) = S(d − 1). Finally, one can use a combination of both horizontal and vertical shifts to create new functions as shown in the next example. Example 40 Let f (x) = x2 . Let g(x) be the function obtained by shifting the graph of f (x) 2 units to the right and then up three units. Find a formula for g(x) and then draw its graph. Solution. The formula of g(x) is g(x) = f (x−2)+3 = (x−2)2 +3 = x2 −4x+7. The graph of g(x) (in green color) consists of a horizontal shift of two units to the right followed by a vertical shift of three units upward of the graph of f (x) (red color). 10 8 6 y 4 2
–4
–2
0
2
x
4
• Vertical Stretches and Compressions We have seen that for a positive k, the graph of f (x) + k is a vertical shift of
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
43
the graph of f (x) upward and the graph of f (x) − k is a vertical shift down. In this section we want to study the effect of multiplying a function by a constant k. This will result by either a vertical stretch or vertical compression. A vertical stretching is the stretching of the graph away from the x-axis. A vertical compression is the squeezing of the graph towards the x-axis.
Example 41 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = 2x2
x -3 -2 -1 0 1 2 3
y = 3x2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice?
Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = 2x2 18 8 2 0 2 8 18
x -3 -2 -1 0 1 2 3
y = 3x2 27 12 3 0 3 12 27
(b) The picture below shows that the graphs of 2f (x) (in green color) and 3f (x) (in yellow color) are vertical stretches of the graph of f (x) (in red color) by a factor of 2 and 3 respectively.
44
CHAPTER 1. LIBRARY OF FUNCTIONS 5 4 3 y 2 1
–2
–1
0
1 x
2
Example 42 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = 12 x2
x -3 -2 -1 0 1 2 3
y = 13 x2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice?
Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = 12 x2 4.5 2 0.5 0 0.5 2 4.5
x -3 -2 -1 0 1 2 3
y = 13 x2 3 4 3 1 3
0
1 3 4 3
3
(b) The picture below shows that the graphs of 12 f (x) (in green color) and 1 3 f (x) (in yellow color) are vertical compressions of the graph of f (x) (in red color) by a factor of 12 and 13 respectively.
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
45
5 4 3 y 2 1
–2
0
–1
1 x
2
It follows that if a function f (x) is given, then the graph of kf (x) is a vertical stretch of the graph of f (x) by a factor of k for k > 1, and a vertical compression for 0 < k < 1. What about k < 0? If |k| > 1 then the graph of kf (x) is a vertical stretch of the graph of f (x) followed by a reflection about the x-axis. If 0 < |k| < 1 then the graph of kf (x) is a vertical compression of the graph of f (x) by a factor of k followed by a reflection about the x-axis. Example 43 (a) Use a graphing calculator to graph the functions f (x) = x2 , −2f (x), and −3f (x) on the same axes. (b) Use a graphing calculator to graph the functions f (x) = x2 , − 12 f (x), and - 13 f (x) on the same axes. Solution. (a) The picture below shows that the graphs of −2f (x) (in green color) and −3f (x) (in yellow color) are vertical stretches followed by reflections about the x-axis of the graph of f (x) (in red color) 4 y
–2
–1
2 0
1 x
2
–2 –4
(b) The picture below shows that the graphs of − 12 f (x) (in green color) and − 13 f (x) (in yellow color) are vertical compressions of the graph of f (x) (in red
46
CHAPTER 1. LIBRARY OF FUNCTIONS
color) 4 y
–2
–1
2 0
1 x
2
–2 –4
• Horizontal Stretches and Compressions A vertical stretch or compression results from multiplying the outside of a function by a constant k. In this section we will see that multiplying the inside of a function by a constant k results in either a horizontal stretch or compression. A horizontal stretching is the stretching of the graph away from the y-axis. A horizontal compression is the squeezing of the graph towards the y-axis. Example 44 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = (2x)2
x -3 -2 -1 0 1 2 3
y = (3x)2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice? Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = (2x)2 36 16 4 0 4 16 36
x -3 -2 -1 0 1 2 3
y = (3x)2 81 36 9 0 9 36 81
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
47
(b) The picture below shows that the graphs of f (2x) = (2x)2 = 4x2 (in green color) and f (3x) = (3x)2 = 9x2 (in yellow color) are horizontal compressions of the graph of f (x) (in red color) by a factor of 21 and 13 respecitvely. 4 y
–2
–1
2 0
1 x
2
–2 –4
Example 45 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = ( 12 x)2
x -3 -2 -1 0 1 2 3
y = ( 13 x)2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice? Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = ( 12 x)2 9 4
1
1 4
0
1 4
1
9 4
x -3 -2 -1 0 1 2 3
y = ( 13 x)2 1 4 9 1 9
0
1 9 4 9
1
(b) The picture below shows that the graphs of f ( x2 ) (in green color) and f ( x3 ) (in yellow color) are horizontal stretches of the graph of f (x) (in red color) by a factor of 2 and 3 respectively.
48
CHAPTER 1. LIBRARY OF FUNCTIONS 10 8 6 y 4 2 –10
–6 –4
0 –2 –4 –6 –8 –10
2 4 6 8 10 x
It follows that if a function f (x) is given, then the graph of f (kx) is a horizontal stretch of the graph of f (x) by a factor of k1 for 0 < k < 1, and a horizontal compression for k > 1. What about k < 0? If |k| > 1 then the graph of f (kx) is a horizontal compression of the graph of f (x) followed by a reflection about the y-axis. If 0 < |k| < 1 then the graph of f (kx) is a horizontal stretch of the graph of f (x) by a factor of k1 followed by a reflection about the y-axis. Example 46 (a) Use a graphing calculator to graph the functions f (x) = x3 , f (−2x), and f (−3x) on the same axes. (b) Use a graphing calculator to graph the functions f (x) = x3 , f (− x2 ), and f (− x3 ) on the same axes. Solution. (a) The picture below shows that the graphs of f (−2x) (in green color) and f (−3x) (in yellow color) are vertical stretches followed by reflections about the y-axis of the graph of f (x) (in red color) 1 0.8 0.6 y 0.4 0.2 –1
–0.6
0 –0.2 –0.4 –0.6 –0.8 –1
0.20.40.60.8 1 x
(b) The picture below shows that the graphs of f (− x2 ) (in green color) and f (− x3 ) (in yellow color) are horizontal stretches of the graph of f (x) (in red color)
1.9. POLYNOMIAL AND RATIONAL FUNCTIONS
49
0.1 0.08 0.06 y 0.04 0.02 –1
–0.6
0 –0.02 –0.04 –0.06 –0.08 –0.1
0.20.40.60.8 1 x
Recommended Problems (pp. 54 - 56): 1, 5, 7, 13, 14, 15, 16, 17, 18, 19, 31.
1.9
Polynomial and Rational Functions
A function f (x) is a power function of x if there is a constant k such that f (x) = kxn for some constant k. If n > 0, then we say that f (x) is proportional to the nth power of x. If n < 0 then f (x) is said to be inversely proportional to the nth power of x. We call k the constant of proportionality.
Polynomial Functions Polynomial functions are among the simplest, most important, and most commonly used mathematical functions. These functions consist of one or more terms of variables with whole number exponents. (Whole numbers are positive integers and zero.) All such functions in one variable (usually x) can be written in form f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , an 6= 0 where an , an−1 , · · · , a1 , a0 are all real numbers, called the coefficients of f (x). The number n is a non-negative integer. It is called the degree of the polynomial. A polynomial of degree zero is just a constant function. A polynomial of degree one is a linear function, of degree two a quadratic function, etc. The number an is called the leading coefficient and a0 is called the constant term. Note that the terms in a polynomial are written in descending order of the exponents. Polynomials are defined for all values of x. Example 47 Find the leading coefficient, the constant term and the degreee of the polynomial f (x) = 4x5 − x3 + 3x2 + x + 1.
50
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. The given polynomial is of degree 5, leading coefficient 4, and constant term 1. A polynomial function will never involve terms where the variable occurs in a denominator, underneath a radical, as an input of either an exponential or logarithmic function. Example 48 Determine whether the function is a polynomial function or not: (a) (b) (c) (d) (e)
f (x) = 3x4 − 4x2 + 5x − 10 g(x) = x3 − ex + 3 h(x) = x2 − √ 3x + x1 + 4 2 i(x) = x − x − 5 j(x) = x3 − 3x2 + 2x − 5 ln x − 3.
Solution. (a) f (x) is a polynomial function of degree 4. (b) g(x) is not a ploynomial degree because one of the terms is an exponential function. (c) h(x) is not a polynomial because x is in the denominator of a fraction. (d) i(x) is not a polynomial because it contains a radical sign. (e) j(x) is not a olynomial because one of the terms is a logarithm of x. • Graphs of a Polynomial Function Polynomials are continuous and smooth everywhere: • A continuous function means that it can be drawn without picking up your pencil. There are no jumps or holes in the graph of a polynomial function. • A smooth curve means that there are no sharp turns (like an absolute value) in the graph of the function. • The y-intercept of the polynomial is the constant term a0 . The shape of a polynomial depends on the degree and leading coefficient: • If the leading coefficient, an , of a polynomial is positive, then the right hand side of the graph will rise towards +∞. • If the leading coefficient, an , of a polynomial is negative, then the right hand side of the graph will fall towards −∞. • If the degree, n, of a polynomial is even, the left hand side will do the same as the right hand side. • If the degree, n, of a polynomial is odd, the left hand side will do the opposite of the right hand side.
1.9. POLYNOMIAL AND RATIONAL FUNCTIONS
51
Example 49 According to the graphs given below, indicate the sign of an and the parity of n for each curve.
(a)
(b)
(c)
(d)
Solution. (a) an < 0 and n is odd. (b) an > 0 and n is odd. (c) an > 0 and n is even. (d) an < 0 and n is even. • Long-Run Behavior of a Polynomial Function If f (x) and g(x) are two functions such that f (x) − g(x) ≈ 0 as x increases without bound then we say that f (x) resembles g(x) in the long run. For example, if n is any positive integer then x1n ≈ 0 in the long run. Now, if f (x) = an xn + an−1 xn + · · · + a1 x + a0 then an−1 an−2 a1 a0 f (x) = xn (an + + 2 + · · · + n−1 + n x x x x Since
1 xk
≈ 0 in the long run, for each 0 ≤ k ≤ n − 1 then f (x) ≈ an xn
in the long run. Example 50 The polynomial function f (x) = 1−2x4 +x3 resembles the function g(x) = −2x4 in the long run. • Zeros of a Polynomial Function If f is a polynomial function in one variable, then the following statements are equivalent: • x = a is a zero or root of the function f. • x = a is a solution of the equation f (x) = 0. • (a, 0) is an x-intercept of the graph of f. That is, the point where the graph crosses the x-axis.
52
CHAPTER 1. LIBRARY OF FUNCTIONS
Example 51 Find the x-intercepts of the polynomial f (x) = x3 − x2 − 6x. Solution. Factoring the given function to obtain f (x) = =
x(x2 − x − 6) x(x − 3)(x + 2)
Thus, the x-intercepts are the zeros of the equation x(x − 3)(x + 2) = 0 That is, x = 0, x = 3, or x = −2. Rational Functions A rational function is a function that is the ratio of two polynomials functions f (x) g(x) . The domain consists of all numbers such that g(x) 6= 0. Example 52 Find the domain of the function f (x) =
x−2 x2 −x−6 .
Solution. The domain consists of all numbers x such that x2 − x − 6 6= 0. But this last quadratic expression is 0 when x = −2 or x = 3. Thus, the domain is set (−∞, −2) ∪ (−2, 3) ∪ (3, ∞). Geometrically, the values of x for which the denominator of a rational function is zero are called vertical asymptotes. Thus, if x = a is a vertical asymptote then as x approaches a from either sides the function becomes either positively large or negatively large. The graph of a function never crosses its vertical asymptotes. Example 53 Find the vertical asymptotes of the function f (x) =
2x−11 x2 +2x−8
Solution. Factoring x2 + 2x − 8 = 0 we find (x − 2)(x + 4) = 0. Thus, the vertical asymptotes are the lines x = 2 and x = −4. If f (x) approaches a value b as x → ∞ or x → −∞ then we call y = b a horizontal asymptote. The graph of a rational function may cross its horizontal asymptote. Example 54 Find the horizontal asymptote, if it exists, for each of the following functions: (a) f (x) = (b) f (x) = (c) f (x) =
3x2 +2x−4 2x2 −x+1 . 2x+3 x3 −2x2 +4 . 2x2 −3x−1 . x−2
1.9. POLYNOMIAL AND RATIONAL FUNCTIONS
53
Solution. (a) In the long run, we have f (x)
= = ≈
3x2 +2x−4 2x2 −x+1 2 4 x2 3+ x − x2 · 1 2 x 2− x + 12 x 3 2
So the line y = 32 is the horizontal asymptote. (b) In the long run 2x+3 f (x) = x3 −2x2 +4 =
x x3
·
3 2+ x 2 1− x +
≈
0
4 x3
So the x-axis is the horizontal asymptote. (c) In the long run 2x2 −3x−1 f (x) = x−2 = ≈
x2 x
·
3 2− x −
2 1− x
1 x2
∞
So the function has no horizontal asymptote. If ((mx + b) − f (x)) ≈ 0 in the long run then we call the line y = mx + b an oblique asymptote. Example 55 Find the oblique asymptote of the function f (x) =
2x2 −3x−1 . x−2
Solution. Using long division of polynomials we can write f (x) = 2x + 1 + Thus, f (x) − (2x + 1) = asymptote.
1 x−2
1 x−2
≈ 0 in the long run. Thus, y = 2x + 1 is the oblique
To graph a rational function h(x) =
f (x) g(x) :
1. Find the domain of h(x) and therefore sketch the vertical asymptotes of h(x). 2. Sketch the horizontal or the oblique asymptote if they exist. 3. Find the x − intercepts of h(x) by solving the equation f (x) = 0. 4. Find the y-intercept: h(0) 5. Draw the graph Example 56 Sketch the graph of the function f (x) =
2 x+3
54
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. 1. 2. 3. 4. 5.
Domain = (−∞, −3) ∪ (−3, ∞). The vertical asymptote is x = −3. In the long run, f (x) ≈ 0 so the axis is the horizontal asymptote. No x-intercepts. The y-intercept is y = 13 . The graph is given below 2 1 y –6
–4 x –2
0
2
–1 –2
Recommended Problems (pp. 60 - 2): 1, 4, 7, 9, 13, 15, 17, 21, 25, 27, 30, 37.
Chapter 2
The Derivative of a Function In Chapter 1, we introduced the average rate of change of a function on a closed interval. In this chapter, we want to introduce the rate of change of a function at a point. This is defined to be the value obtained by taking the average rate of change over smaller and smaller intervals. The rate of change at a point is called the instantaneous rate of change or the derivative.
2.1
Instantaneous Rate of Change
In this section, we discuss the concept of the instantaneous rate of change of a given function. As an application, we use the velocity of a moving object. The motion of an object along a line at a particular instant is very difficult to define precisely. The modern approach consists of computing the average velocity over smaller and smaller time intervals. To be more precise, let s(t) be the position function of a moving object at time t. We would like to compute the velocity of the object at the instant t = t0 : • Average Velocity We start by finding the average velocity of the object over the time interval t0 ≤ t ≤ t0 + ∆t given by the expression v=
Distance T raveled s(t0 + ∆t) − s(t0 ) = Elapsed T ime ∆t
Geometrically, the average velocity over the time interval [t0 , t0 + ∆t] is just the slope of the line joining the points (t0 , s(t0 )) and (t0 + ∆t, s(t0 + ∆t)) on the graph of s(t).(See Fig. 1) 55
56
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
t
t0+ t
0
Fig. 1
Example 57 A freely falling body experiencing no air resistance falls s(t) = 16t2 feet in t seconds. Complete the following table time interval Average velocity
[1.8,2]
[1.9,2]
[1.99,2]
[1.999,2]
[1.8,2] 60.8
[1.9,2] 62.4
[1.99,2] 63.84
[1.999,2] 63.98
[2,2.0001]
[2,2.001]
[2,2.01]
Solution. time interval Average velocity
[2,2.0001] 64.0016
[2,2.001] 64.016
• Instantaneous Velocity and Speed The next step is to calculate the average velocity on smaller and smaller time intervals ( that is, make ∆t close to zero). The average velocity in this case approaches what we would intuitively call the instantaneous velocity at time t = t0 which is defined using the limit notation by
v(t0 ) = lim
∆t→0
s(t0 + ∆t) − s(t0 ) ∆t
Geometrically, the instantaneous velocity at t0 is the slope of the tangent line to the graph of s(t) at the point (t0 , s(t0 )).(See Fig. 2)
[2,2.01] 64.16
2.1. INSTANTANEOUS RATE OF CHANGE
57
tangent line t
t0+ t
0
Fig. 2 Example 58 For the distance function in Example ??, find the instantaneous velocity at t = 2. Solution. Examining the bottom row of the table in Example ??, we see that the average velocity seems to be approaching the value 64 as we shrink the time intervals. Thus, it is reasonable to expect the velocity to be v(2) = 64 f t/sec. In general, we define the instantaneous rate of change of a function y = f (x) at x = a to be f (x) − f (a) lim . x→a x−a We define the speed of a moving object to be the absolute value of the velocity function. Sometimes there is confusion between the words ”speed” and ”velocity”. Speed is a nonnegative number that indicates how fast an object is moving, whereas velocity indicates both speed and direction(relative to a coordinate system). For example, if the object is moving along a vertical line we define a positive velocity when the object is going upward and a negative velocity when the object is going downward. The instantaneous rate of change of a function f (x) from x = a to x = a + h is called the derivative of f at a and we denote it by f 0 (a) : f 0 (a) = lim
h→0
f (a + h) − f (a) h
If the derivative exists, i.e. can be found, then we say that the function is differentiable at x = a. The process of finding the derivative of a function is called differentiation. If a function has no derivative at a point then we say
58
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
that it is non-differentiable there. Since the instantaneous rate of change represents the slope of a tangent line then f 0 (a) is the slope of the tangent line to the graph of y = f (x) at the point (a, f (a)). The equation of the tangent line is given by y − f (a) = f 0 (a)(x − a). Exercise 26 (a) Find f 0 (1) for f (x) = x2 . (b) Find the equation of the tangent line to the graph of f (x) at the point (1, f (1)). Solution. Completing the following chart x f (b)−f (a) b−a
[0.9,1] 1.9
[0.99,1] 1.99
[0.999,1] 1.999
[1,1.0001] 2.0001
[1,1.001] 2.001
[1,1.01] 2.01
[1,1.1] 2.1
we see that f 0 (1) = 2. (b) The equation of the tangent line is y − f (1) = f 0 (1)(x − 1) or y − 1 = 2(x − 1) In point-intercept form, we have y = 2x − 1. Example 59 (Numerical Estimation of the Derivative) Find approximate values for f 0 (x) at each of the x-values given in the following table x f(x)
0 100
5 70
10 55
15 46
20 40
Solution. The derivative can be estimated by the average rate of change or the difference quotient f (a + h) − f (a) f 0 (a) ≈ . h Thus, f 0 (0) f 0 (5) f 0 (10) f 0 (15)
≈ ≈ ≈ ≈
f (5)−f (0) 5 f (10)−f (5) 5 f (15)−f (10) 5 f (20)−f (15) 5
= −6 = −3 = −1.8 = −1.2
2.2. THE DERIVATIVE FUNCTION
59
Note that in the above example, we used the points (5, 70) and (10, 55) to estimate f 0 (5). This is known as right slope estimation. Similarly, we can estimate f 0 (5) by using a left slope estimation,i.e. f 0 (5) ≈
f (5) − f (0) = −6 5−0
An improved estimation consists of taking the average of the left slope and the right slope, that is, −3 − 6 f 0 (5) ≈ = −4.5. 2 Example 60 Use the average of the right slope estimation and left slope estimation is estimating the values of f 0 of the previous example. Solution. f 0 (0) 0
f (5) 0
≈ ≈
f (10)
≈
f 0 (15)
≈
0
f (20)
≈
f (5)−f (0) 5 ´ f (5)−f (0) 1 f (10)−f (5) + 2³ 5 5 ´ f (10)−f (5) 1 f (15)−f (10) + 2³ 5 5 ´ f (15)−f (10) 1 f (20)−f (15) + 2 5 5 f (20)−f (15) 5
³
=
−6
= −4.5 = −2.4 = −1.5 = −1.2
Recommended Problems (pp. 99 - 101): 1, 3, 4, 6, 8, 10, 11, 12, 13, 15, 18, 19, 24.
2.2
The derivative Function
Recall that a function f is differentiable at x if the following limit exists f (x + h) − f (x) h→0 h
f 0 (x) = lim
(2.1)
Thus, we associate with the function f , a new function f 0 whose domain is the set of points x at which the limit (??) exists. We call the function f 0 the derivative function of f. Since the derivative at a point represents the slope of the tangent line then one can obtain the graph of the derivative function from the graph of the original function. It is important to keep in mind the relationship between the graphs of f and f 0 . If f 0 (x) > 0 on an interval I then the tangent line must be tilted upward and the graph of f is rising or increasing over that interval. Similarly, if f 0 (x) < 0 on I then the tangent line is tilted downward and the graph of f is falling or decreasing over I. If f 0 (x) = 0 for all x in I then the tangent line is always horizontal and this occurs only when f is a constant function over I.
60
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
Example 61 Sketch the graph of the derivative of the function shown below Fig. 9 30 20 –2
10
2
x
4
6
–10 –20 –30 –40
Solution. Note that for x < −0.36 or x > 3.7 the derivative is positive whereas for −0.36 < x < 3.7 the derivative is negative. Also, f 0 (−0.3) = f 0 (3.7) ≈ 0. Thus, a possible graph of f 0 is given in Fig. 10 Fig. 10 50 40 30 20 10 0
–2
–10
2
x
4
6
Example 62 (Numerical Estimation of the Derivative) Find approximate values for f 0 (x) at each of the x-values given in the following table x f(x)
0 100
5 70
10 55
15 46
20 40
Solution. The derivative can be estimated by the average rate of change or the difference quotient f (a + h) − f (a) f 0 (a) ≈ . h Thus, f (5)−f (0) = −6 f 0 (0) ≈ 5 (5) f 0 (5) ≈ f (10)−f = −3 5 (10) = −1.8 f 0 (10) ≈ f (15)−f 5 (15) f 0 (15) ≈ f (20)−f = −1.2 5
2.2. THE DERIVATIVE FUNCTION
61
Note that in the above example, we used the points (5, 70) and (10, 55) to estimate f 0 (5). This is known as right slope estimation. Similarly, we can estimate f 0 (5) by using a left slope estimation,i.e. f 0 (5) ≈
f (5) − f (0) = −6 5−0
An improved estimation consists of taking the average of the left slope and the right slope, that is, −3 − 6 f 0 (5) ≈ = −4.5. 2 Exercise 27 Estimate f 0 (20) in the previous exercise. Solution. Since 20 is a right endpoint then we will use a left slope estimation to obtain f 0 (20) ≈
f (20) − f (15) = −1.2 20 − 15
Now, if a formula for f is given then by applying the definition of f 0 (x) as the limit of the difference quotient we can find a formula of f as shown in the following two examples. Example 63 (Derivative of a Constant Function) Suppose that f (x) = k for all x. Find a formula for f 0 (x). Solution. (x) f 0 (x) = limh→0 f (x+h)−f h = limh→0 k−k h =0
Thus, f 0 (x) = 0. Example 64 (Derivative of a Linear Function) Find the derivative of the linear function f (x) = mx + b. Solution. f 0 (x) = = =
(x) limh→0 f (x+h)−f h m(x+h)+b−(mx+b) limh→0 h limh→0 mh h =m
Thus, f 0 (x) = m. Recommended Problems (pp. 104 - 6): 5, 6, 7, 9, 11, 13, 15, 19, 21.
62
2.3
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
Leibniz Notation for The Derivative
When dealing with mathematical models that involve derivatives it is convenient dy to denote the prime notation of the derivative of a function y = f (x) by dx . That is, dy = f 0 (x) dx This notation is called Leibniz notation (due to W.G. Leibniz). For example, dy we can write dx = 2x for y 0 = 2x. When using Leibniz notation to denote the value of the derivative at a point a we will write ¯ dy ¯¯ dx ¯x=a Thus, to evaluate
dy dx
= 2x at x = 2 we would write ¯ dy ¯¯ = 2x|x=2 = 2(2) = 4. dx ¯x=2
dy appears as a fraction but it is not. It is just an alRemark. Even though dx ternative notation for the derivative. A concept called differential will provide meaning to symbols like dy and dx.
One of the advantages of Leibniz notation is the recognition of the units of the derivative. For example, if the position function s(t) is expressed in meters and the time t in seconds then the units of the velocity function ds dt is meters/sec. In general, the units of the derivative are the units of the dependent variable divided by the units of the independent variable. Example 65 The cost, C ( in dollars) to produce x gallons of ice cream can be expressed as C =¯ f (x). What are the units of measurements and the meaning of the statement dC ¯ dx x=200 = 1.4? Solution. dC dx is measured in dollars per gallon. The notation ¯ dC ¯¯ = 1.4 dx ¯x=200 means that if 200 gallons of ice cream have already been produced then the cost of producing the next gallon will be roughly 1.4 dollars. Exercise 28 The derivative of the velocity function v is called acceleration and is denoted by a. Suppose that v is measured in meters/seconds, what are the units of a?
2.4. THE SECOND DERIVATIVE
63
Solution. The units of a are meters/seconds/seconds = meters/seconds2 . Finally, one can use the derivative at a point to approximate values of the function at nearby points. For example, if we know the values of f (a) and f 0 (a) then for a nearby point b the value of f (b) is found by the formula f (b) ≈ f 0 (a)(b − a) + f (a). Exercise 29 Climbing health care costs have been a source of concern for some time. Use the data in the table below to estimate the average per capita expenditure in 1991 and 2010 assuming that the costs climb at the same rate since 1990. Year Per capita expenditure ($)
1970 349
1975 591
1980 1055
1985 1596
1990 2714
Solution. = $223.60 Between 1985 and 1990 the rate of increase in the costs is 2714−1596 5 per year. Since we are assuming that the costs continue to increase at the same rate then C(1991) ≈ C(1990) + C 0 (1990)(1991 − 1990) = 2714 + 223.60 = $2937.60 and C(2010) = 2714 + 223.60(10) = $7, 186. Recommended Problems (pp. 111 - 3): 1, 3, 6, 8, 9, 12, 13, 14, 17, 23.
2.4
The Second Derivative
Let f (x) be a differentiable function. If the limit f 0 (x + h) − f 0 (x) h→0 h lim
exists then we say that the function f 0 (x) is differentiable and we denote its derivative by f 00 (x) or using Leibniz notation µ ¶ d2 y d dy = . dx2 dx dx We call f 00 (x) the second derivative of f (x). Now, recall that if f 0 (x) > 0 ( resp. f 0 (x) < 0) over an interval I then the function f (x) is increasing (resp. decreasing on I). So if f 00 (x) > 0 on I then f 0 (x) is increasing on I. This occurs only when the graph of f is bending up
64
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
on the interval I. In this case, we say that f is concave up on I. Similarly, if f 00 (x) < 0 on I then f 0 (x) is decreasing and this happens when the graph of f is bending down. In this case, we say that f is concave down. Remark. Note that when a curve is concave up then the tangent lines lie below the curve whereas when it is concave down then the tangent lines lie above the curve. Example 66 Give the signs of f 0 and f 00 for the following function Fig. 11. 8 6 4 2 0
0.5
1
1.5 x
2
2.5
3
Solution. Since f is always increasing then f 0 is always positive. Since the graph is concave up then f 00 is always positive. Example 67 Find where the graph of f (x) = x3 +3x+1 is concave up and where it is concave down. Solution. Finding the first and second derivatives of f we obtain f 0 (x) = 3x2 + 3 and f 00 (x) = 6x. Thus, the graph of f is concave up for x > 0 and concave down for x < 0. As an application to the second derivative, we consider the motion of an object determined by the position function s(t). Recall that the velocity of the object is defined to be the first derivative of s(t), i.e. ds dt and the absolute value of v(t) is the speed. When the object speeds up we say that he/she accelerates and when the object slows down we say that he/she decelerates. We define the acceleration of an object as the derivative of the velocity function and consequently as the second derivative of the position function ds d2 s . a(t) = 2 = dt dt v(t) = s0 (t) =
2.5. MARGINAL COST AND REVENUE
65
Example 68 A particle is moving along a straight line. If its distance, s, to the right of a fixed point is given by Fig. 12, estimate: (a) When the particle is moving to the right and when it is moving to the left. (b) When the particle has positive acceleration and when it has negative acceleration. Fig. 12.
3 2.5 2 1.5 1 0.5 0
0.5
1
1.5 x
2
2.5
3
Solution. (a) When s is increasing then the particle moves to the right. This occurs when 0 < t < 23 and for t > 2. On the other hand, the particle moves to the left when s is decreasing. This happens when 23 < t < 2. (b) Positive acceleration occurs when the graph is concave up. This occurs when t > 43 . The particle has negative acceleration when the curve is concave down, i.e for t < 43 . Recommended Problems (pp. 115 - 7): 2, 3, 8, 9, 11, 12, 13, 15, 17, 18, 23.
2.5
Marginal Cost and Revenue
Marginal analysis is an area of economics concerned with estimating the effect on quantities such as cost, revenue, and profit when the level of production is changed by a unit amount. For example, if C(q) is the cost of producing q units of a certain commodity, then the marginal cost, MC(q), is the additional cost of producing one more unit and is given by the difference M C(q) = C(q + 1) − C(q). Using the estimation C 0 (q) ≈ we find that
C(q + 1) − C(q) = C(q + 1) − C(q) (q + 1) − q M C(q) ≈ C 0 (q)
and for this reason, we will compute the marginal cost by the derivative C 0 (q). Similarly, if R(q) is the revenue obtained from producing q units of a commodity,
66
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
then the marginal revenue, MR(q), is the additional revenue obtained from producing one more unit, and we compute M R(q) by the derivative R0 (q). Example 69 Let C(q) represent the cost, R(q) the revenue, and P (q) the total profit, in dollars, of producing q units. (a) If C 0 (50) = 75 and R0 (50) = 84, approximately how much profit is earned by the 51st item? (b) If C 0 (90) = 71 and R0 (90) = 68, approximately how much profit is earned by the 91st item? Solution. (a) P 0 (50) = R0 (50) − C 0 (50) = 84 − 75 = 9. (b) P 0 (90) = R0 (90) − C 0 (90) = 68 − 71 = −3. A loss by 3 dollars. We will now consider the question of maximizing profit. To solve this problem, we want to maximize the profit function P (q) = R(q) − C(q). We will see later that the profit function attains its maximum for the level of production q for which P 0 (q) = 0. That is, when M R(q) = M C(q). Example 70 A manufacturer estimates that when q units of a particular commodity are produced each month, the total cost (in dollars) will be C(q) =
1 2 q + 4q + 200 8
and all units are sold at a price p = 49 − q dollars per unit. Determine the price that corresponds to the maximum profit. Solution. The revenue function is given by R(q) = pq = q(49 − q) = 49q − q 2 . Thus, M R(q) = 49 − 2q and M C(q) = 41 q + 4. The profit is maximized when M C(q) = M R(q). That is, when 14 q + 4 = 49 − 2q. Solving this equation for q we find q = 20 units. In this case, the price is p(20) = 49 − 20 = $29. Exercise 30 If the revenue and cost functions are given by the figure below, sketch graphs of M C and M R.
2.5. MARGINAL COST AND REVENUE
67
C
R
100 Solution.
MC
MR
100 Recommended Problems (pp. 122 - 4): 1, 3, 5, 7, 8, 9, 13.
68
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
Chapter 3
Rules of Differentiation In this chapter we derive differentiation formulas for common functions such as exponential, logarithmic, polynomial, and rational functions. Also, we discuss general rules such as the product, quotient, and chain rules which allow us to differentiate combinations of the above list of functions.
3.1
Derivative Formulas for Power and Polynomials
Finding the derivative function by using the limit of the difference quotient is sometimes difficult for functions with complicated expressions. Fortunately, there is an indirect way for computing derivatives that does not compute limits but instead uses formulas which we will derive in this section and in the coming sections. We first derive a couple of formulas of differentiation. Theorem 1 If f is differentiable and k is a constant then the new function kf (x) is differentiable with derivative given by [kf (x)]0 = kf 0 (x). Proof. (x) (x)) [kf 0 (x) = limh→0 kf (x+h)−kf = limh→0 k(f (x+h)−f h h f (x+h)−f (x) 0 = kf (x) = k limh→0 h
where we used the fact that a constant can be taking across the limit sign by the properties of limits. Theorem 2 If f (x) and g(x) are two differentiable functions then the functions f + g and 69
70
CHAPTER 3. RULES OF DIFFERENTIATION
f − g are also differentiable with derivatives [f (x) ± g(x)]0 = f 0 (x) ± g 0 (x) Proof. Again by using the definition of the derivative and the fact that the limit of a sum/difference is the sum/difference of limits we find [f (x) + g(x)]0
= = = limh→0
(f (x+h)+g(x+h))−(f (x)+g(x)) h (f (x+h)−f (x))+(g(x+h)−g(x)) h f (x+h)−f (x) + limh→0 g(x+h)−g(x) = f 0 (x) h h
limh→0 limh→0
+ g 0 (x)
The same proof is valid for the difference formula. Next, we state without proof a rule for finding the derivative of a power function of the form f (x) = xn . Theorem 3 (Power Rule) For any real number n, the derivative of the function y = xn is given by the formula dy = nxn−1 dx Example 71 Use the power rule to differentiate the following: 4
(a) y = x 3
(b) y =
1 √ 3 x
(c) y = xπ .
Solution. 1 (a) Using the power rule with n = 43 to obtain y 0 = 43 x 3 . 1 4 (b) Since y = x− 3 then using the power rule with n = − 13 to obtain y 0 = − 13 x− 3 . (c) Using the power rule with n = π to obtain y 0 = πxπ−1 . Combining the results discussed above, we can find the derivative of functions that are combinations of power functions of the form axn . In particular, the derivative of a polynomial function f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 is given by the formula f 0 (x) = nan xn−1 + (n − 1)an−1 xn−2 + · · · + a1 . Example 72 √ Find the derivative of the function y = 3x7 −
x5 5
+ π.
Solution. √ The derivative is f 0 (x) = 7 3x6 − x4 . Example 73 √ Find the second derivative of y = 5 3 x −
10 x4
+
1 √ . 2 x
3.2. DERIVATIVE FORMULAS FOR EXPONENTIAL AND LOGARITHMIC FUNCTIONS71 Solution. 1 1 Note that the given function can be written in the form y = 5x 3 −10x−4 + 12 x− 2 . Thus, the first derivative is y0 =
1 3 5 −2 x 3 + 40x−5 − x− 2 . 3 4
The second derivative is y 00 = −
10 − 5 3 5 x 3 − 200x−6 + x− 2 . 9 8
Recommended Problems (pp. 141 - 2): 1, 3, 6, 11, 18, 29, 32, 34, 36, 38, 41, 42.
3.2
Derivative Formulas for Exponential and Logarithmic Functions
We start this section by looking at the limit eh − 1 . h→0 h lim
The chart below suggests that the limit is 1. h eh −1 h
-0.01 0.995
-0.001 0.9995
-0.0001 0.99995
0 undefined
0.0001 1.0000
0.001 1.0005
0.01 1.005
Now, let’s try and find the derivative of the function f (x) = ex at any number x. By the definition of the derivative and the limit above we see that f 0 (x) = = = =
(x) limh→0 f (x+h)−f h ex+h −ex limh→0 h x h limh→0 e (eh −1) h ex limh→0 e h−1 = ex
This means that ex is its own derivative: d x (e ) = ex . dx Example 74 Find the derivative of f (x) = ex+2 . Solution. Rewriting f to get f (x) = e2 · ex . Thus, f 0 (x) = e2 (ex )0 = e2 · ex = ex+2 . • Derivative of eu
72
CHAPTER 3. RULES OF DIFFERENTIATION
Now, suppose that the x in ex is replaced by a differentiable function of x, say u(x). We would like to find the derivative of eu with respect to x. That is, what d is dx (eu )? d u dx (e )
= = = = = =
eu(x+h) −eu(x) h u(x+h) −eu(x) u(x+h)−u(x) · limh→0 eu(x+h)−u(x) h eu(x+h) −eu(x) limh→0 u(x+h)−u(x) limh→0 u(x+h)−u(x) h u(x)+w −eu(x) limw→0 e limh→0 u(x+h)−u(x) ww h eu limw→0 e w−1 limh→0 u(x+h)−u(x) h d eu dx (u)
limh→0
where w = u(x + h) − u(x). Note that since u is differentiable then it is continuous so that limh→0 (u(x + h) − u(x)) = 0. Also, we have used the limit discussed at the beginning of this section. • Derivative of ax We can now use the above discussion to find the derivative of the function f (x) = ax . First, note that f (x) = ax = ex ln a since in general aloga x = x. Thus, by the previous result we see that d x d x ln a d (a ) = (e ) = ex ln a (x ln a) = ln aex ln a = ax ln a. dx dx dx Example 75 Find the derivative of each of the following functions: (a) f (x) = 3x
(b) y = 2 · 3x + 5 · e3x−4 .
Solution. (a) f 0 (x) = 3x ln 3. (b) y 0 = 2(3x )0 + 5(e3x−4 )0 = 2 · 3x ln 3 + 5(3)e3x−5 = 2 · 3x ln 3 + 15 · e3x−4 . • Derivative of ln x We end this section by finding the derivative of f (x) = ln x. Since eln x = x then taking the derivative of both sides to obtain eln x (ln x)0 = 1 or (ln x)0 = eln1 x = x1 . Using Leibnitz notation we have d 1 (ln x) = dx x Example 76 Find the equation of the tangent line to the graph of f (x) = ln x at x = 2. Solution. Since f 0 (x) = x1 then the slope of the tangent line is f 0 (2) = 21 . Thus, y = x2 + b. Since the line crosses the point (2 ln 2) then ln 2 = 22 + b or b = ln 2 − 1. Hence,
3.3. DERIVATIVES OF COMPOSITE FUNCTIONS: THE CHAIN RULE73 y=
x 2
+ ln 2 − 1.
Recommended Problems (pp. 28, 31, 32.
3.3
145 - 6): 1, 5, 11, 15, 16, 21, 25,
Derivatives of Composite Functions: The Chain Rule
In this section we want to find the derivative of a composite function f (g(x)) where f (x) and g(x) are two differentiable functions: d dx [f (g(x))]
= = = = =
f (g(x+h))−f (g(x)) h (g(x)) g(x+h)−g(x) limh→0 f (g(x+h))−f · g(x+h)−g(x) h f (g(x+h))−f (g(x)) limh→0 g(x+h)−g(x) · limh→0 g(x+h)−g(x) h (g(x)) · limh→0 g(x+h)−g(x) limw→0 f (w+g(x))−f w h 0 0
limh→0
f (g(x))g (x)
where w = g(x + h) − g(x) and limh→0 (g(x + h) − g(x)) = 0 since g(x) is continuous(since it is differentiable). This result is known as the chain rule. Thus, the derivative of f (g(x)) is the derivative of f (x) evaluated at g(x) times the derivative of g(x). Example 77 Let’s find the derivative of y = (4x2 + 1)7 . First note that y = f (g(x)) where f (x) = x7 and g(x) = 4x2 + 1. Thus, f 0 (x) = 7x6 , f 0 (g(x)) = 7(4x2 + 1)6 and g 0 (x) = 8x. So according to the chain rule, y 0 = 7(4x2 +1)6 (8x) = 56x(4x2 +1)6 . As a result of this rule we can have the following differentiation formulas: d u du (e ) = eu , dx dx
d n du (u ) = nun−1 , dx dx
d u0 (ln u) = dx u
where u is a function of x. Exercise 31 2 Find the derivative of the function f (x) = (3x2 − 5)3 + ln (x10 + 1) − 5e−x . Solution. Using the chain rule we have 2
f 0 (x) = [(3x2 − 5)3 ]0 + [ln (x10 + 1)]0 − 5[e−x ]0 10 0 −x2 = 3(3x2 − 5)2 (3x2 − 5)0 + (xx10+1) (−x2 )0 +1 − 5e 10x9 2 −x2 = 9(3x − 5) + x10 +1 + 10xe
74
CHAPTER 3. RULES OF DIFFERENTIATION
Exercise 32 Suppose that $1,000 is deposited into a bank account that pays 8% annual interest, compounded continuously. Find a formula f (t) for the balance t years after the initial deposit. Find f (10) and f 0 (10) and explain what your answers mean in terms of money. Solution. The formula is given by
f (t) = 1, 000e0.08t .
The balance after 10 years is f (10) = 1, 000e0.08(10) ≈ $2, 225.54 Next, we compute f 0 (10). Since f 0 (t) = 1, 000(0.08)e0.08t = 80e0.08t then f 0 (10) = 80e0.8 ≈ $178.04 This means that after 10 years, the balance is growing at the rate of about $178 per year. Recommended Problems (pp. 149 - 150): 3, 15, 23, 29, 30, 32, 33, 35, 36, 38.
3.4
The Product and Quotient Rules
At this point we don’t have the tools to find the derivative of either the function 2 2 f (x) = x3 ex or the function g(x) = xex . Looking closely at the function f (x) 2 we notice that this function is the product of two functions, namely, x3 and ex . On the other hand, the function g(x) is the ratio of two functions. Thus, we hope to have a rule for differentiating a product of two functions and one for differentiating the ratio of two functions. We start by finding the derivative of the product u(x)v(x), where u and v are differentiable function: 0
(u(x)v(x))
= limh→0 u(x+h)v(x+h)−u(x)v(x) h = limh→0 u(x+h)(v(x+h)−v(x))+v(x)(u(x+h)−u(x)) h = limh→0 u(x + h) limh→0 v(x+h)−v(x) + v(x) limh→0 u(x+h)−u(x) h h 0 0 = u(x)v (x) + u (x)v(x)
Note that since u is differentiable so it is continuous and therefore lim u(x + h) = u(x).
h→0
The formula d d d (u(x)v(x)) = u(x) (v(x)) + (u(x))v(x) dx dx dx is called the product rule.
(3.1)
3.4. THE PRODUCT AND QUOTIENT RULES
75
Example 78 2 2 Let u(x) = x3 and v(x) = ex . Then u0 (x) = 3x2 and v 0 (x) = 2xex . Thus, by the product rule we have 2
2
2
2
f 0 (x) = x3 (2x)ex + 3x2 ex = 2x4 ex + 3x2 ex . The quotient rule is obtained from the product rule as follows: Let f (x) = u(x) 0 0 v(x) . Then u(x) = f (x)v(x). Using the product rule, we find u (x) = f (x)v (x)+ 0 0 f (x)v(x). Solving for f (x) to obtain u0 (x) − f (x)v 0 (x) . v(x)
f 0 (x) = Now replace f (x) by
u(x) v(x)
³
to obtain u(x) v(x)
´0
u(x)
= = =
Example 79 Consider the function g(x) = quotient rule we have
x2 ex .
f 0 (x)
u0 (x)− v(x) v 0 (x) v(x) u0 (x)v(x)−u(x)v 0 (x) v(x)
v(x) u0 (x)v(x)−u(x)v 0 (x) (v(x))2
Let u(x) = x2 and v(x) = ex . Then by the
= =
(x2 )0 ex −x2 (ex )0 (ex )2 2xex −x2 ex e2x
Recommended Problems (pp. 152 - 3): 3, 5, 12, 22, 25, 27, 33, 34, 35, 38.
76
CHAPTER 3. RULES OF DIFFERENTIATION
Chapter 4
Applications of the Derivative In this chapter, we will apply the notion of derivative in solving optimization problems. By that we mean, finding the maximum and the minimum (if they exist) of a given function.
4.1
Using First and Second Derivatives
We start this section by reviewing what the first and second derivatives of a function tell us about its graph: • • • •
If If If If
f 0 (x) > 0 on an open interval I then f (x) is increasing on I. f 0 (x) < 0 on an open interval I then f (x) is decreasing on I. f 00 (x) > 0 on an open interval I then f (x) is concave up on I. f 00 (x) < 0 on an open interval I then f (x) is concave down on I.
Exercise 33 Consider the function f (x) = x3 − 9x2 − 48x + 52. (a) Find the intervals where the function is increasing and decreasing. (b) Find the intervals where is the function is concave up and concave down. Solution. (a) Finding the first derivative we obtain f 0 (x) = 3x2 −18x−48 = 3(x−8)(x+2). Constructing the chart of signs below
77
78
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
(-)(-)
(-)(+) -2
(+)(+) 8
(+)
(-)
(+)
we see that f (x) is increasing on (−∞, −2) ∪ (8, ∞) and decreasing on (−2, 8). (b) Finding the second derivative, we obtain f 00 (x) = 6x − 18 = b(x − 3). So, f (x) is concave up on (3, ∞) and concave down on (−∞, 3). • Local Maxima and Minima Points of interest on the graph of a function are those points that are the highest on the curve, or the lowest, in a specific interval. Such points are called local extrema. The highest point, say f (a), is called a local maximum and satisfies f (x) ≤ f (a) for all x in an interval I. A local minimum is a point f (a) such that f (a) ≤ f (x) for all x in an interval I. Exercise 34 Find the local maxima and the local minima of the given function. 500 –10
10 x 20
30
–500 –1000
Solution. The local maxima occur at x = −2 and x = 14 whereas the local minimum occurs at x = 8. Next we will discuss two procedures for finding local extrema. We notice from the previous exercise that local extrema occur at points p where the derivative is either zero or undefined. We call (p, f (p)) a critical point and f (p) a critical value. Also, the graph in the previous exercise suggests the following test for finding local extrema: First-Derivative Test Suppose that a continuous function f has a critical point at p. • If f 0 changes sign from negative to positive at p, then f has a local minimum
4.1. USING FIRST AND SECOND DERIVATIVES
79
at p. • If f 0 changes sign from positive to negative at p, then f has a local maximum at p. Exercise 35 (a) Find the local extrema of the function f (x) = x3 − 9x2 − 48x + 52. (b) Find the local extrema of the function f (x) = sin x + ex , x ≥ 0. Solution. (a) Using the chart of signs of f 0 discussed in the first exercise of this section, we find that f (x) has a local maximum at x = −2 and a local minimum at x = 8. (b) Finding the derivative to obtain f 0 (x) = cos x + ex . But for x ≥ 0, 1 ≤ ex . Since −1 ≤ cos x ≤ 1 then adding the two inequalities we see that 0 ≤ cos x+ex . This implies that f 0 does not change sign for x ≥ 0. Therefore, there are no local maxima or minima. We have seen that local extrema are always critical points. The converse of this statement is not true in general. That is, there are critical points that are not local extrema of a function. An example of this situation is discussed in the next exercise. Exercise 36 Show that f (x) = x3 has a critical point at x = 0 but 0 is neither a local maximum nor a local minimum. Solution. Finding the derivative to obtain f 0 (x) = 3x2 . Setting this to 0 we find the critical point x = 0. Since f 0 (x) does not change sign at 0 then 0 is neither a local maximum nor a local minimum. Looking again closely to the graph of the second problem of this section we can use concavity in finding the local extrema: Second-Derivative Test Let f be a continuous function such that f 0 (p) = 0. • if f 00 (p) > 0 then f has a local minimum at p. • if f 00 (p) < 0 then f has a local maximum at p. • if f 00 (p) = 0 then the test fails. In this case, it is recommended that you use the first derivative test. Exercise 37 Use the second derivative test to find the local extrema of the function f (x) = x3 − 9x2 − 48x + 52. Solution. The second derivative of f (x) is given by f 00 (x) = 6(x − 3). The critical numbers are −2 and 8. Since f (−2) = −30 < 0 then x = −2 is a local maximum. Since f (8) = 30 > 0 then x = 8 is a local minimum.
80
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
Exercise 38 Find the local extrema of the function f (x) = x4 . Solution. Let’s try and find the local extrema by using the second derivative test. Since f 0 (x) = 4x3 then x = 0 is the only critcal number. Since f 00 (x) = 12x2 then f 00 (0) = 0. So the second derivative test is inconclusive. Now, using the first derivative test, we see that f 0 (x) changes sign from negative to positive at x = 0. so that x = 0 is a local minimum. Recommended Problems (pp. 170 - 1): 1, 3, 5, 7, 9, 12, 15, 17, 19, 21.
4.2
Concavity and Points of Inflection
We have seen that a local extremum is a point where the first derivative changes sign. In this section we will discuss points where the second derivative changes sign. That is, the points where the graph of the function changes concavity. We call such points points of inflection. How do you find the points of inflection? Well, since f 00 changes sign on the two sides of an inflection point then it makes sense to say that points of inflection occur at points where either the second derivative is 0 or undefined. Exercise 39 Find the point(s) of inflection of the function f (x) = xe−x . Solution. Using the product rule to obtain f 0 (x) = e−x − xe−x . Using the product rule for the seond time we find f 00 (x) = e−x (x − 2). Thus, a candidate for a point of inflection is x = 2. Since f 00 (x) > 0 for x > 2 and f 00 (x) < 0 for x < 2 then x = 2 is a point of inflection. Remark. We have seen that not every value of x where the derivative is zero or undefined is a local maximum or minimum. The same thing applies for points of inflection. That is, it is not always true that if the second derivative is 0 or undefined then automatically you have a point of inflection. It is critical that f 00 changes sign at such a point in order to have a point of inflection. Exercise 40 Consider the function f (x) = x4 . Show that f 00 (0) = 0 but 0 is not a point of inflection. Solution. The second derivative is given by the formula f 00 (x) = 12x2 . Clearly, f 00 (0) = 0. Since f 00 (x) ≥ 0, that is, f 00 (x) does not change sign then 0 is not a point of inflection.
4.2. CONCAVITY AND POINTS OF INFLECTION
81
Exercise 41 Graph the derivative of the function f (x) = x + sin x. Determine where f is increasing most rapidly, and least rapidly.
Solution. The derivative of f (x) is given by the expression f 0 (x) = cos x + 1 ≥ 0 so that f (x0 is always increasing. Now, f (x) increases most rapidly at the maximum values of f 0 (x) and increases least rapidly at the minimum values of f 0 (x). Graphing the function f 0 (x) we find Plot of 1+cos(x) 2
±3π ±2π −π
π
2π
3π
Thus, f increases most rapidly at x = 2nπ and least rapidly at x = (2n + 1)π where n is an integer.
Exercise 42 Graph a function with the following properties: • f has a critical point at x = 4 and an inflection point at x = 8. • f 0 < 0 for x < 4 and f 0 > 0 for x > 4. • f 00 > 0 for x < 8 and f 00 < 0 for x > 8.
Solution.
82
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
4
8
Recommended Problems (pp. 175 - 6): 1, 3, 7, 9, 15, 16, 26, 27, 28, 29.
4.3
Global Maxima and Minima
In this section we will look for the largest or the smallest values of a function on its domain. Such points are called global extrema. If f (a) is the largest value then it satisfies the inequality f (x) ≤ f (a) for all x in the domain of f. We call f (a) the global or absolute maximum value of f and the point (a, f (a)) the global maximum point. Similarly, if f (a) is the smallest value of f (x) then f (a) ≤ f (x) for all x in the domain of f. We call f (a) the absolute or global minimum value of f and the point (a, f (a)) the global minimum. The process of finding the global extrema is called optimization. Problems that involve finding the global extrema are called optimization problems. How do we find the global extrema? • If the function is continuous on a closed interval then the global extrema occur at either the critical points or the endpoints of the interval. Example 80 Find the global extrema of the function f (x) = x3 − 9x2 − 48x + 52 on the closed interval [−5, 12]. Solution. Finding the derivative of f (x) we get f 0 (x) = 3x2 −18x−48. Solving the equation f 0 (x) = 0 that is, x2 − 6x − 16 = 0 we find the critical points at x = 8 and
4.3. GLOBAL MAXIMA AND MINIMA x = −2. Now, evaluating the function find f (−5) f (−2) f (8) f (12)
83
at these points and at the endpoints we = = = =
−58 104 −396 −92
It follows that (−2, 104) is the global maximum point and (8, −396) is the global minimum point. • If a function is continuous on an open interval or on all real numbers then it is recommended to find the global extrema by graphing the function. Example 81 Find the global extrema of the function f (x) = 100(e−0.02x − e−0.1x ) for x ≥ 0. Solution. Let’s sketch the graph of this function. The standard process of graphing consists of the following steps: Step 1. Find the critical numbers. Setting f 0 (x) = 0 to obtain 100(−0.02e−0.02x + 0.1e−0.1x ) = 0 0.02e−0.02x = 0.1e−0.1x e−0.02x 0.1 = e−0.1x 0.02 0.08x e = 5 0.08x = ln 5 ln 5 x = 0.08
= 20.12
Step 2. We construct the following chart:
x f’(x) f(x)
+ %
20.12 0 53.50
&
Step 3. Find the second derivative to obtain f 00 (x) = 100(0.0004e−0.02x −0.01e−0.1x ). Setting this to zero and solving for x as in Step 2 we find x ≈ 40.25. Now we construct the table
x f”(x) f(x) Step 4.Graph
∩
40.25 0 f(40.25)
+ ∪
84
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE 50 40 30 20 10 0
20
40
x
60
80
100
Thus, from the graph we see that (20.12, 53.50) is a global maximum. The function has a global minimum at x = 0. Recommended Problems (pp. 21, .
4.4
179 - 182): 2, 3, 5, 9, 11, 13, 19,
Applications of Optimization to Marginality
Management of most businesses always aim to maximizing profit. In this section we will use the derivative to optimizie profit and revenue functions. • Optimizing Profit Recall that the profit resulting from producing and selling q items is defined by P (q) = R(q) − C(q) where C(q) is the total cost of producing a quantity q and R(q) is the total revenue from selling a quantity q of some good. To maximize or minimize profit over a closed interval, we optimize the profit function P. We know that global extrema occur at the critical numbers of P or at the endpoints of the interval. Thus, the process of optimization requires finding the critical numbers which are the zeros of the marginal profit function P 0 (q) = R0 (q) − C 0 (q) = 0 where R0 (q) is the marginal revenue function and C 0 (q) is the marginal cost function. Thus, the global maximum or the global minimum of P occurs when M R(q) = M C(q) or at the endpoints of the interval. Example 82 Find the quantity q which maximizes profit given the total revenue and cost
4.4. APPLICATIONS OF OPTIMIZATION TO MARGINALITY
85
functions R(q) = 5q − 0.003q 2 C(q) = 300 + 1.1q. where 0 ≤ q ≤ 800 units. What production level gives the minimum profit? Solution. The profit function is given by P (q) = R(q) − C(q) = −0.003q 2 + 3.9q − 300. The critical numbers of P are the solutions to the equation P 0 (q) = 0. That is, 3.9 − 0.006q = 0 or q = 650 units. Since P (0) = −$300, P (800) = $900 and P (650) = $967.50 then the maximum profit occurs when q = 800 units and the minimum profit(or loss) occurs when q = 0, i.e. when there is no production. Exercise 43 The total revenue and total cost curves for a product are given below
C(q) S
R(q)
q
q 1
q 3
q 2
q 4
(a) Sketch the curves for the marginal revenue and marginal cost on the same axes. Show on this graph the quantities where marginal revenue equals marginal cost. What is the significance of these two quantities? At which quantity is profit maximum? (b) Graph the profit function π(q). Solution. Since R is a straight line with positive slope then its derivative is a positive constant. That is, the graph of the margnal revenue is a horizontal line at some value a > 0. Since C is always increasing then its derivative M C is always positive. For 0 < q < q3 the curve is concave down so that M C is decreasing. For
86
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
q > q3 the graph of C is concave up and so M C is increasing. Thus, the graphs of C and R are shown below.
MC MR
q
1
q
q
3
2
According to the graph, marginal revenue equals marginal costs at the values q = q1 and q = q2 . So maximum profit occurs either at q1 , q2 or at the endpoints. Notice that the production levels q1 and q2 correspond to the two points where the tangent line to C is parallel to the tangent line to R. Now, for 0 < q < q1 we have M R < M C so that P 0 = M R − M C < 0 and this shows that P is decreasing. For q1 < q < q2 , M R > M C so that P 0 >) and hence P is increasing. So P changes from decreasing to increasing at q1 which means that P has a minimum at q1 . Now, for q > q2 we have that M R < M C so that P 0 < 0 and P is decreasing. Thus, P changes from increasing to decreasing at q2 so q2 is a local maximum for P. So maximum profit occurs either at the endpoint or at q2 . Since profit is negative for q < q1 and q > q2 then the profit is maximum for q = q2 . (b)
q -C(0)
• Optimizing Revenue Example 83
q
1
4
q
3
q
2
4.5. AVERAGE COST
87
The demand equation for a product is p = 45−0.01q. Write the revenue function as a function of q and find the quantity that maximizes revenue. What price corresponds to this quantity? What is the total revenue at this price? Solution. The revenue function is given by R(q) = pq = 45q − 0.01q 2 . This is a parabola that opens down so its vertex is the global maximum. The maximum then occurs at the critical number of R(q). That is, at the solution of R0 (q) = 0 or 45 − 0.02q = 0. Solving for q we find q = 2250 units. The maximum revenue is R(2250) = $50, 625. The unit price in this case is p = 45−0.01(2250) = $22.50 Recommended Problems (pp. 187 - 8): 1, 2, 3, 5, 7, 8, 9, 16, 17.
4.5
Average Cost
We have seen that an important principle in economics is the problem of maximizing profit. A second general principle involves the relationship between the marginal cost and the average cost a(q) =
C(q) . q
Example 84 The cost of producing q items is C(q) = 2500 + 12q dollars. (a) What is the marginal cost of producing the 100th item? (b) What is the average cost of producing 100 items? Solution. (a) The marginal cost is given by M C(q) = 12. This means that after producing the 99 items, it costs an additional $12 to produce the 100th item. 2500+12(100) (b) a(100) = C(100) = $37 per item. 100 = 100 Since a(q) = C(q) = C(q)−0 then a(q) is the slope of the line passing through q q−0 the points (q, C(q)) and the origin (0, 0). The important question in this section is the question of minimizing the average cost function a(q). Let’s try to find the derivative of a(q). Using the quotient rule of differentiation we obtain a0 (q) =
C 0 (q)q − C(q) C 0 (q) − a(q) = . q2 q
Thus, a0 (q) = 0 when C 0 (q) = a(q). So critical numbers of a(q) satisfy the relationship C 0 (q) = a(q). In economics theory the global minimum occurs at a critical number. Graphically, the minimum average cost occurs at the point on the graph of C(q) where the line passing through the origin is tangent to the graph of C(q). Thus, if q0 is a critical number of a then for q < q0 the
88
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
marginal cost is less than the average cost. This means, increasing production will decrease the average cost. If, on the other hand, q > q0 then the marginal cost is greater than the average cost. This means that increasing production will increase the average cost.
Example 85 A total cost function, in thousands of dollars, is given by C(q) = q 3 − 6q 2 + 15q, where q is in thousands and 0 ≤ q ≤ 5. (a) Graph C(q). Estimate the quantity at which average cost is minimized. (b) Graph the average cost function. Use it to estimate the minimum average cost. (c) Determine analytically the exact value of q at which average cost is minimized.
Solution. (a) A graph of C(q) is given below. The average cost is minimized at the point where the line going through the origin is tangent to the graph of C(q). This occur at approximately q = 3. 50 40 30 20 10 0
1
2
x
3
4
5
(b) The average cost function is given by a(q) = C(q) = q 2 − 6q + 15. The q graph of this function is given below. Notice that the minimum occurs at approximately q = 3.
4.6. ELASTICITY OF DEMAND
89
14 12 10 8 6 0
1
2
x
3
4
5
(c) The minimum average cost occurs when C 0 (q) = a(q). That is, 3q 2 − 12q + 15 = q 2 − 6q + 15. This gives 2q 2 − 6q = 0. Solving for q we find q = 0 or q = 3. Since the average cost is not defined when q = 0 then the average cost is minimum at q = 3. Recommended Problems (pp. 192 - 3): 1, 3, 5, 9, 11, 13.
4.6
Elasticity of Demand
An important quantity in economics theory is the price elasticity of demand which measures the responsiveness of demand to a given change in price and is found using the formula ¯ ¯ ¯ change in quantity demanded ¯ E = ¯ percentage ¯ percentage ¯change ¯ in price ¯ dq ¯ q ¯ ¯ dp = ¯ ¯ ¯ p ¯ ¯ p dq ¯ = ¯ q · dp ¯ Changing the price of an item by 1% causes a change of E% in the quantity sold. If E > 1 then this means that an increase (or decrease) of 1% in price causes demand to drop (increase) by more than one percent. In this case, we say that the demand is elastic. If 0 ≤ E < 1 then an increase (decrease) of 1% in price causes demand to drop (increase) by less than one percent and in this case we say that the demand is inelastic. Note that we are assuming that increasing the price usually decreases demand dp and decreasing the price will increase demand so that dq q and p have opposite sign, that is their ratio is always negative.
Example 86 Raising the price of hotel rooms from $75 to $80 per night reduces weekly sales from 100 rooms to 90 rooms.
90
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
(a) What is the elasticity of demand for rooms at a price of $75? (b) Should the owner raise the price? Solution. (a) The percent change in price is ∆q q
∆p 80−75 = 0.067 = 6.7% p = 75 90−100 = −0.1 = −10%. Thus, 100
and the percent
change in demand is = the elasticity of 0.1 demand is E = 0.067 = 1.5. (b) The weekly revenue per room at the price of $75 is 100 · 75 = $7500 whereas at the price of $80 the weekly revenue is 90 · 80 = $7200. A price increase results in loss of revenue, so the price should not be raised. Example 87 The demand for a product is q = 2000 − 5p where q is units sold at a price of p dollars. Find the elasticity if the price is $10, and interpret your answer in terms of demand. Solution. Using Leibniz Notation we find
¯
dq ¯ dp ¯
p=10
= −5 and for p = 10 the corresponding
quantity is q = 2000 − 50 = 1950 so that the elasticity is ¯ ¯ ¯ p dq ¯ 10 · 5 ¯= ¯ = 0.03. E=¯ q dp ¯ 1950 The demand is inelastic at the given price; a 1% increase in price will result in a decrease of 0.03% in demand. Finally, we would like to know the price that maximizes revenue. That is, the price that brings the greatest revenue. Recall that the revenue function is dq p dq given by R = pq so that dR dp = q + p dp = q(1 + q dp )
dq dq If E > 1 then pq dp < −1 so that 1 + pq dp −1 so that 1 + pq dp > 0 and consequently dR dp > 0. This means that increasing price will increase revenue. Finally, note that if dR dp = 0 then E = 1. That is, E = 1 at the critical points of the revenue function.
R E=1 E1 R(p) p
4.6. ELASTICITY OF DEMAND
91
Recommended Problems (pp. 196 - 8): 1, 2, 3, 7, 8, 10, 12, 13, 17, 18.
92
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
Chapter 5
The Definite Integral Integration is the reverse process of differentiation. That is, given a function f (x) we would like to find a function F (x) such that F 0 (x) = f (x). In particular, given f 0 (x) we want to find the original function f (x). In this chapter, we introduce the concept of definite integral. This concept is used in many applications such as finding the area of a region, distance traveled, etc.
5.1
Measuring The Distance Traveled
We have seen that the velocity of an object moving along the curve s(t) is obtained by taking the average rate of change on smaller and samller intervals , that is finding the derivative of s, i..e v(t) = s0 (t). In this and the following sections we want to go the opposite direction. That is, given the velocity function v(t) we want to find the position function s(t). That is we want to estimate the total change of the distance using the average rate of change. To be more precise, suppose that we want to estimate the distance s traveled by a car after 10 seconds of departure. Assume for example, that we are given the velocity of the car every two seconds as shown in the table below Time (sec) Velocity (ft/sec)
0 20
2 30
4 38
6 44
8 48
10 50
Since we don’t know the instantaneous velocity of the car then what we can do is to estimate the distance traveled. For the first two seconds, the velocity is at least 20 miles per second so that the distance traveled is at least 20 × 2 = 40 feet. Likewise, at least 30 × 2 = 60 feet has been traveled the next two seconds and so on. Thus, we obtain an underestimate to the exact distance traveled 20 × 2 + 30 × 2 + 38 × 2 + 44 × 2 + 48 × 2 = 360 f eet. However, we can reason differently and get an overestimate to the total distance traveled as follows: For the first two seconds the car’s velocity is at most 30 feet so that the car travels at most 30 × 2 = 60 feet. In the next two seconds, it 93
94
CHAPTER 5. THE DEFINITE INTEGRAL
travels 38 × 2 = 76 feet and so on. So an overestimate of the total distance traveled is 30 × 2 + 38 × 2 + 44 × 2 + 48 × 2 + 50 × 2 = 420 f eet Hence, 360 f eet ≤ T otal distance traveled ≤ 420 f eet. Notice that the difference between the upper and lower estimates is 60 feet. The graph below shows both the lower estimate and the upper estimate. The graph of the velocity is obtained by plotting the points given in the above table and then connect them with a smooth curve. The area of the lower rectangles represent the lower estimate and the larger rectangles represent the upper estimate.
50 30 20 0
2
4
6
8
10
To visualize the difference between the upper and lower estimates, look at the above figure, and imagine that all the unshaded rectangles are pushed to the right and stacked on top of each other. This gives a rectangle of width 2 and height 30 so its area is the difference between the estimates. Exercise 44 Suppose that the velocity of the car is given every second instead as shown in the table below. Find the lower and upper estimates of the total distance traveled. What is the difference between the lower and upper estimates? Do you think that knowing the velocity at every second is a better estimate than knowing the velocity every two seconds? Time (sec) Velocity (ft/sec)
0 20
1 26
2 30
3 35
4 38
5 42
6 44
7 46
8 48
9 49
10 50
Solution. The lower estimate is (20)(1)+(26)(1)+(30)(1)+(35)(1)+(38)(1)+(42)(1)+(44)(1)+(46)(1)+(48)(1)+(49)(1) = 378 f eet
5.1. MEASURING THE DISTANCE TRAVELED
95
and the upper estimate is (26)(1)+(30)(1)+(35)(1)+(38)(1)+(42)(1)+(44)(1)+(46)(1)+(48)(1)+(49)(1)+(50)(1) = 408 f eet Hence, 378 f eet ≤ T otal distance traveled ≤ 408 f eet. So the difference between the upper and lower estimates is 408 − 378 = 30 f eet. This shows that by increasing the parition points we get better and better estimate. Remark. Once the upper estimate and the lower estimate are found then one can get an even better estimate by taking the average of the two estimates. The use of the average rate of change of the distance leads to finding the total distance traveled. This same method can be used to find the total change from the rate of change of other quantities. Example 88 The following table gives world oil consumptions, in billions of barrels per year. Estimate the total oil consumption during this 20-year period. Year Oil (barrels/yr)
1980 22.3
1985 23.0
1990 23.9
1995 24.9
2000 27.0
Solution. We underestimate the total oil consumption as follows: 22.3 × 5 + 23.0 × 5 + 23.9 × 5 + 24.9 × 5 = 470.5 billion barrels The overestimate is 23.0 × 5 + 23.9 × 5 + 24.9 × 5 + 27.0 × 5 = 494 billion barrels A good single estimate of the total oil consumption is the average of the above estimates. That is T otal oil consumption ≈
470.5 + 494 = 482.25 billion barrels 2
• Finding the Exact Distance Traveled Suppose that we want to find the total distance traveled over the time interval a ≤ t ≤ b. We take measurements of the velocity v(t) at equally spaced times, a = t0 , t1 , t2 , · · · , tn = b. This means that we devide the interval [a, b] into n equal pieces each of length ∆t = b−a n . We first use the left-end point of each interval [ti−1 , ti ] and construct the left-hand sum L(v, n) = v(t0 )∆t + v(t1 )∆t + · · · + v(tn−1 )∆t.
96
CHAPTER 5. THE DEFINITE INTEGRAL
Geometrically, this sum represents the sum of areas of rectangles constructed by taking the height to be the value of the function at the left-endpoint of each subinterval. 4 3 2 1
0 1
1.2
1.4
x
1.6
1.8
2
Secondly, we use the right-end point of each interval [ti−1 , ti ] and construct the right-hand sum R(v, n) = v(t1 )∆t + v(t2 )∆t + · · · + v(tn )∆t. Geometrically, this sum represents the sum of areas of rectangles constructed by taking the height to be the value of the function at the right-endpoint of each subinterval. 4 3 2 1
0 1
1.2
1.4
x
1.6
1.8
2
Notice that for an increasing function the left-hand sum is an underestimate whereas the right-hand sum is an overestimate. This role is reversed for a decreasing function. Now, the exact distance traveled lies between the two estimates. As we have seen earlier, by making the time interval smaller and smaller we can make the difference between the two estimates as small as we like. This is equivalent to letting n → ∞. If the function v(t) is continuous then the following two limits
5.2. THE DEFINITE INTEGRAL
97
are equal to the exact distance traveled from t = a to t = b. T otal distance traveled = lim L(v, n) = lim R(v, n). n→∞
n→∞
Geometrically, each of the above limit represents the area under the graph of v(t) bounded by the lines t = a, t = b and the horizontal axis. 16 14 12 10 8 6 4 2 0 1
1.5
2
2.5 x
3
3.5
4
Recommended Problems (pp. 223 - 5): 1, 3, 7, 9, 10, 13, 18.
5.2
The Definite Integral
We discussed in the previous section how to find the exact distance traveled by taking the limit of both the left-hand sum and the right-hand sum as the number of subintervals increases without bound. In this section, we will construct these sums for any continuous function on a closed interval [a, b]. We start by dividing the interval [a,b] into n subintervals each of length ∆x =
b−a . n
Let a = x0 , x1 , · · · , xn−1 , xn = b be the endpoints of the subdivisions. We construct the left-hand sum or the left Riemann sum L(f, n) = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x =
n−1 X
f (xi )∆x
i=0
and the right-hand sum or the right Riemann sum R(f, n) = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x =
n X
f (xi )∆x
i=1
It is shown in advanced calculus that for a continuous function on a closed interval [a, b] that as n → ∞ both the left-hand sum and the right-hand sum
98
CHAPTER 5. THE DEFINITE INTEGRAL
Rb exist and are equal. We denote the common value by the notation a f (x)dx. Thus, Rb f (x)dx = limn→∞ L(f, n) a = limn→∞ R(f, n) Rb We call a f (x)dx the definite integral of f from x = a to x = b. We call a the lower limit and b the upper limit. The function f is called the integrand. Example 89 (a) On a sketch of y = ln x, represent the left Riemann sum with n = 2 approxR2 imating 1 ln xdx. Write out the terms in the sum, but do not evaluate. (b) On a another sketch of y = ln x, represent the right Riemann sum with n = 2 R2 approximating 1 ln xdx. Write out the terms in the sum, but do not evaluate. (c) Which sum is an underestimate? Which sum is an overestimate? Solution. (a) The left Riemann sum is the sum L(ln x, 2) = ln 1(0.5) + ln (1.5)(0.5) = 0.5 ln (1.5). The sum is represented by the rectangle shaded in the figure to the left
1
1.5
2
Left Riemann Sum
1
1.5
2
Right Riemann Sum
(b) The right Riemann sum is the sum R(ln x, 2) = ln (1.5)(0.5) + ln 2(0.5) = 0.5 ln (2)(1.5) = 0.5 ln 3. The sum is represented by the rectangles shaded in the figure to the right. R2 (c) L(ln x, 2) < 1 ln xdx < R(ln x, 2). In the next section, we will see that a definite integral represents an area under the graph of a function. Recommended Problems (pp. 231 - 2): 1, 3, 5, 7, 9, 11, 15, 19, 23.
5.3. THE DEFINITE INTEGRAL AS AREA
5.3
99
The Definite Integral as Area
In this section, we will see how definite integrals are used to find areas. Case 1:f (x) ≥ 0 Looking closely to either the left Riemann sum or the right Riemann sum we see that if f (x) ≥ 0 then a term of the form f (x)∆x represents the area of a rectangle. As n increases without bound, that is, the width ∆t of the rectangles approaches zero, the rectangles fit the curve of the graph more exactly, and the sum of their areas gets closer and closer to the area under the graph, bounded by the vertical lines x = a and x = b and the x-axis. Thus, Z
b
f (x)dx = Area under graph of f between a and b. a
Example 90 R1 √ Consider the integral −1 1 − x2 dx. (a) Interpret the integral as an area, and find its exact value. (b) Estimate the integral using a calculator. Solution. (a) Note that the equation of a circle centered at the origin and with radius √ 1 is √ given by x2 + y 2 = 1. Solving for y we find y = ± 1 − x2 . The function √ y = 1 − x2 corresponds to the upper semicircle and the function y = − 1 − x2 corresponds to the lower semicircle.
100
CHAPTER 5. THE DEFINITE INTEGRAL 1 0.8 0.6 0.4 0.2
–1
0
–0.6
0.20.40.60.8 1 x
It follows that the given integral represents the area of the upper semicircle and therefore is equal to π2 . That is, Z
p π 1 − x2 dx = . 2 −1 1
(b) Using a TI-83 calculator we find p f nInt( 1 − x2 , x, −1, 1) ≈ 1.571. Case 2:f (x) ≤ 0 In this case, since each product of the form f (x)∆x is less than or equal to zero then the area gets counted negatively. That is, the absolute value of the integral gives the area above the curve between x = a and x = b. Example 91 Find the area above the graph of y = x2 − 1 from x = −1 to x = 1. Solution. The graph of y = x2 − 1 is given by –1
–0.6 –0.2 0 –0.2 –0.4 y –0.6 –0.8 –1
x 0.20.40.60.8 1
5.4. INTERPRETATIONS OF THE DEFINITE INTEGRAL
101
R1 The area is given by | −1 (x2 − 1)dx| ≈ | − 1.33| = 1.33. Case 3:f changes sign In this case, the integral is the sum of the areas above the x-axis, counted positively, and the areas below the x-axis, counted negatively. If the integral is positive then the region above the x-axis has larger area than the region below the x-axis. If the integral is negative then the region below the x-axis has a larger area then the region above the x-axis. Example 92 Find the area between the graph of y = x3 and the x-axis from x = −1 to x = 1. Solution. The area is shown below 1 0.5
–1
–0.6 –0.2
0.20.40.60.8 1 x
–0.5 –1
It follows that the area is given by ¯Z 0 ¯ Z ¯ ¯ 3 ¯ x dx¯¯ + ¯ −1
1
x3 dx = 0.5
0
Recommended Problems (pp. 235 - 7): 1, 3, 5, 7, 11, 13, 17, 19, 22, 25. 27, 30, 31, 32.
5.4
Interpretations of the Definite Integral
We start this section by showing that the definite integral of a rate of change gives the total change of the function. We define the total change of a function F (t) from t = a to t = b to be the difference F (b) − F (a). Suppose that F (t) is continuous in [a,b] and differentiable in (a, b). Divide the interval [a, b] into n equal subintervals each of length ∆t = b−a n . Let a = t0 , t1 , · · · , tn = b be the partition points of the subdivision. Then on the interval [t0 , t1 ] the change in F can be estimated by the formula F 0 (t0 ) ≈
F (t0 + ∆t) − F (t0 ) ∆t
102 or that is
CHAPTER 5. THE DEFINITE INTEGRAL
F (t0 + ∆t) − F (t0 ) ≈ F 0 (t0 )∆t F (t1 ) − F (t0 ) ≈ F 0 (t0 )∆t
On the interval [t1 , t2 ] we get the estimation F (t2 ) − F (t1 ) = F 0 (t1 )∆t Continuing in this fashion we find that on the interval [tn−1 , tn ] we have F (tn−1 ) − F (tn ) ≈ F 0 (tn−1 )∆t. Adding all these approximations we find that F (tn ) − F (t0 ) ≈
n−1 X
F 0 (ti )∆t
i=0
Letting n → ∞ we see that Z
b
F (b) − F (a) =
F 0 (t)dt.
a
Example 93 The amount of waste a company produces, W, in metric tons per week, is approximated by W = 3.75e−0.008t , where t is in weeks since January 1, 2000. Waste removal for the company costs $15/ton. How much does the company pay for waste removal during the year 2000? Solution. The R 52 amount of tons produced during the year 2000 is just the definite integral W (t)dt. Using a calculator we find that 0 Z T otal waste during the year =
52
3.75e−0.008t dt ≈ 159 tons
0
The cost to remove this quantity is 159 × 15 = $2385. Rb Remark. When using a f (x)dx in applications then its units is the product of the units of f (x) with the units of x. Recommended Problems (pp. 240 - 3): 1, 3, 5, 8, 9, 15, 18.
5.5
The Fundamental Theorem of Calculus
The following result is considered among the most important result in calculus.
5.5. THE FUNDAMENTAL THEOREM OF CALCULUS
103
The Fundamental Theorem of Calculus If f (x) is a continuous function on [a,b] and F 0 (x) = f (x) then Z
b
f (x)dx = F (b) − F (a) a
We call the function F (x) an antiderivative of f (x). Proof. Partition the interval [a,b] into n subintervals each of length ∆x = b−a n and let {a = x0 , x1 , · · · , xn = b} be the partition points. Applying the Mean Value Theorem on the interval [x0 , x1 ] we can find a number x0 < c1 < x1 such that F (x1 ) − F (x0 ) = F 0 (c1 )∆x. Continuing this process on the remaining intervals we find = .. .
F 0 (c1 )∆x
F (xn ) − F (xn−1 ) =
F 0 (cn )∆x
F (x2 ) − F (x1 )
Adding these equalities we find n X
F (xn ) − F (x0 ) =
f (ci )∆x
i=1
Letting n → ∞ to obtain Z F (b) − F (a) =
b
f (x)dx a
Example 94 R2 Use FTC to compute 1 2xdx. Use a calculator to find the answer to the integral and compare. Solution. Since the derivative of x2 is 2x then F (x) = x2 . Thus, by the FTC we have Z 2 2xdx = F (2) − F (1) = 4 − 1 = 3. 1
Using a calculator we find
R2 1
2xdx = 3.
Example 95 Let F (t) represent a bacteria population which is 5 million at time t = 0. After t hours, the population is growing at an instantaneous rate of 2t million bacteria per hour. Estimate the total increase in the bacteria population during the first hour, and the population at t = 1.
104
CHAPTER 5. THE DEFINITE INTEGRAL
Solution. Since total change is the definite integral of F 0 (t) = 2t from t = 0 to t = 1 then Z 1 Change in population = F (1) − F (0) = 2t dt ≈ 1.44 million bacteria 0
Since F (0) = 5 then Z
1
F (1) = F (0) +
2t dt ≈ 5 + 1.44 = 6.44 million.
0
If C(q) is the total cost to produce a quantity q of a certain comodity then we can use the Fundamental Theorem of Calculus and compute the total cost of producing b units as follows Z b C(b) − C(0) = C 0 (q)dq 0
or
Z
b
C(b) = C(0) + We call the quantity
Rb 0
C 0 (q)dq
0
C 0 (q)dq the total variable cost.
Example 96 The marginal cost function for a company is given by C 0 (q) = q 2 − 16q + 70 dollars/unit, where q is the quantity produced. If C(0) = 500, find the total cost of producing 20 units. What is the fixed cost and what is the total variable cost for this quantity? Solution. We find C(20) as follows: Z C(20) = C(0) +
20
Z C 0 (q)dq = 500 +
0
20
(q 2 − 16q + 70)dq.
0
where C(0) = 500 is the fixed cost. Using a calculator we find total variable cost to be Z 20 (q 2 − 16q + 70)dq ≈ 866.7 0
Thus, the total cost of producing 20 units is C(20) ≈ 500 + 866.7 = 1366.7 Recommended Problems (pp. 245 - 6): 2, 3, 5, 7, 9, 11, 16, 17, 20, 25.
5.6. THE DEFINITE INTEGRAL AS AN AVERAGE
5.6
105
The Definite Integral as an Average
We know that the average of n given numbers is just the sum divided by n. What is the average in the continuous case. That is, what is the average of a continuous function on a closed interval [a, b]? Partition the interval into n equal subintervals each of length ∆t = b−a n and let a = t0 , t1 , t2 , · · · , tn be the division points. Then Average of f (t) on [a, b] ≈ But n =
b−a ∆t
f (t0 ) + f (t1 ) + · · · + f (tn−1 ) n
so that
Average of f (t) on [a, b] ≈ =
1 b−a (f (t0 )
+ f (t1 ) + · · · + f (tn−1 ))∆t Pn−1 1 i=0 f (ti )∆t b−a
Letting n → ∞ we see that 1 Average of f (t) on [a, b] = b−a
Z
b
f (x)dx. a
Example 97 A bar of metal is cooling from 1000◦ C to room temperature, 20◦ C. The temperature, H, of the bar t minutes after it starts cooling is given by H = 20 + 980e−0.1t . Find the average temperature over the first hour. Solution. The average temperature is given by Average temperature f or the f irst hour =
1 60
Z
60
(20 + 980e−0.1t )dt ≈ 183◦ C.
0
Example 98 Suppose that C(t) represent the daily cost of heating your house, in dollars per day, where t isRtime in days and t =R0 corresponds to January 1, 2002. Interpret 90 90 1 C(t)dt. the quantities 0 C(t)dt and 90−0 0 Solution. R 90 The integral 0 C(t)dt represents the total cost in dollars to heat your house for the first 90 days of 2002. The second expression represents the average cost per day to heat your house during the first 90 days of 2002. Remark. From the definition of the average value we can write Z b (average value of f ) × (b − a) = f (x)dx a
106
CHAPTER 5. THE DEFINITE INTEGRAL
Geometrically, this says that the area of the rectangle of dimensions (average value of f )× (b − a) is equal to the area under the graph of f (x) from x = a to x = b. Recommended Problems (pp. 258 - 9): 1, 4, 5, 9, 11, 14, 18.
Chapter 6
Antiderivatives and Indefinite Integrals In this chapter we discuss the procedure of computing definite integrals using the Fundamental Theorem of Calculus. This requires finding the so-called an antiderivative of the integrand. We will discuss how to construct antiderivatives analytically, numerically, and graphically.
6.1
Finding Antiderivatives Graphically and Numerically
In this section we study the process of graphing a function f (x) given the graph of its derivative f 0 (x). Recall first that if a function f (x) is given then the new function f 0 (x) is called the derivative of f (x). For example, if f (x) = x2 then f 0 (x) = 2x. This process is referred to as differentiation. Now, if instead f 0 (x) is known then the process of finding f (x) is called integration or antidifferentiation. In this case, we call f (x) an antiderivative of f 0 (x). In general, if f and F are two functions such that F 0 = f then we say that F (x) is an antiderivative of f (x). For example, x2 is an antiderivative of 2x since (x2 )0 = 2x. Note that, there are infinitely many antiderivatives of 2x, namely, x2 + C where C is a constant. We call x2 + C the general antiderivative ( or the indefinite integral) of x2 and we represent this symbolically by Z 2xdx = x2 + C. Example 99 The graph of f 0 (x) is given.
107
108
CHAPTER 6. ANTIDERIVATIVES AND INDEFINITE INTEGRALS Graph of f’(x) 2 1
–2
–1
1 x
2
–1 –2
Sketch a graph of the function f (x) satisfying f (0) = 1. Solution. Note that since f 0 (x) is always increasing then the graph of f (x) is concave up since f 00 (x) > 0. Since f 0 (x) < 0 for x < 0 then f (x) is decreasing there. Similarly, since f 0 (x) > 0 for x > 0 then f (x) is increasing there. Since f 0 (0) = 0 and f (x) is decreasing and then increasing we conclude that x = 0 is a minimum. A graph of f (x) is given below. Graph of f(x) 3 2 y 1 –2
–1
0 –1
1 x
2
1 x
2
–2 –3
Example 100 2 The graph of f 0 (x) = e−x is given below. Graph of f’(x) 2 1.8 1.6 1.4 1.2 y 1 0.8 0.6 0.4 0.2 –2
–1
0
Sketch the graph of f (x) satisfying f (0) = 0. Solution.
6.1. FINDING ANTIDERIVATIVES GRAPHICALLY AND NUMERICALLY109 Since f 0 (x) is always positive then the graph of f (x) is always increasing. Now, for x < 0, f 0 (x) is increasing so that f 00 (x) > 0 and therefore f (x) is concave up. For x > 0 the function f 0 (x) is decreasing and so f 00 (x) < 0. That is, f (x) is concave down there. Thus, x = 0 is a point of inflection. Finally, since limx→±∞ f 0 (x) = 0 then the graph of f (x) levels off at both ends (See picture below) Graph of f(x) 4 y
–3
–2
2
–1 0
1 x 2
3
–2 –4
Next, we will estimate numerically the value of an antiderivative at a given point. For this purpose we remind the reader of the Fundamental Theorem of Rb Calculus (abbreviated by FTC): If F 0 (x) = f (x) then a f (x)dx = F (b) − F (a). In particular, we have Z
b
f 0 (x)dx = f (b) − f (a).
a
Rb Now, recall that for f (x) ≥ 0 the definite integral a f (x)dx represents the area under the graph of f (x) between the lines x = a and x = b. If the region is Rb below the x-axis then a f (x)dx is the negative of the area of that region. Example 101 Given below is the graph of f 0 (x). Suppose that f (−1) = −2. Find f (0), f (1), and f (3). 2
Graph of f’(x)
1.5 1 0.5
–1
0
1
x
2
3
110
CHAPTER 6. ANTIDERIVATIVES AND INDEFINITE INTEGRALS
Solution. By the FTC we have Z
b
f (b) = f (−1) +
f 0 (x)dx.
−1
Thus,
R0 f (0) = f (−1) + −1 f 0 (x)dx = −2 + 12 (1 · 2) = −1 R1 f (1) = f (0) + 0 f 0 (x)dx = −1 + 1 · 2 = 1 R3 0 f (3) = f (1) + 1 f (x)dx = 1 + 12 (2 · 2) = 3 Rt where we compute a f 0 (x)dx by determining the area between f 0 and the horizontal axis for a ≤ x ≤ t. Recommended Problems (pp. 294 - 6): 1, 3, 5, 9, 11, 17, 21, 22, 23, 24, 25.
6.2
Analytical Construction of Antiderivatives
In this section we will find analytical expressions of antiderivatives. As discussed in the previous section, there are infinitely many antiderivatives of a given function f (x). They all differ by a constant and the family of antiderivatives is represented by F (x) + C. The notation of the general antiderivative is called an indefinite integral and is written Z f (x)dx = F (x) + C. R The symbol is the symbol of integration, f (x) is called the integrand and C is called the constant of integration. Keep in mind the relationship between f (x) and F (x) which is given by F 0 (x) = f (x). Warning: The indefinite integral is a short-hand notation for a family of functions F (x) + C with the property F 0 (x) = f (x) for all x. It is not to be confused Rb with the definite integral a f (x)dx which is a real number. Example R102 Show that 0dx = C. Solution. Since the derivative of a constant function is always zero then Z 0dx = C.
6.2. ANALYTICAL CONSTRUCTION OF ANTIDERIVATIVES
111
Example R103 Show that kdx = kx + C where k is a constant. Solution. Since the derivative of kx is just k then Z kdx = kx + C. Example 104 R Show that xn dx =
xn+1 n+1
+ C for n 6= −1.
Solution. By the power rule, if F (x) =
xn+1 n+1
then F 0 (x) = xn . Thus,
Z xn dx =
xn+1 + C. n+1
Note that this formula is valid only if n 6= −1 for if n = −1 we would have which doesn’t make sense. The case n = −1 is treated in the next example. Example 105 Show that
Z
x0 0
dx = ln |x| + C. x
Solution. Suppose first that x > 0 so that ln |x| = ln x. Then (ln |x|)0 = (ln x)0 = x1 . Now, if −1 x < 0 then ln |x| = ln (−x) and by the chain rule (ln |x|)0 = (ln (−x))0 = −x = x1 . 1 0 Thus, in both cases (ln |x|) = x . Example 106 R Show that for a 6= 0, eax dx =
eax a
+ C.
Solution. ax If a is a nonzero constant and F (x) = ea then F 0 (x) = eax . This shows that Z eax eax dx = + C. a • Properties of Indefinite Integrals Z Z Z [f (x) ± g(x)]dx = f (x)dx ± g(x)dx. To see why this property is true, let F (x) be an antiderivative of f (x) and G(x) d [F (x) ± be an antiderivative of g(x). The result follows from the fact that dx G(x)] = f (x) ± g(x). Z Z cf (x)dx = c
f (x)dx.
112
CHAPTER 6. ANTIDERIVATIVES AND INDEFINITE INTEGRALS
R To see this, suppose that F (x) is an antiderivative of f (x). Then f (x)dx = d F (x) + C. R But dx (cF (x)) = cf0(x) so that cF (x) is an antiderivative of cf (x), that is, cf (x)dx = cF (x) + C . This implies Z Z Z Z 0 0 0 cf (x)dx = cF (x)+C = c( f (x)dx−C)+C = c f (x)dx−cC+C = c f (x)dx. 0 Note that the R constant −cC + C is ignored since a constant of integration will result from f (x)dx.
Once we have found an antiderivative of f (x), computing definite integrals is easy by the Fundamental Theorem of Calculus. ExampleR 107 2 Compute 1 3x2 dx. Solution. Since F (x) = x3 is an antiderivative of f (x) = 3x2 , then by FTC Z
2 1
3x2 dx = x3 |21 = 23 − 13 = 7.
Recommended Problems: pp. 281 - 2: 6, 8, 10, 13, 15, 22, 24, 32, 36, 38, 40, 41, 48. pp. 289 - 90: 1, 3, 5, 11, 13, 15, 17, 24, 25, 27, 28.
2
PREFACE This supplement consists of my lectures of a freshmen-level mathematics class offered at Arkansas Tech University. The lectures are designed to accompany the textbook ”Applied Calculus” written by the Harvard Consortium. The lectures cover sections from Chapters 1, 2, 3, 4, 5, 6, and 7. These chapters are basically well suited for a one semester course in Business Calculus. Marcel B. Finan January 2003
Contents 1 Library of Functions 1.1 What is a Function? . . . . . . . . . . . 1.2 Linear Functions . . . . . . . . . . . . . 1.3 Rate of Change . . . . . . . . . . . . . . 1.4 Applications of Functions to Economics 1.5 Exponential Functions . . . . . . . . . . 1.6 Logarithms and Their Properties . . . . 1.7 Exponential Growth and Decay . . . . . 1.8 Building New Functions from Old Ones 1.9 Polynomial and Rational Functions . . . 2 The 2.1 2.2 2.3 2.4 2.5
Derivative of a Function Instantaneous Rate of Change . . . . The derivative Function . . . . . . . Leibniz Notation for The Derivative The Second Derivative . . . . . . . . Marginal Cost and Revenue . . . . .
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55 55 59 62 63 65
3 Rules of Differentiation 69 3.1 Derivative Formulas for Power and Polynomials . . . . . . . . . . 69 3.2 Derivative Formulas for Exponential and Logarithmic Functions . 71 3.3 Derivatives of Composite Functions: The Chain Rule . . . . . . . 73 3.4 The Product and Quotient Rules . . . . . . . . . . . . . . . . . . 74 4 Applications of the Derivative 4.1 Using First and Second Derivatives . . . . . 4.2 Concavity and Points of Inflection . . . . . 4.3 Global Maxima and Minima . . . . . . . . . 4.4 Applications of Optimization to Marginality 4.5 Average Cost . . . . . . . . . . . . . . . . . 4.6 Elasticity of Demand . . . . . . . . . . . . . 3
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77 77 80 82 84 87 89
4 5 The 5.1 5.2 5.3 5.4 5.5 5.6
CONTENTS Definite Integral Measuring The Distance Traveled . . . . The Definite Integral . . . . . . . . . . . The Definite Integral as Area . . . . . . Interpretations of the Definite Integral . The Fundamental Theorem of Calculus . The Definite Integral as an Average . .
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93 . 93 . 97 . 99 . 101 . 102 . 105
6 Antiderivatives and Indefinite Integrals 107 6.1 Finding Antiderivatives Graphically and Numerically . . . . . . . 107 6.2 Analytical Construction of Antiderivatives . . . . . . . . . . . . . 110
Chapter 1
Library of Functions This chapter lays the foundation for the topics that we will discuss in this course. It consists of a brief review of the following family of functions: linear, exponential, logarithmic, trigonometric, polynomial, and rational functions.
1.1
What is a Function?
Functions play a crucial role in mathematics. The goal of this section is to introduce this important concept. Before giving the definition of a function we first consider an application.
Example 1 The sales tax on an item is 6%. So if p denotes the price of the item and C the total cost of buying the item then if the item is sold at $ 1 then the cost is 1 + (0.06)(1) = $1.06. If the item is sold at $2 then the cost of buying the item is 2 + (0.06)(2) = $2.12 and so on. Thus we have a relationship between the quantities C and p such that each value of p determines exactly one value of C. In this case, we say that C is a function of p. The relationship between C and p can be expressed by a table, a graph, or a formula. The chart below gives the total cost of buying an item at price p as a function of p. p C
1 1.06
2 2.12
3 3.18
4 4.24
5 5.30
6 6.36
7 7.42
The graph of the function C is obtained by plotting the data in the above table to obtain 5
6
CHAPTER 1. LIBRARY OF FUNCTIONS 7 6 5 4 3 2 1
1
2
3
4
5
6
7
The formula that describes the relationship between C and p is given by C(p) = 1.06p. By a function we mean a relationship f between two quantities x and y such that each value of x yields a unique value y. We call x the input and y the output. In function notation we write y = f (x) Note that y depends on x so that we call x the independent variable and y the dependent variable. Exercise 1 Using the following table, determine which variable is a function of what variable. x y
17 73
18 77
19 69
20 73
21 75
22 75
23 70
Solution. Since each value of x determines exactly one value of y then y is a function of x. On the contrary, x is not a function of y since the input 73 has two output values 17 and 20. So x is not a function of y. Now, most of the functions that we will encounter in this course have formulas. For example, the area A of a circle is a function of its radius r. In function notation, we write A = πr2 . However, there are functions that can not be represented by a formula. For example, if D is a day in July and T is lowest temperature in that day, then T is a function of D. Note that T can not be expressed as a formula in D. Functions of this nature, are mostly represented by either a graph or a table of numerical data. Exercise 2 The table below shows the daily low temperature for a one-week period in New
1.1. WHAT IS A FUNCTION?
7
York City during July. (a) What was the low temperature on July 19? (b) When was the low temperature 73◦ F ? (c) Is the daily low temperature a function of the date?Explain. (d) Can you express T as a formula? D T
17 73
18 77
19 69
20 73
21 75
22 75
23 70
Solution. (a) The low temperature on July 19 was 69◦ F. (b) On July 17 and July 20 the low temperature was 73◦ F. (c) T is a function of D since each value of D determines exactly one value of T. (d) T can not be expressed as a formula in terms of D otherwise there would be no need for meteorologists. Next, we present a relationship between two quantities x and y that is not a function. Exercise 3 Let x and y be two quantities related by the equation x2 + y 2 = 4. (a) Is x a function of y? Explain. (b) Is y a function of x? Explain. Solution. (a) For y = 0 we have two values of x, namely, x = −2 and x = 2. So x is not a function of y. (b) For x = 0 we have two values of y, namely, y = −2 and y = 2. So y is not a function of x. Suppose that the graph of a relationship between two quantities x and y is given. To say that y is a function of x means that for each value of x there is exactly one value of y. Graphically, this means that each vertical line must intersect the graph at most once. Hence, to determine if a graph is a function one uses the following test: Vertical Line Test: A graph is a function if and only if every vertical line crosses the graph at most once. According to the vertical line test and the definition of a function, if a vertical line cuts the graph for example twice, the graph could not be the graph of a function since we have two y values for the same x − value and this violates the definition of a function.
8
CHAPTER 1. LIBRARY OF FUNCTIONS
Exercise 4 Which of the graphs (a) through (i) represent y as a function of x? (Note that an open circle indicates a point that is not included in the graph; a solid dot indicates a point that is included in the graph.)
c
a
b
d
g
e
..
f
.. h
. . .. ..
j
..
Solution. (a), (d), and (g) are functions. The rest are not by the vertical line test. • Domain and Range of a Function If we√ try to find the possible input values that can be used in the function y = x − 2 we see that we must restrict x to the interval [2, ∞), that is x ≥ 2. Similarly, the function y = x12 takes only certain values for the output, namely, y > 0. The above discussion leads to the following definitions: By the domain of a function we mean all possible input values. Graphically, the domain is part of the horizontal axis. The range of a function is the collection of all possible output values. The range is part of the vertical axis. The domain and range of a function can be found either algebraically or graphically. • Finding the Domain and the Range Algebraically When finding the domain of a function, ask yourself what values can’t be used. Your domain is everything else. There are simple basic rules to consider: • The domain of all polynomial functions, i.e. functions of the form f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , is the Real numbers IR.
1.1. WHAT IS A FUNCTION?
9
• Square root functions can not contain a negative underneath the radical. Set the expression under the radical greater than or equal to zero and solve for the variable. This will be your domain. • Rational functions, i.e. ratios of polynomials, can not have zeros in the denominator. Determine which values of the input cause the denominator to equal zero, and set your domain to be everything else. Exercise 5 Find, algebraically, the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx2
(b) y =
√1 x−4
(c) y = 2 + x1 .
Solution. (a) Since the function is a polynomial then its domain is the interval (−∞, ∞). To find the range, note that y ≥ 0 for all x. So the range is the interval [0, ∞). 1 (b) The domain of y = √x−4 consists of all numbers such that x − 4 > 0 or x > 4. That is, the interval (4, ∞). To find the range, notice that y > 0 for all values of x in the domain of y. Thus, the range is the interval (0, ∞). (c) The domain of y = 2 + x1 is the interval (−∞, 0) ∪ (0, ∞). To find the range, 1 write x in terms of y to obtain x = y−2 . The values of y for which this later formula is defined is the range of the given function, that is, (−∞, 2) ∪ (2, ∞). Remark. Note that the domain of the function y = πx2 of the previous problem consists of all real numbers. If this function is used to model a real-world situation, that is, if the x stands for the radius of a circle and y is the corresponding area then the domain of y in this case consists of all numbers x ≥ 0. In general, for a word problem the domain is the set of all x values such that the problem makes sense. • Finding the Domain and the Range Graphically We often use a graphing calculator to find the domain and range of functions. In general, the domain will be the set of all x values that has corresponding points on the graph. We note that if there is an asymptote (shown as a vertical line on the TI series) we do not include that x value in the domain. To find the range, we seek the top and bottom of the graph. The range will be all points from the top to the bottom (minus the breaks in the graph). Exercise 6 Use a graphing calculator to find the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx2
(b) y =
√1 x−4
(c) y = 2 + x1 .
Solution. (a) The graph of y = πx2 is given below
10
CHAPTER 1. LIBRARY OF FUNCTIONS
2 1.8 1.6 1.4 1.2 y 1 0.8 0.6 0.4 0.2 –2
0
–1
1 x
2
The domain is the set (−∞, ∞) and the range is [0, ∞). 1 (b) The graph of y = √x−4 is given below 8 7 6 5 y4 3 2 1 4
5
6 x
7
8
The domain is the set (4, ∞) and the range is (0, ∞). (c) The graph of y = 2 + x1 is given below 20
y 10
–2
–1
0
1 x
2
–10
–20
The domain is the set (−∞, 0) ∪ (0, ∞) and the range is (−∞, 2) ∪ (2, ∞).
1.2. LINEAR FUNCTIONS
11
Recommended Problems (pp. 4 - 6): 1, 3, 5, 10, 13, 15, 17, 24, 25.
1.2
Linear Functions
This section is designed to introduce students to the concept of linear functions. So, the first question that comes to mind is the question of recognizing a linear function. For any function f (x), the average rate of change of f (x) from x = a to x = b is the difference quotient f (b) − f (a) b−a In general, the average rate of change of a function is different on different intervals. A linear function is a function with the property that the average rate of change on any interval is the same. We say that y is changing at a constant rate with respect to x. In this case, the graph is a straight line ( and thus the term linear). It follows, that if f is linear and if [a, b] and [c, d] are two intervals of equal length, i.e. b − a = d − c then f (b) − f (a) f (d) − f (c) = b−a d−c and this implies that f (b) − f (a) = f (d) − f (c). That is, if a function y is such that equal increments in x corresponds to equal increments in y then y is a linear function of x. Exercise 7 Which of the following tables could represent a linear function? x 0 5 10 15
f(x) 10 20 30 40
x 0 10 20 30
g(x) 20 40 50 55
Solution. Since equal increments in x yield equal increments in y then f (x) is a linear 50−40 function. On the contrary, since 40−20 10−0 6= 20−10 then g(x) is not linear. Now, suppose that f (x) is a linear function of x. Then f changes at a constant rate m. That is, if we pick two points (0, f (0)) and (x, f (x)) then m=
f (x) − f (0) . x−0
12
CHAPTER 1. LIBRARY OF FUNCTIONS
That is, f (x) = mx + f (0). This is the function notation of the linear function f (x). Another notation is the equation notation, y = mx + f (0). We will denote the number f (0) by b. In this case, the linear function will be written as f (x) = mx + b or y = mx + b. Since b = f (0) then the point (0, b) is the point where the line crosses the vertical line. We call it the y-intercept. So the y-intercept is the output corresponding to the input x = 0, sometimes known as the initial value of y. If we pick any two points (x1 , y1 ) and (x2 , y2 ) on the graph of f (x) = mx + b then we must have y2 − y1 m= . x2 − x1 We call m the slope of the line. Exercise 8 The value of a new computer equipment is $20,000 and the value drops at a constant rate so that it is worth $ 0 after five years. Let V (t) be the value of the computer equipment t years after the equipment is purchased. (a) Find the slope m and the y-intercept b. (b) Find a formula for V (t). Solution. (a) Since V (0) = 20, 000 and V (5) = 0 then the slope of V (t) is m=
0 − 20, 0002 = −4, 000 5−0
and the vertical intercept is V (0) = 20, 000. (b) A formula of V (t) is V (t) = −4, 000t + 20, 000. Exercise 9 The table below shows the cost of selling various amounts of coffee per day from a cart. (a) Using the table, show that the relationship appears to be linear. (b) Plot the data in the table. (c) Find the slope of the line. Explain what this means in the context of the given situation. x(cups per day) C(dollars)
0 50.00
5 51.25
10 52.50
50 62.50
100 75.00
200 100.00
Solution. (a) Since an increment of 5 in x yields an increment of 1.25 in y then the table shows that equal increments in x yield equal increments in y so that y is linear. (b) The graph is given below
1.2. LINEAR FUNCTIONS
13
100 90 80 70 60 50 0
50
100
150
200
(c) The slope of the line is m = 51.25−50.00 = 0.25. This number means that an 5−0 increase in production by one cup will increase cost by 25 cents. • Formulas for Linear Functions Recall that f is linear if and only if f (x) can be written in the form f (x) = mx + b. So the problem of finding the formula of f is equivalent to finding the quantities m and b. Suppose that we know two points on the graph of f (x), say (x1 , f (x1 )) and (x2 , f (x2 )). Since the slope m is just the average rate of change of f (x) on the interval [x1 , x2 ] then f (x2 ) − f (x1 ) m= . x2 − x1 To find b, we use one of the points in the formula of f (x); say we use the first point. Then f (x1 ) = mx1 + b. Solving for b we find b = f (x1 ) − mx1 . Example 2 Let’s find the formula of a linear function given by a table of data values. The table below gives data for a linear function. Find the formula. x f(x)
1.2 0.736
1.3 0.614
1.4 0.492
1.5 0.37
Solution. We use the first two points to find the value of m : m=
f (1.3) − f (1.2) 0.614 − 0.736 = = −1.22 1.3 − 1.2 1.3 − 1.2
To find b we can use the first point to obtain 0.736 = −1.22(1.2) + b. Solving for b we find b = 2.2 Thus, f (x) = −1.22x + 2.2
14
CHAPTER 1. LIBRARY OF FUNCTIONS
Example 3 Suppose that the graph of a linear function is given and two points on the graph are known. For example, the graph below is the graph of a linear function going through the points (100, 1) and (160, 1.6). Find the formula.
(160,6)
(100,1)
Solution. The slope m is found as follows: m=
6−1 = 0.083. 160 − 100
To find b we use the first point to obtain 1 = 0.083(100) + b. Solving for b we find b = −7.3. So the formula for f (x) = −7.3 + 0.083x. Example 4 Sometimes a linear function is given by a verbal description as in the following problem: In a college meal plan you pay a membership fee; then all your meals are at a fixed price per meal. If 30 meals cost $152.50 and 60 meals cost $250 then find the formula for the cost C of a meal plan in terms of the number of meals n. Solution. We find m first:
250 − 152.50 = $3.25/meal. 60 − 30 To find b or the membership fee we use the point (30, 152.50) in the formula C = mn + b to obtain 152.50 = 3.25(30) + b. Solving for b we find b = $55. Thus, C = 3.25n + 55. m=
So far we have represented a linear function by the expression y = mx + b. This is known as the slope-intercept form of the equation of a line. Now, if the slope m of a line is known and one point (x0 , y0 ) is given then by taking any point (x, y) on the line and using the definition of m we find y − y0 = m. x − x0 Cross multiply to obtain: y − y0 = m(x − x0 ). This is known as the point-slope form of a line.
1.3. RATE OF CHANGE
15
Example 5 Find the equation of the line passing through the point (100, 1) and with slope m = 0.01. Solution. Using the above formula we have: y − 1 = 0.01(x − 100) or y = 0.01x. Note that the form y = mx + b can be rewritten in the form Ax + By = C
(1.1)
where A = m, B = −1, and C = b. The form (??) is known as the standard form of a linear equation. Example 6 Rewrite in standard form: 3x + 2y + 40 = x − y. Solution. Subtracting x − y + 40 from both sides to obtain 2x + 3y = −40.
Recommended Problems (pp. 11 - 3): 1, 3, 7, 9, 11, 12, 15, 17, 23, 25, 29.
1.3
Rate of Change
In this section we will introduce the concept of the average rate of change of a function. We start with an example. Example 7 (Average Speed) During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car; that is, it shows your speed at a particular instant in time. The instantaneous speed of an object is not to be confused with the average speed. Average speed is a measure of the distance traveled in a given period of time. That is, Average Speed =
Distance traveled Time elapsed
So if the trip to school takes 0.2 hours (i.e. 12 minutes) and the distance traveled is 5 miles then the average speed of your car is 5 miles = 25miles/hour 0.2 hours This says that on the average, your car was moving with a speed of 25 miles per hour. During your trip, there may have been times that you were stopped Ave. Speed =
16
CHAPTER 1. LIBRARY OF FUNCTIONS
and other times that your speedometer was reading 50 miles per hour; yet on the average you were moving with a speed of 25 miles per hour. Looking at the above example, we see that the average speed over the interval [0,0.2] is s(0.2) − s(0) Average Speed = 0.2 − 0 where s(t) is the position function of the car at time t hours. Example 8 A freely falling body experiencing no air resistance falls s(t) = 16t2 feet in t seconds. Complete the following table time interval Average velocity
[1.8,2]
[1.9,2]
[1.99,2]
[1.999,2]
[2,2.0001]
[2,2.001]
[2,2.01]
[1.8,2] 60.8
[1.9,2] 62.4
[1.99,2] 63.84
[1.999,2] 63.98
[2,2.0001] 64.0016
[2,2.001] 64.016
[2,2.01] 64.16
Solution. time interval Average velocity
The average speed is a example of the average rate of a function. In general, the average rate of change of a function f (x) from x = a to x = b is the difference quotient f (b) − f (a) . b−a Geometrically, this quantity represents the slope of the secant line going through the points (a, f (a)) and (b, f (b)) on the graph of f (x). Exercise 10 Calculate the average rate of change of f (x) = x2 between x = −2 and x = 1. Solution. Ave. rate of change =
f(1) − f(−2) −3 = = −1 1 − (−2) 3
• Average Rate of Change and Monotone Functions We say that a function is increasing if its graph climbs as x moves from left to right. That is, the function values increase as x increases. It is said to be decreasing if its graph falls as x moves from left to right. This means that the function values decrease as x increases. As an application of the average rate of change, we can use such quantity to decide whether a function is increasing or decreasing. A function is monotone if it is either increasing or decreasing.
1.3. RATE OF CHANGE
17
If a function f is increasing on an interval I then by taking any two points in the interval I, say a < b, we see that f (a) < f (b) and in this case f (b) − f (a) > 0. b−a Thus, if the average rate of change is positive in an interval then the function is increasing in that interval. Similarly, if the average rate of change is negative in an interval I then the function is decreasing. Exercise 11 The table below gives values of a function w = f (t). Is this function increasing or decreasing? t 0 4 8 12 16 20 24 w 100 58 32 24 20 18 17 Solution. The average of w over the interval [0, 4] is w(4) − w(0) 58 − 100 = = −10.5 4−0 4−0 The average rate of change of the remaining intervals are given in the chart below time interval Average
[0,4] -10.5
[4,8] -6.5
[8,12] -2
[12,16] -1
[16, 20] -0.5
[20,24] -0.25
Since the average rate of change is always negative on [0, 24] then the function is decreasing over there. Exercise 12 Determine the intervals where the function is increasing and decreasing 10 8 6 y 4 2 –4
–2
0 –2 –4 –6 –8 –10
2
x
4
Solution. The function is increasing on (−∞, −1) ∪ (1, ∞) and decreasing on the interval (−1, 1).
18
CHAPTER 1. LIBRARY OF FUNCTIONS
• Average Rate of Change and Concavity We say that the graph of a function is concave up or opens up if the average rate of change is increasing. The graph is concave down or opens down if the average rate of change is decreasing. Exercise 13 Find the intervals over which: (a) The function is increasing; decreasing. (b) The graph is concave up; concave down. 4
Fig. 2
3 y 2 1 –1
0
1
2 x
3
4
–1
Solution. The graph is concave up on the interval (1, ∞) and concave down on the interval (−∞, 1). Exercise 14 The table below gives values of a function w = f (t). Is this function concave up or concave down? t 0 4 8 12 16 20 24 w 100 58 32 24 20 18 17 Solution. Completing the following table Interval
[0,4] [4,8] [8,12] [12,16] [16,20] [20,24] -10.5 -6.5 -2 -1 -0.5 -0.25 we see that the average rate of change is increasing so that the graph is concave up. f (b)−f (a) b−a
Recommended Problems (pp. 19 - 21): 1, 2, 3, 4, 5, 7, 10, 12, 13, 16, 17, 25, 27, 30.
1.4
Applications of Functions to Economics
The goal of this section is to exhibit some functions used in business and economics.
1.4. APPLICATIONS OF FUNCTIONS TO ECONOMICS
19
• The Cost Function The cost function C gives the cost C(q) of manufacturing a quantity q of some good. A linear cost function has the form C(q) = mq + b, where the y-intercept b is called the fixed cost, i.e. the costs incurred even if nothing is produced, and the slope m is called the marginal cost or the variable costs per unit. Exercise 15 What is the cost function of manufacturing a product with fixed cost of $400 and variable costs of $40 per item, assuming the function is linear? Solution. The cost function is C(q) = 40q + 400. Exercise 16 Values of a linear cost function are shown below. What are the fixed costs and the marginal cost? Find a formula for the cost function. q C(q)
0 5000
5 5020
10 5040
15 5060
20 5080
Solution. The fixed costs are b = C(0) = $5, 000, the marginal cost is m=
5020 − 5000 =4 5−0
The cost function is C(q) = 4q + 5, 000 • The Revenue Function A revenue function R gives the total revenue R(q) from the sale of a quantity q at a unit price p dollars. Thus, R(q) = pq. Exercise 17 A company that makes a certain brand of chairs has fixed costs of $5, 000 and marginal cost of $30 per chair. The company sells the chairs for $50 each. Find formulas for the cost and revenue functions. Solution. The cost function is C(q) = 30q+5000. The revenue function is R(q) = pq = 30q.
20
CHAPTER 1. LIBRARY OF FUNCTIONS
• The Profit Function Profit is defined to be revenue minus cost. That is P (q) = R(q) − C(q). The break-even point is the point where the profit is zero, i.e. R(q) = C(q). Exercise 18 A company has cost and revenue functions, in dollars, given by C(q) = 6, 000 + 10q and R(q) = 12q. (a) Graph the functions C(q) and R(q) on the same coordinate axes. (b) Find the break-even point and illustrate it graphically. (c) When does the company make a profit? Loses money? Solution. (a) The graph is given below, where the red line is the graph of the cost function and the green one is the graph of the revenue function. 60000 50000 40000 y30000 20000 10000 0
2000 4000 6000 8000 10000 x
(b) The break-even point is the point of intersection of the two lines. To find the point, set 12q = 10q + 6000 and solve for q to find q = 3000. Thus, the break-even point is the point (3000, 36000). (c) The company makes profit for q > 3000 and loses money for q < 3000. • Marginal Cost, Marginal Revenue, and Marginal Profit The marginal cost is the approximate cost of producing an additional item. The marginal revenue is the approximate revenue from selling an additional item. The marginal profit is the approximate profit from selling an additional item. Exercise 19 Consider the cost and revenue functions of the previous exercise. Find the marginal cost, marginal revenue, and marginal profit.
1.5. EXPONENTIAL FUNCTIONS
21
Solution. The marginal cost is the slope of C(q). That is, $10 dollars per item. The marginal cost is $12 dollar per item and since P (q) = R(q) − C(q) = 2q − 6000 then the marginal profit is $2 dollars per item. • Supply and Demand Curves The quantity q manufactured and sold depends on the unit price p. In general, when the price goes up then manufacturers are willing to supply more of the product whereas consumers are going to reduce their buyings. Since consumers and manufacturers react differently to changes in price, there are two curves relating p and q. The supply curve is the quantity that producers are willing to make at a given price. Thus, increasing price will increase quantity. The demand curve is the amount that will be bought by consumers at a given price. Thus, decreasing price will increase quantity. Even though quantity is a function of price, it is the tradition to use the vertical y-axis for the variable p and the horizontal x-axis for the variable q. The supply and demand curves intersect at point (q ∗ , p∗ ) called the point of equilibrium. We call p∗ the equilibrium price and q ∗ the equilibrium quantity. Exercise 20 Find the equilibrium point for the supply function S(p) = 3p−50 and the demand function D(p) = 100 − 2p. Solution. Setting the equation S(p∗ ) = D(p∗ ) to obtain 3p∗ − 50 = 100 − 2p∗ . By adding 2p∗ + 50 to both sides we obtain 5p∗ = 150. Solving for p∗ we find p∗ = 30. Substitute this value in S(p) we find q ∗ = 3(30) − 50 = 40. recommended Problems (pp. 29 - 32): 1, 3, 4, 5, 7, 8, 11, 14, 18, 20, 24.
1.5
Exponential Functions
Linear functions are functions that change at a constant rate. For example, if f (x) = mx + b then f (x + 1) = m(x + 1) + b = f (x) + m. Exponential functions are functions that change at a constant percent rate. That is, if f (x) is an exponential function then f (x + 1) = f (x) + r%f (x). Before we go on to discuss exponential problems, we should pause and discuss some of the terms that are frequently used in these problems. First let’s talk about the terms factor and percent. For example, suppose you invested $100, and after a time, your investment was worth $300. The final value ($300) would be three (3) times the initial value. We would say that your investment had increased by a factor of 3.
22
CHAPTER 1. LIBRARY OF FUNCTIONS
The general equation for percentage changes is Nf inal − Ninitial × 100 = % change Ninitial Thus, a $100 investment that increased to $300 (increased by a factor of 3), had a percentage increase of: 300 − 100 × 100 = 200% 100 Example 9 A quantity increases from 10 to 12. By what percent has it increased? Solution. The increase is
2 10
= .2 = 20%.
Exercise 21 If we multiply a quantity A by 1.15 then A is increased by what percent? Solution. Since 1.15A = A + 0.15A = A + 15%A then multiplying A by 1.15 is equivalent to increasing A by 15%. Exponential functions are used to model increasing quantities such as population problems. Example 10 Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells doubled every minute. That is, the population cell is increasing at the constant rate of 100%. Write an equation to determine the number (population) of cells after one hour. Solution. The table below shows the number of cells for the first 5 minutes. Let P (t) be the number of cells after t minutes. t 0 1 2 3 4 5
P(t) 1 2 4 8 16 32
At time 0, i.e t=0, the number of cells is 1 or 20 = 1. After 1 minute, when t = 1, there are two cells or 21 = 2. After 2 minutes, when t = 2, there are 4 cells or 22 = 4. Therefore, one formula to estimate the number of cells (size of population) after t minutes is the equation (model) f (t) = 2t .
1.5. EXPONENTIAL FUNCTIONS
23
Now, to determine the number of cells after one hour we convert to minutes to obtain t = 60 minutes so that f (60) = 260 = 1.15 × 1018 cells. Exponential functions can also model decreasing quantities. Example 11 If you start a biology experiment with 5,000,000 cells and 45% of the cells are dying every minute, how long will it take to have less than 50,000 cells? Solution. Let P (t) is the number of cells after t minutes. Then P (t + 1) = P (t) − 45%P (t) or P (t + 1) = 0.55P (t). By constructing a table data we find t 0 1 2 3 4 5 6 7 8
P(t) 5,000,000 2,750,000 1,512,500 831,875 457,531.25 251,642.19 138,403.20 76,121.76 41,866.97
So it takes 8 minutes for the population to reduce to less than 50,000 cells. A formula of P (t) is P (t) = 5, 000, 000(0.55)t . From the previous two examples, we see that an exponential function has the general form P (t) = b · at , a > 0 a 6= 1. Since b = P (0) then we call b the initial value. Since P (t + 1) = kat+1 = a(kat ) = aP (t) then we call a the growth factor. Also, since P (t) = P (t) + r%P (t) where r = a−1 then we call r the percent growth rate. Note that for a > 1 (or r < 1) the exponential function represents a growth model whereas for 0 < a < 1 ( or r > 1) the function represents a decay model. Remark. Why a is restricted to a > 0 and a 6= 1? Since t is allowed√ to have any value then a negative a will create meaningless expressions such as a (if t = 12 ). Also, for a = 1 the function P (t) = b is called a constant function and its graph is a horizontal line. Exercise 22 Suppose you are offered a job at a starting salary of $40,000 per year. To strengthen the offer, the company promises annual raises of 6% per year for the first 10 years. Let P (t) be your salary after t years. Find a formula for P (t) and then compute your projected salary after 4 years from now.
24
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. A formula of P (t) is P (t) = 40, 000(1.06)t . After four years, the projected salary is P (3) = 40, 000(1.06)4 ≈ 50, 499.08. Exercise 23 The amount in milligrams of a drug in the body t hours after taking a pill is given by A(t) = 25(0.85)t . (a) What is the initial dose given? (b) What percent of the drug leaves the body each hour? (c) What is the amount of drug left after 10 hours? Solution. (a) Initial dose give is A(0) = 25 mg. (b) r = a − 1 = 0.85 − 1 = −.15 so that 15% of the drug leaves the body each hour. (c) A(10) = 25(0.85)10 ≈ 4.92 mg. • Graphs of Exponential Functions Recall that an exponential function with base a and initial value b is a function of the form f (x) = b · ax . Since b = f (0) then (0, b) is the vertical intercept of f (x). In this section we consider graphs of exponential functions. Let’s see the effect of the parameter b on the graph of f (x) = bax . Example 12 Graph, on the same axes, the exponential functions f1 (x) = 2 · (1.1)x , f2 (x) = (1.1)x , and f3 (x) = 0.75(1.1)x . Solution. Using a graphing calculator, we find 18 16 14 12 10 8 6 4 2 0 0.20.40.60.8 1 1.21.41.61.8 2 x
where f1 is in red, f2 in green and f3 in yellow. Note that these functions have the same growth factor but different b and therefore different vertical intercepts.
1.5. EXPONENTIAL FUNCTIONS
25
We know that the slope of a linear function measures the steepness of the graph. Similarly, the parameter a measures the steepness of the graph of an exponential function.
Example 13 Graph, on the same axes, the exponential functions f( x) = 2 · 2x , f2 (x) = 2 · 3x , and f3 (x) = 2 · 4x . Solution. Using a graphing calculator we find
6 5 4 3 2 0 0.05 0.1 0.15 0.2 0.25 0.3 x
where f1 is in red, f2 in green, and f3 in yellow. • Observations. (i) Note that the greater the value of a, the more rapidly the graph rises. That is, the growth factor a affects the steepness of an exponential function. (ii) Since a > 1 then the graphs represent exponential growths. (iii) Note that as x decreases, the function values approach the x-axis. Symbolically, as x → −∞, y → 0.
Example 14 Graph, on the same axes, the exponential functions f1 (x) = 2·2−x = 2 · 3−x , and f3 (x) = 2 · 4−x = 32x . Solution. Using a graphing calculator we find
2 2x , f2 (x)
=
26
CHAPTER 1. LIBRARY OF FUNCTIONS 2 1.5 1 0.5
0
1
2 x
3
4
where f1 is in red, f2 in green, and f3 in yellow. • Observations. (i) Note that the smaller the value of a, the more rapidly the graph falls. (ii) Since 0 < a < 1 then the graphs represent exponential decays. (iii) Note that as x increases, the function values approach the x-axis. Symbolically, as x → ∞, y → 0. • General Observations (i) For a > 1, as x decreases, the function values get closer and closer to 0. Symbolically, as x → −∞, y → 0. For 0 < a < 1, as x increases, the function values gets closer and closer to the x-axis. That is, as x → ∞, y → 0. We call the x-axis, a horizontal asymptote. (ii) The domain of an exponential function consists of the set of all real numbers whereas the range consists of the set of all positive real numbers. (iii) Since the average rate of change is the slope of a secant line connecting two points on a graph then the average rate of change of the exponential functions discussed above is always increasing. This means that the graph opens up. We also say that the graph is concave up ( If the average rate of change of a function is decreasing then the graph opens down and we say that the graph is concave down). Exponential Functions Versus Linear Functions Next, we consider the question of recognizing whether a function given by a table of values is exponential or linear. We know that for a linear function, equal increments in x correspond to equal increments in y. For an exponential function let us first assume that we have a formula for the function, say f (x) = bax . Then f (x+1) f (x) = a. Thus, if the ratio of consecutive y-values of a function is the same then the function is an exponential function. Example 15 Decide if the function is linear or exponential?Find a formula for each case.
1.5. EXPONENTIAL FUNCTIONS
x 0 1 2 3 4
f(x) 12.5 13.75 15.125 16.638 18.301
x 0 1 2 3 4
27
g(x) 0 2 4 6 8
Solution. 15.125 16.638 18.301 Since 13.75 12.5 ≈ 13.75 ≈ 15.125 ≈ 16.638 ≈ 1.1 then f (x) is an exponential function. To find a formula for f (x) = bax we use the first two points obtaining 12.5 = f (0) = b and 13.75 = f (1) = ba = 12.5a. Hence, a = 13.75 12.5 ≈ 1.1 so that f (x) = 12.5(1.1)x . On the other hand, equal increments in x corresponds to equal increments in the g-values so that g(x) is linear, say g(x) = mx + b. Since g(0) = 0 then b = 0. Also, 2 = g(1) = m so that g(x) = 2x. Finally, we consider the question of finding a formula for an exponential function. The above example shows how to find linear and exponential functions using two data points. The next example illustrates how to find the formula of an exponential function given two points on its graph. Example 16 Find a formula for the exponential function whose graph is given below
(-1,2.5) (1,1.6)
Solution. Write f (x) = bax . Since f (−1) = 2.5 then ba−1 = 2.5. Similarly, ba = 1.6. 2 Taking the ratio we find baba−1 = 1.6 2.5 or a = .64 or a = 0.8. From ba = 1.6 we x find that b = 1.6 = 2 so that f (x) = 2(0.8) . 0.8 Recommended Problems (pp. 37 - 9): 1, 3, 7, 8, 10, 11, 13, 15, 20, 25.
28
1.6
CHAPTER 1. LIBRARY OF FUNCTIONS
Logarithms and Their Properties
In this section we find an algebraic way to solve equations of the form ax = b where a and b are positive constants. For this, we introduce two functions that are found in today’s calculators, namely, the functions log x and ln x. If x > 0 then we define log x to be a number y that satisfies the equality 10y = x. For example, log 100 = 2 since 102 = 100. Similarly, log 0.01 = −2 since 10−2 = 0.01. We call log x the common logarithm of x. Thus, y = log x if and only if 10y = x. Example 17 (a) Rewrite log 30 = 1.477 using exponents instead of logarithms. (b) Rewrite 100.8 = 6.3096 using logarithms instead of exponents. Solution. (a) log30 = 1.477 is equivalent to 101.477 = 30. (b) 100.8 = 6.3096 is equivalent to log 6.3096 = 0.8. Example 18 Without a calculator evaluate the following expressions: (a) log 1
(b) log 100
(c) log ( √110 )
(d) 10log 100
(e) 10log (0.01)
Solution. (a) log 1 = 0 since 100 = 1. (b) log 100 = log 1 = 0 by (a). 1 (c) log ( √110 ) = log 10− 2 = − 21 . (d) 10log 100 = 102 = 100. (e) 10log (0.01) = 10−2 = 0.01. • Properties of Logarithms (i) Since 10x = 10x we can write log 10x = x (ii) Since log x = log x then 10log x = x (iii) log 1 = 0 since 100 = 1. (iv) log 10 = 1 since 101 = 10. (v) Suppose that m = log a and n = log b. Then a = 10m and b = 10n . Thus, a · b = 10m · 10n = 10m+n . Rewriting this using logs instead of exponents, we see that log (a · b) = m + n = log a + log b.
1.6. LOGARITHMS AND THEIR PROPERTIES (vi) If, in (v), instead of multiplying we divide, that is using logs again we find ³a´ log = log a − log b. b
29 a b
=
10m 10n
= 10m−n then
(vii) It follows from (vi) that if a = b then log a − log b = log 1 = 0 that is log a = log b. (viii) Now, if n = log b then b = 10n . Taking both sides to the power k we find bk = (10n )k = 10nk . Using logs instead of exponents we see that log bk = nk = k log b that is log bk = k log b. Example 19 Solve the equation: 4(1.171)x = 7(1.088)x . Solution. ¡ ¢x Rewriting the equation into the form 1.171 = 74 and then using properties 1.088 (vii) and (viii) to obtain µ ¶ 1.171 7 x log = log . 1.088 4 Thus, x=
log 74 ¡ ¢. log 1.171 1.088
Exercise 24 Solve the equation log (2x + 1) + 3 = 0. Solution. Subtract 3 from both sides to obtain log (2x + 1) = −3. Switch to exponential form to get 2x + 1 = 10−3 = 0.001. Subtract 1 and then divide by 2 to obtain x = −0.4995. Remarks. • All of the above arguments are valid for the function ln x for which we replace the number 10 by the number e = 2.718 · · · . That is, y = ln x is equivalent to x = ey , ln (a · b) = ln a + ln b, etc. We call ln x the natural logarithm of x. • Keep in mind the following: log (a + b) 6= log a + log b log (a − b) 6= log a − log b log (ab) ¡ ¢ 6= logaa · log b log ab 6= log ¡ ¢ log b log a1 6= log1 a (Give an example for each case!) • The Logarithmic Function
30
CHAPTER 1. LIBRARY OF FUNCTIONS
We have seen that the notation y = log x is equivalent to 10y = x. Since 10 raised to any power is always positive then the domain of the function log x consists of all positive numbers. That is, log x cannot be used with negative numbers. Now, let us sketch the graph of this function by first constructing the following chart: x 0 0.001 0.01 0.1 1 10 100 1000
log x underfined -3 -2 -1 0 1 2 3
Average Rate of Change 111.11 11.11 1.11 0.11 0.011 0.0011
From the chart we see that the graph is always increasing. Since the average rate of change is decreasing then the graph is always concave down. Now plotting these points and connecting them with a smooth curve to obtain 4 2
0
20
40
x
60
80
100
–2
We observe the following: (a) The range of log x consists of all real numbers. (b) The graph never crosses the y-axis since a positive number raised to any power is always positive. (c) The graph crosses the x-axis at x = 1. (d) As x gets closer and closer to 0 from the right the function log x decreases without bound. That is, as x → 0+ , x → −∞. We call the y-axis a vertical asymptote. In general, if a function increases or decreases without bound as x gets closer to a number a then we say that the line x = a is a vertical asymptote. Next, let’s graph the function y = 10x by using the above process:
1.6. LOGARITHMS AND THEIR PROPERTIES x -3 -2 -1 0 1 2 3
10x 0.001 0.01 0.1 1 10 100 1000
31
Average Rate of Change 0.009 0.09 0.9 9 90 900
Note that this chart can be obtained from the chart of log x discussed above by interchanging the variables x and y. This means, that the graph of y = 10x (in red) is a reflection of the graph of y = log x (in blue) about the line y = x (in green) as seen in the picture below. 3 2 y 1 –2
–1
0
1 x
2
–1 –2 –3
Example 20 Sketch the graphs of the functions y = ln x and y = ex on the same axes.
Solution. The functions y = ln x and y = ex are inverses of each other like the functions y = log x and y = 10x . So their graphs are reflections of one another across the line y = x as shown below
32
CHAPTER 1. LIBRARY OF FUNCTIONS
• Continuous Growth Rate and the Number e For any positive number c, eln c = c. Thus, any positive number a can be written in the form a = ek where k = ln a. If a > 1 then ln a > 0 and therefore k > 0. If 0 < a < 1 then ln a < 0 and thus k < 0. Now, suppose that Q(t) = bat . Then Q(t) = b(ek )t = bekt . Note that for k > 0 ek > 1 so that Q(t) represents an exponential growth and if k < 0 then ek < 1 so that Q(t) is an exponential decay. We call the constant k the continuous growth rate. Example 21 If f (t) = 3(1.072)t is rewritten as f (t) = 3ekt , find k. Solution. By comparison of the two functions we find ek = 1.072. Take the ln of both sides to obtain k = ln (1.072). Example 22 A population increases from its initial level of 7.3 million at the continuous rate of 2.2% per year. Find a formula for the population P (t) as a function of the yeat t. When does the population reach 10 million? Solution. We are given the intial value 7.3 million and the continuous growth rate k = 0.022. Therefore, P (t) = 7.3e0.022t . Next,we want to find the time when P (t) =
1.7. EXPONENTIAL GROWTH AND DECAY
33
10. That is , 7.3e0.022t = 10. Divide both sides by 7.3 and then take the ln of ln 10 both sides to obtain 0.022t = ln 7.3 . Thus, t=
1 ln 10 ≈ 14.3. 0.022 ln 7.3
Next, in order to convert from Q(t) = bekt to Q(t) = bat we let a = ek . For example, to convert the formula Q(t) = 7e0.3t to the form Q(t) = bat we let b = 7 and a = e0.3 ≈ 1.35. Thus, Q(t) = 7(1.35)t . Example 23 Find the annual percent rate and the continuous percent growth rate of Q(t) = 200(0.886)t . Solution. The annual percent of decrease is r = a − 1 = 0.886 − 1 = −0.114 = −11.4%. To find the continuous percent growth rate we let ek = 0.886 and solve for k by taking the ln of both sides and obtain k = ln (0.886) ≈ −0.121 = −12.1%. Recommended Problems (pp. 42 - 3): 3, 5, 9, 11, 15, 17, 19, 22, 23, 26, 27, 31.
1.7
Exponential Growth and Decay
In this section, we consider some applications of exponential functions. • Doubling Time In some exponential models one is interested in finding the time for an exponential growing quantity to double. We call this time the doubling time. To find it, we start with the equation b · at = 2b or at = 2. Solving for t we find log 2 t = log a. Example 24 Find the doubling time of a population growing according to P = P0 e0.2t . Solution. Setting the equation P0 20.2t = 2P0 and dividing both sides by P0 to obtain 2 20.2t = 2. Take ln of both sides to obtain 0.2t = ln 2. Thus, t = ln 0.2 ≈ 3.47. • Half-Life On the other hand, if a quantity is decaying exponentially then the time required for the quantity to reduce into half is called the half-life. To find it, we start with the equation bat = 2b and we divide both sides by b to obtain at = 0.5. Take the log of both sides to obtain t log a = log (0.5). Solving for t we find t = loglog(0.5) a .
34
CHAPTER 1. LIBRARY OF FUNCTIONS
Example 25 The half-life of Iodine-123 is about 13 hours. You begin with 50 grams of this substance. What is a formula for the amount of Iodine-123 remaining after t hours? Solution. Since the problem involves exponential decay then if Q(t) is the quantity remaining after t hours then Q(t) = 50at with 0 < t < 1. But Q(13) = 25. That 1 is, 50a13 = 25 or a13 = 0.5. Thus a = (0.5) 13 ≈ 0.95 and Q(t) = 50(0.95)t . • Compound Interest The term compound interest refers to a procedure for computing interest whereby the interest for a specified interest period is added to the original principal. The resulting sum becomes a new principal for the next interest period. The interest earned in the earlier interest periods earn interest in the future interest periods. The future value is the total amount due on the maturity date of the loan or the investment. The maturity value is calculated by using the compound amount formula which is given by A = P (1 +
r nt ) n
where P is the present value or principal, r is the nominal annual interest rate, n is the number of periods in a year, and t is the time in years. Possible values of n are: 1 (annually), 2 (semiannually), 4 (quarterly), 12 (monthly), and 365 (daily). When interest is compounded more frequently than once a year, the account effectively earns more than the nominal rate. Thus, we distinguish between nominal rate and effective rate. The effective annual rate tells how much interest the investment actually earns. The quantity (1 + nr )n − 1 is known as the effective interest rate. Example 26 Translating a value to the future is referred to as compounding. What will be the maturity value of an investment of $15, 000 invested for four years at 9.5% compounded semi-annually? Solution. Using the formula for compound interest with P = $15, 000, t = 4, n = 2, and r = .095 we obtain µ ¶8 0.95 A = 15, 000 1 + ≈ $21, 743.20 2 Example 27 Translating a value to the present is referred to as discounting. We call (1 + nr )−nt the discount factor. What principal invested today will amount to $8, 000 in 4 years if it is invested at 8% compounded quarterly?
1.7. EXPONENTIAL GROWTH AND DECAY
35
Solution. The present value is found using the formula µ ¶−16 ³ r ´−nt 0.08 P =A 1+ = 8, 000 1 + ≈ $5, 827.57 n 4 Example 28 What is the effective rate of interest corresponding to a nominal interest rate of 5% compounded quarterly? Solution. effective rate =
µ ¶4 0.05 1+ − 1 ≈ 0.051 = 5.1% 4
Continuous Compound Interest When the compound formula is used over smaller time periods the interest becomes slightly larger and larger. That is, frequent compounding earns a higher effective rate, though the increase is small. This suggests compounding more and more, or equivalently, finding the value of A in the long run. The example at previous section shows ¡ ¢n the end of the that as n → ∞ the expression 1 + nr approaches er so that A = P ert . This formula is known as the continuous compound formula. In this case, the annual effective interest rate is found using the formula er − 1. Example 29 Find the effective rate if $1000 is deposited at 5% annual interest rate compounded continuously. Solution. The effective interest rate is e0.05 − 1 ≈ 0.05127 = 5.127% Example 30 Which is better: An account that pays 8% annual interest rate compounded quarterly or an account that pays 7.95% compounded continuously? Solution. The effective rate corresponding to the first option is µ ¶4 0.08 1+ − 1 ≈ 8.24% 4 That of the second option e0.0795 − 1 ≈ 8.27% Thus, we see that 7.95% compounded continuously is better than 8% compounded quarterly. Recommended Problems (pp. 48 - 51): 5, 7, 8, 13, 19, 22, 28, 29, 30, 33.
36
1.8
CHAPTER 1. LIBRARY OF FUNCTIONS
Building New Functions from Old Ones
In this section we discuss various ways for building new functions from old ones. New functions can be obtained by composing functions, using arithmetic combinations, and finally by making changes to either the input or the output of a function. • Composition of Functions In this section we will discuss a procedure for building new functions from old ones known as the composition of functions. We start with an example of a real-life situation where composite functions are applied. Example 31 You have two money machines, both of which increase any money inserted into them. The first machine doubles your money. The second adds five dollars. The money that comes out is described by f (x) = 2x in the first case, and g(x) = x + 5 in the second case, where x is the number of dollars inserted. The machines can be hooked up so that the money coming out of one machine goes into the other. Find formulas for each of the two possible composition machines. Solution. Suppose first that x dollars enters the first machine. Then the amount of money that comes out is f (x) = 2x. This amount enters the second machine. The final amount coming out is given by g(f (x)) = f (x) + 5 = 2x + 5. Now, if x dollars enters the second machine first, then the amount that comes out is g(x) = x + 5. If this amount enters the second machine then the final amount coming out is f (g(x)) = 2(x + 5) = 2x + 10. The function f (g(x)) is called the composition of the functions f and g; the function g(f (x)) is called the composition of the functions g and f. In general, suppose we are given two functions f and g such that the range of g is contained in the domain of f so that the output of g can be used as input for f. We define a new function, called the composition of f and g, by the formula (f ◦ g)(x) = f (g(x)). In a similar way, we can define the composition of g and f to be the function (g ◦ f )(x) = g(f (x)) so that the output of f is the input of g. Using a Venn diagram we have
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
37
• Composition of Functions Defined by Tables Example 32 Complete the following table x f(x) g(x) f(g(x))
0 1 5
1 0 2
2 5 3
3 2 1
4 3 4
5 4 8
0 1 5 4
1 0 2 5
2 5 3 2
3 2 1 0
4 3 4 3
5 4 8 undefined
Solution. x f(x) g(x) f(g(x))
• Composition of Functions Defined by Formulas Example 33 Suppose that f (x) = 2x + 1 and g(x) = x2 − 3. (a) Find f ◦ g and g ◦ f. (b) Calculate (f ◦ g)(5) and (g ◦ f )(−3). (c) Are f ◦ g and g ◦ f equal? Solution. (a) (f ◦ g)(x) = f (g(x)) = f (x2 − 3) = 2(x2 − 3) + 1 = 2x2 − 5. Similarly,
38
CHAPTER 1. LIBRARY OF FUNCTIONS
(g ◦ f )(x) = g(f (x)) = g(2x + 1) = (2x + 1)2 − 3 = 4x2 + 4x − 2. (b) (f ◦ g)(5) = 2(5)2 − 5 = 45 and (g ◦ f )(−3) = 4(−3)2 + 4(−3) − 2 = 22. (c) f ◦ g 6= g ◦ f. With only one function you can build new functions using composition of the function with itself. Also, there is no limit on the number of functions that can be composed. Example 34 Suppose that f (x) = 2x + 1 and g(x) = x2 − 3. (a) Find (f ◦ f )(x). (b) Find [f ◦ (f ◦ g)](x). Solution. (a) (f ◦ f )(x) = f (f (x)) = f (2x + 1) = 2(2x + 1) + 1 = 4x + 3. (b) [f ◦ (f ◦ g)](x) = f (f (g(x))) = f (f (x2 − 3)) = f (2x2 − 5) = 2(2x2 − 5) + 1 = 4x2 − 9. • Decomposition of Functions If a formula for (f ◦ g)(x) is given then the process of finding the formulas for f and g is called decomposition. Example 35 √ Decompose (f ◦ g)(x) = 1 − 4x2 . Solution. √ 2 One possible √ answer is f (x) = x and g(x) = 1 − 4x . Another possible answer 2 is f (x) = 1 − x and g(x) = 2x. • Combinations of Functions In this section we are going to construct new functions from old ones using the operations of addition, subtraction, multiplication, and division. Let f (x) and g(x) be two given functions. Then for all x in the common domain of these two functions we define new functions as follows: • Sum: (f + g)(x) = f (x) + g(x). • Difference: (f − g)(x) = f (x) − g(x). • Product: (f ³ ·´g)(x) = f (x) · g(x). • Division:
f g
(x) =
f (x) g(x)
provided that g(x) 6= 0.
In the following example we see how to construct the four functions discussed above when the individual functions are defined by formulas. Example 36 √ Let f (x) = x + 1 and g(x) = x + 3. Find the common domain and then find a formula for each of the functions f + g, f − g, f · g, fg .
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
39
Solution. The domain of f (x) consists of all real numbers whereas the domain of g(x) consists of all numbers x ≥ 3. Thus, the common domain is the interval [−3, ∞). For any x in this domain we have √ (f + g)(x) = x + 1 + √x + 3 (f − g)(x) = x+1− √ x+3 √ (f x x+3+ x+3 ³ · ´g)(x) = f √x+1 provided x > −3. (x) = g x+3 In the next example, we see how to evaluate the four functions when the individual functions are given in numerical forms. Example 37 Suppose the functions following table: x -1 -1 f(x) 8 2 g(x) -1 -5 (f + g)(x) (f − g)(x) (f · g)(x) ( fg )(x)
f and g are given in numerical forms. Complete the 0 7 -11
1 -1 7
1 -5 8
3 -3 9
Solution. x f(x) g(x) (f + g)(x) (f − g)(x) (f · g)(x) ( fg )(x)
-1 8 -1 7 9 -8 -8
-1 2 -5 -3 7 -10 - 25
0 7 -11 -4 18 -77 7 - 11
1 -1 7 6 -8 -7 - 17
1 -5 8 3 -13 -40 - 85
3 -3 9 6 -12 -27 - 13
• Vertical and Horizontal Translations In this section we want to discuss the effect of change to the input of a function as well as to the output. More precisely, we will discuss graphing new functions of the form f (x + k) and f (x) + k where k is a given constant. First, let’s look at the following example. Example 38 Let f (x) = x2 . (a) Use a calculator y3 = f (x) + 3. What (b) Use a calculator y3 = f (x) − 3. What
to graph the functions y1 = f (x) + 1, y2 = f (x) + 2 and can you say about these graphs? to graph the functions y1 = f (x) − 1, y2 = f (x) − 2 and can you say about these graphs?
40
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. (a) The graph of y1 = f (x) + 1 = x2 + 1 (green color) is obtained by lifting the graph of f (x) (red color) 1 unit upward. Similarly, the graph of y2 = f (x) + 2 = x2 + 2 (yellow color) is the graph of f (x) lifted two units up and the graph of y3 = f (x) + 3 = x2 + 3 (blue color) is the graph of f (x) lifted up three units. 10 8 6 y 4 2
–4
–2
0
2
x
4
(b) The graph of y1 = f (x) − 1 = x2 − 1 (green color) is obtained by shifting the graph of f (x) (red color) 1 unit downward. Similarly, the graph of y2 = f (x) − 2 = x2 − 2 (yellow color) is the graph of f (x) shifted two units down and the graph of y3 = f (x) − 3 = x2 − 3 (blue color) is the graph of f (x) shifted down three units. 5 4 y
3 2 1
–4
–2
0 –1
2
x
4
–2 –3
It follows that the graph of f (x) + k with k > 0 is a vertical translation of the graph of f (x), k units upward, whereas for k < 0 it is a shift by k units downward. Next, we discuss horizontal shifts. Example 39 Let f (x) = x2 .
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES (a) Use a calculator y3 = f (x + 3). What (b) Use a calculator y3 = f (x − 3). What
41
to graph the functions y1 = f (x + 1), y2 = f (x + 2) and can you say about these graphs? to graph the functions y1 = f (x − 1), y2 = f (x − 2) and can you say about these graphs?
Solution. (a) The graph of y1 = f (x + 1) = (x + 1)2 (green color) is obtained by shifting the graph of f (x) (red color) 1 unit to the left. Similarly, the graph of y2 = f (x + 2) = (x + 2)2 (yellow color) is the graph of f (x) shifted two units to the left and the graph of y3 = f (x + 3) = (x + 3)2 (blue color) is the graph of f (x) shifted to the left by three units. 10 8 6 y 4 2
–4
0
–2
2
x
4
(b) The graph of y1 = f (x − 1) = (x − 1)2 (green color) is obtained by shifting the graph of f (x) (red color) 1 unit to the right. Similarly, the graph of y2 = f (x − 2) = (x − 2)2 (yellow color) is the graph of f (x) shifted two units to the right and the graph of y3 = f (x − 3) = (x − 3)2 (blue color) is the graph of f (x) shifted to the right by three units. 10 8 6 y 4 2
–4
–2
0
2
x
4
It follows that the graph of f (x + k) with k > 0 is a horizontal translation
42
CHAPTER 1. LIBRARY OF FUNCTIONS
of the graph of f (x) k units to the left whereas for k < 0 it is a shift by k units to the right. Exercise 25 Suppose S(d) gives the height of high tide in Seattle on a specific day, d, of the year. Use shifts of the function S(d) to find formulas of each of the following functions: (a) T (d), the height of high tide in Tacoma on day d, given that high tide in Tacoma is always one foot higher than high tide in Seattle. (b) P (d), the height of high tide in Portland on day d, given that high tide in Portland is the same height as the previous day’s hight in Seattle. Solution. (a) T (d) = S(d) + 1. (b) P (d) = S(d − 1). Finally, one can use a combination of both horizontal and vertical shifts to create new functions as shown in the next example. Example 40 Let f (x) = x2 . Let g(x) be the function obtained by shifting the graph of f (x) 2 units to the right and then up three units. Find a formula for g(x) and then draw its graph. Solution. The formula of g(x) is g(x) = f (x−2)+3 = (x−2)2 +3 = x2 −4x+7. The graph of g(x) (in green color) consists of a horizontal shift of two units to the right followed by a vertical shift of three units upward of the graph of f (x) (red color). 10 8 6 y 4 2
–4
–2
0
2
x
4
• Vertical Stretches and Compressions We have seen that for a positive k, the graph of f (x) + k is a vertical shift of
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
43
the graph of f (x) upward and the graph of f (x) − k is a vertical shift down. In this section we want to study the effect of multiplying a function by a constant k. This will result by either a vertical stretch or vertical compression. A vertical stretching is the stretching of the graph away from the x-axis. A vertical compression is the squeezing of the graph towards the x-axis.
Example 41 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = 2x2
x -3 -2 -1 0 1 2 3
y = 3x2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice?
Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = 2x2 18 8 2 0 2 8 18
x -3 -2 -1 0 1 2 3
y = 3x2 27 12 3 0 3 12 27
(b) The picture below shows that the graphs of 2f (x) (in green color) and 3f (x) (in yellow color) are vertical stretches of the graph of f (x) (in red color) by a factor of 2 and 3 respectively.
44
CHAPTER 1. LIBRARY OF FUNCTIONS 5 4 3 y 2 1
–2
–1
0
1 x
2
Example 42 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = 12 x2
x -3 -2 -1 0 1 2 3
y = 13 x2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice?
Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = 12 x2 4.5 2 0.5 0 0.5 2 4.5
x -3 -2 -1 0 1 2 3
y = 13 x2 3 4 3 1 3
0
1 3 4 3
3
(b) The picture below shows that the graphs of 12 f (x) (in green color) and 1 3 f (x) (in yellow color) are vertical compressions of the graph of f (x) (in red color) by a factor of 12 and 13 respectively.
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
45
5 4 3 y 2 1
–2
0
–1
1 x
2
It follows that if a function f (x) is given, then the graph of kf (x) is a vertical stretch of the graph of f (x) by a factor of k for k > 1, and a vertical compression for 0 < k < 1. What about k < 0? If |k| > 1 then the graph of kf (x) is a vertical stretch of the graph of f (x) followed by a reflection about the x-axis. If 0 < |k| < 1 then the graph of kf (x) is a vertical compression of the graph of f (x) by a factor of k followed by a reflection about the x-axis. Example 43 (a) Use a graphing calculator to graph the functions f (x) = x2 , −2f (x), and −3f (x) on the same axes. (b) Use a graphing calculator to graph the functions f (x) = x2 , − 12 f (x), and - 13 f (x) on the same axes. Solution. (a) The picture below shows that the graphs of −2f (x) (in green color) and −3f (x) (in yellow color) are vertical stretches followed by reflections about the x-axis of the graph of f (x) (in red color) 4 y
–2
–1
2 0
1 x
2
–2 –4
(b) The picture below shows that the graphs of − 12 f (x) (in green color) and − 13 f (x) (in yellow color) are vertical compressions of the graph of f (x) (in red
46
CHAPTER 1. LIBRARY OF FUNCTIONS
color) 4 y
–2
–1
2 0
1 x
2
–2 –4
• Horizontal Stretches and Compressions A vertical stretch or compression results from multiplying the outside of a function by a constant k. In this section we will see that multiplying the inside of a function by a constant k results in either a horizontal stretch or compression. A horizontal stretching is the stretching of the graph away from the y-axis. A horizontal compression is the squeezing of the graph towards the y-axis. Example 44 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = (2x)2
x -3 -2 -1 0 1 2 3
y = (3x)2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice? Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = (2x)2 36 16 4 0 4 16 36
x -3 -2 -1 0 1 2 3
y = (3x)2 81 36 9 0 9 36 81
1.8. BUILDING NEW FUNCTIONS FROM OLD ONES
47
(b) The picture below shows that the graphs of f (2x) = (2x)2 = 4x2 (in green color) and f (3x) = (3x)2 = 9x2 (in yellow color) are horizontal compressions of the graph of f (x) (in red color) by a factor of 21 and 13 respecitvely. 4 y
–2
–1
2 0
1 x
2
–2 –4
Example 45 (a) Complete the following tables x -3 -2 -1 0 1 2 3
y = x2
x -3 -2 -1 0 1 2 3
y = ( 12 x)2
x -3 -2 -1 0 1 2 3
y = ( 13 x)2
(b) Use the tables of values to graph and label each of the 3 functions on the same axes. What do you notice? Solution. (a) x -3 -2 -1 0 1 2 3
y = x2 9 4 1 0 1 4 9
x -3 -2 -1 0 1 2 3
y = ( 12 x)2 9 4
1
1 4
0
1 4
1
9 4
x -3 -2 -1 0 1 2 3
y = ( 13 x)2 1 4 9 1 9
0
1 9 4 9
1
(b) The picture below shows that the graphs of f ( x2 ) (in green color) and f ( x3 ) (in yellow color) are horizontal stretches of the graph of f (x) (in red color) by a factor of 2 and 3 respectively.
48
CHAPTER 1. LIBRARY OF FUNCTIONS 10 8 6 y 4 2 –10
–6 –4
0 –2 –4 –6 –8 –10
2 4 6 8 10 x
It follows that if a function f (x) is given, then the graph of f (kx) is a horizontal stretch of the graph of f (x) by a factor of k1 for 0 < k < 1, and a horizontal compression for k > 1. What about k < 0? If |k| > 1 then the graph of f (kx) is a horizontal compression of the graph of f (x) followed by a reflection about the y-axis. If 0 < |k| < 1 then the graph of f (kx) is a horizontal stretch of the graph of f (x) by a factor of k1 followed by a reflection about the y-axis. Example 46 (a) Use a graphing calculator to graph the functions f (x) = x3 , f (−2x), and f (−3x) on the same axes. (b) Use a graphing calculator to graph the functions f (x) = x3 , f (− x2 ), and f (− x3 ) on the same axes. Solution. (a) The picture below shows that the graphs of f (−2x) (in green color) and f (−3x) (in yellow color) are vertical stretches followed by reflections about the y-axis of the graph of f (x) (in red color) 1 0.8 0.6 y 0.4 0.2 –1
–0.6
0 –0.2 –0.4 –0.6 –0.8 –1
0.20.40.60.8 1 x
(b) The picture below shows that the graphs of f (− x2 ) (in green color) and f (− x3 ) (in yellow color) are horizontal stretches of the graph of f (x) (in red color)
1.9. POLYNOMIAL AND RATIONAL FUNCTIONS
49
0.1 0.08 0.06 y 0.04 0.02 –1
–0.6
0 –0.02 –0.04 –0.06 –0.08 –0.1
0.20.40.60.8 1 x
Recommended Problems (pp. 54 - 56): 1, 5, 7, 13, 14, 15, 16, 17, 18, 19, 31.
1.9
Polynomial and Rational Functions
A function f (x) is a power function of x if there is a constant k such that f (x) = kxn for some constant k. If n > 0, then we say that f (x) is proportional to the nth power of x. If n < 0 then f (x) is said to be inversely proportional to the nth power of x. We call k the constant of proportionality.
Polynomial Functions Polynomial functions are among the simplest, most important, and most commonly used mathematical functions. These functions consist of one or more terms of variables with whole number exponents. (Whole numbers are positive integers and zero.) All such functions in one variable (usually x) can be written in form f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , an 6= 0 where an , an−1 , · · · , a1 , a0 are all real numbers, called the coefficients of f (x). The number n is a non-negative integer. It is called the degree of the polynomial. A polynomial of degree zero is just a constant function. A polynomial of degree one is a linear function, of degree two a quadratic function, etc. The number an is called the leading coefficient and a0 is called the constant term. Note that the terms in a polynomial are written in descending order of the exponents. Polynomials are defined for all values of x. Example 47 Find the leading coefficient, the constant term and the degreee of the polynomial f (x) = 4x5 − x3 + 3x2 + x + 1.
50
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. The given polynomial is of degree 5, leading coefficient 4, and constant term 1. A polynomial function will never involve terms where the variable occurs in a denominator, underneath a radical, as an input of either an exponential or logarithmic function. Example 48 Determine whether the function is a polynomial function or not: (a) (b) (c) (d) (e)
f (x) = 3x4 − 4x2 + 5x − 10 g(x) = x3 − ex + 3 h(x) = x2 − √ 3x + x1 + 4 2 i(x) = x − x − 5 j(x) = x3 − 3x2 + 2x − 5 ln x − 3.
Solution. (a) f (x) is a polynomial function of degree 4. (b) g(x) is not a ploynomial degree because one of the terms is an exponential function. (c) h(x) is not a polynomial because x is in the denominator of a fraction. (d) i(x) is not a polynomial because it contains a radical sign. (e) j(x) is not a olynomial because one of the terms is a logarithm of x. • Graphs of a Polynomial Function Polynomials are continuous and smooth everywhere: • A continuous function means that it can be drawn without picking up your pencil. There are no jumps or holes in the graph of a polynomial function. • A smooth curve means that there are no sharp turns (like an absolute value) in the graph of the function. • The y-intercept of the polynomial is the constant term a0 . The shape of a polynomial depends on the degree and leading coefficient: • If the leading coefficient, an , of a polynomial is positive, then the right hand side of the graph will rise towards +∞. • If the leading coefficient, an , of a polynomial is negative, then the right hand side of the graph will fall towards −∞. • If the degree, n, of a polynomial is even, the left hand side will do the same as the right hand side. • If the degree, n, of a polynomial is odd, the left hand side will do the opposite of the right hand side.
1.9. POLYNOMIAL AND RATIONAL FUNCTIONS
51
Example 49 According to the graphs given below, indicate the sign of an and the parity of n for each curve.
(a)
(b)
(c)
(d)
Solution. (a) an < 0 and n is odd. (b) an > 0 and n is odd. (c) an > 0 and n is even. (d) an < 0 and n is even. • Long-Run Behavior of a Polynomial Function If f (x) and g(x) are two functions such that f (x) − g(x) ≈ 0 as x increases without bound then we say that f (x) resembles g(x) in the long run. For example, if n is any positive integer then x1n ≈ 0 in the long run. Now, if f (x) = an xn + an−1 xn + · · · + a1 x + a0 then an−1 an−2 a1 a0 f (x) = xn (an + + 2 + · · · + n−1 + n x x x x Since
1 xk
≈ 0 in the long run, for each 0 ≤ k ≤ n − 1 then f (x) ≈ an xn
in the long run. Example 50 The polynomial function f (x) = 1−2x4 +x3 resembles the function g(x) = −2x4 in the long run. • Zeros of a Polynomial Function If f is a polynomial function in one variable, then the following statements are equivalent: • x = a is a zero or root of the function f. • x = a is a solution of the equation f (x) = 0. • (a, 0) is an x-intercept of the graph of f. That is, the point where the graph crosses the x-axis.
52
CHAPTER 1. LIBRARY OF FUNCTIONS
Example 51 Find the x-intercepts of the polynomial f (x) = x3 − x2 − 6x. Solution. Factoring the given function to obtain f (x) = =
x(x2 − x − 6) x(x − 3)(x + 2)
Thus, the x-intercepts are the zeros of the equation x(x − 3)(x + 2) = 0 That is, x = 0, x = 3, or x = −2. Rational Functions A rational function is a function that is the ratio of two polynomials functions f (x) g(x) . The domain consists of all numbers such that g(x) 6= 0. Example 52 Find the domain of the function f (x) =
x−2 x2 −x−6 .
Solution. The domain consists of all numbers x such that x2 − x − 6 6= 0. But this last quadratic expression is 0 when x = −2 or x = 3. Thus, the domain is set (−∞, −2) ∪ (−2, 3) ∪ (3, ∞). Geometrically, the values of x for which the denominator of a rational function is zero are called vertical asymptotes. Thus, if x = a is a vertical asymptote then as x approaches a from either sides the function becomes either positively large or negatively large. The graph of a function never crosses its vertical asymptotes. Example 53 Find the vertical asymptotes of the function f (x) =
2x−11 x2 +2x−8
Solution. Factoring x2 + 2x − 8 = 0 we find (x − 2)(x + 4) = 0. Thus, the vertical asymptotes are the lines x = 2 and x = −4. If f (x) approaches a value b as x → ∞ or x → −∞ then we call y = b a horizontal asymptote. The graph of a rational function may cross its horizontal asymptote. Example 54 Find the horizontal asymptote, if it exists, for each of the following functions: (a) f (x) = (b) f (x) = (c) f (x) =
3x2 +2x−4 2x2 −x+1 . 2x+3 x3 −2x2 +4 . 2x2 −3x−1 . x−2
1.9. POLYNOMIAL AND RATIONAL FUNCTIONS
53
Solution. (a) In the long run, we have f (x)
= = ≈
3x2 +2x−4 2x2 −x+1 2 4 x2 3+ x − x2 · 1 2 x 2− x + 12 x 3 2
So the line y = 32 is the horizontal asymptote. (b) In the long run 2x+3 f (x) = x3 −2x2 +4 =
x x3
·
3 2+ x 2 1− x +
≈
0
4 x3
So the x-axis is the horizontal asymptote. (c) In the long run 2x2 −3x−1 f (x) = x−2 = ≈
x2 x
·
3 2− x −
2 1− x
1 x2
∞
So the function has no horizontal asymptote. If ((mx + b) − f (x)) ≈ 0 in the long run then we call the line y = mx + b an oblique asymptote. Example 55 Find the oblique asymptote of the function f (x) =
2x2 −3x−1 . x−2
Solution. Using long division of polynomials we can write f (x) = 2x + 1 + Thus, f (x) − (2x + 1) = asymptote.
1 x−2
1 x−2
≈ 0 in the long run. Thus, y = 2x + 1 is the oblique
To graph a rational function h(x) =
f (x) g(x) :
1. Find the domain of h(x) and therefore sketch the vertical asymptotes of h(x). 2. Sketch the horizontal or the oblique asymptote if they exist. 3. Find the x − intercepts of h(x) by solving the equation f (x) = 0. 4. Find the y-intercept: h(0) 5. Draw the graph Example 56 Sketch the graph of the function f (x) =
2 x+3
54
CHAPTER 1. LIBRARY OF FUNCTIONS
Solution. 1. 2. 3. 4. 5.
Domain = (−∞, −3) ∪ (−3, ∞). The vertical asymptote is x = −3. In the long run, f (x) ≈ 0 so the axis is the horizontal asymptote. No x-intercepts. The y-intercept is y = 13 . The graph is given below 2 1 y –6
–4 x –2
0
2
–1 –2
Recommended Problems (pp. 60 - 2): 1, 4, 7, 9, 13, 15, 17, 21, 25, 27, 30, 37.
Chapter 2
The Derivative of a Function In Chapter 1, we introduced the average rate of change of a function on a closed interval. In this chapter, we want to introduce the rate of change of a function at a point. This is defined to be the value obtained by taking the average rate of change over smaller and smaller intervals. The rate of change at a point is called the instantaneous rate of change or the derivative.
2.1
Instantaneous Rate of Change
In this section, we discuss the concept of the instantaneous rate of change of a given function. As an application, we use the velocity of a moving object. The motion of an object along a line at a particular instant is very difficult to define precisely. The modern approach consists of computing the average velocity over smaller and smaller time intervals. To be more precise, let s(t) be the position function of a moving object at time t. We would like to compute the velocity of the object at the instant t = t0 : • Average Velocity We start by finding the average velocity of the object over the time interval t0 ≤ t ≤ t0 + ∆t given by the expression v=
Distance T raveled s(t0 + ∆t) − s(t0 ) = Elapsed T ime ∆t
Geometrically, the average velocity over the time interval [t0 , t0 + ∆t] is just the slope of the line joining the points (t0 , s(t0 )) and (t0 + ∆t, s(t0 + ∆t)) on the graph of s(t).(See Fig. 1) 55
56
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
t
t0+ t
0
Fig. 1
Example 57 A freely falling body experiencing no air resistance falls s(t) = 16t2 feet in t seconds. Complete the following table time interval Average velocity
[1.8,2]
[1.9,2]
[1.99,2]
[1.999,2]
[1.8,2] 60.8
[1.9,2] 62.4
[1.99,2] 63.84
[1.999,2] 63.98
[2,2.0001]
[2,2.001]
[2,2.01]
Solution. time interval Average velocity
[2,2.0001] 64.0016
[2,2.001] 64.016
• Instantaneous Velocity and Speed The next step is to calculate the average velocity on smaller and smaller time intervals ( that is, make ∆t close to zero). The average velocity in this case approaches what we would intuitively call the instantaneous velocity at time t = t0 which is defined using the limit notation by
v(t0 ) = lim
∆t→0
s(t0 + ∆t) − s(t0 ) ∆t
Geometrically, the instantaneous velocity at t0 is the slope of the tangent line to the graph of s(t) at the point (t0 , s(t0 )).(See Fig. 2)
[2,2.01] 64.16
2.1. INSTANTANEOUS RATE OF CHANGE
57
tangent line t
t0+ t
0
Fig. 2 Example 58 For the distance function in Example ??, find the instantaneous velocity at t = 2. Solution. Examining the bottom row of the table in Example ??, we see that the average velocity seems to be approaching the value 64 as we shrink the time intervals. Thus, it is reasonable to expect the velocity to be v(2) = 64 f t/sec. In general, we define the instantaneous rate of change of a function y = f (x) at x = a to be f (x) − f (a) lim . x→a x−a We define the speed of a moving object to be the absolute value of the velocity function. Sometimes there is confusion between the words ”speed” and ”velocity”. Speed is a nonnegative number that indicates how fast an object is moving, whereas velocity indicates both speed and direction(relative to a coordinate system). For example, if the object is moving along a vertical line we define a positive velocity when the object is going upward and a negative velocity when the object is going downward. The instantaneous rate of change of a function f (x) from x = a to x = a + h is called the derivative of f at a and we denote it by f 0 (a) : f 0 (a) = lim
h→0
f (a + h) − f (a) h
If the derivative exists, i.e. can be found, then we say that the function is differentiable at x = a. The process of finding the derivative of a function is called differentiation. If a function has no derivative at a point then we say
58
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
that it is non-differentiable there. Since the instantaneous rate of change represents the slope of a tangent line then f 0 (a) is the slope of the tangent line to the graph of y = f (x) at the point (a, f (a)). The equation of the tangent line is given by y − f (a) = f 0 (a)(x − a). Exercise 26 (a) Find f 0 (1) for f (x) = x2 . (b) Find the equation of the tangent line to the graph of f (x) at the point (1, f (1)). Solution. Completing the following chart x f (b)−f (a) b−a
[0.9,1] 1.9
[0.99,1] 1.99
[0.999,1] 1.999
[1,1.0001] 2.0001
[1,1.001] 2.001
[1,1.01] 2.01
[1,1.1] 2.1
we see that f 0 (1) = 2. (b) The equation of the tangent line is y − f (1) = f 0 (1)(x − 1) or y − 1 = 2(x − 1) In point-intercept form, we have y = 2x − 1. Example 59 (Numerical Estimation of the Derivative) Find approximate values for f 0 (x) at each of the x-values given in the following table x f(x)
0 100
5 70
10 55
15 46
20 40
Solution. The derivative can be estimated by the average rate of change or the difference quotient f (a + h) − f (a) f 0 (a) ≈ . h Thus, f 0 (0) f 0 (5) f 0 (10) f 0 (15)
≈ ≈ ≈ ≈
f (5)−f (0) 5 f (10)−f (5) 5 f (15)−f (10) 5 f (20)−f (15) 5
= −6 = −3 = −1.8 = −1.2
2.2. THE DERIVATIVE FUNCTION
59
Note that in the above example, we used the points (5, 70) and (10, 55) to estimate f 0 (5). This is known as right slope estimation. Similarly, we can estimate f 0 (5) by using a left slope estimation,i.e. f 0 (5) ≈
f (5) − f (0) = −6 5−0
An improved estimation consists of taking the average of the left slope and the right slope, that is, −3 − 6 f 0 (5) ≈ = −4.5. 2 Example 60 Use the average of the right slope estimation and left slope estimation is estimating the values of f 0 of the previous example. Solution. f 0 (0) 0
f (5) 0
≈ ≈
f (10)
≈
f 0 (15)
≈
0
f (20)
≈
f (5)−f (0) 5 ´ f (5)−f (0) 1 f (10)−f (5) + 2³ 5 5 ´ f (10)−f (5) 1 f (15)−f (10) + 2³ 5 5 ´ f (15)−f (10) 1 f (20)−f (15) + 2 5 5 f (20)−f (15) 5
³
=
−6
= −4.5 = −2.4 = −1.5 = −1.2
Recommended Problems (pp. 99 - 101): 1, 3, 4, 6, 8, 10, 11, 12, 13, 15, 18, 19, 24.
2.2
The derivative Function
Recall that a function f is differentiable at x if the following limit exists f (x + h) − f (x) h→0 h
f 0 (x) = lim
(2.1)
Thus, we associate with the function f , a new function f 0 whose domain is the set of points x at which the limit (??) exists. We call the function f 0 the derivative function of f. Since the derivative at a point represents the slope of the tangent line then one can obtain the graph of the derivative function from the graph of the original function. It is important to keep in mind the relationship between the graphs of f and f 0 . If f 0 (x) > 0 on an interval I then the tangent line must be tilted upward and the graph of f is rising or increasing over that interval. Similarly, if f 0 (x) < 0 on I then the tangent line is tilted downward and the graph of f is falling or decreasing over I. If f 0 (x) = 0 for all x in I then the tangent line is always horizontal and this occurs only when f is a constant function over I.
60
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
Example 61 Sketch the graph of the derivative of the function shown below Fig. 9 30 20 –2
10
2
x
4
6
–10 –20 –30 –40
Solution. Note that for x < −0.36 or x > 3.7 the derivative is positive whereas for −0.36 < x < 3.7 the derivative is negative. Also, f 0 (−0.3) = f 0 (3.7) ≈ 0. Thus, a possible graph of f 0 is given in Fig. 10 Fig. 10 50 40 30 20 10 0
–2
–10
2
x
4
6
Example 62 (Numerical Estimation of the Derivative) Find approximate values for f 0 (x) at each of the x-values given in the following table x f(x)
0 100
5 70
10 55
15 46
20 40
Solution. The derivative can be estimated by the average rate of change or the difference quotient f (a + h) − f (a) f 0 (a) ≈ . h Thus, f (5)−f (0) = −6 f 0 (0) ≈ 5 (5) f 0 (5) ≈ f (10)−f = −3 5 (10) = −1.8 f 0 (10) ≈ f (15)−f 5 (15) f 0 (15) ≈ f (20)−f = −1.2 5
2.2. THE DERIVATIVE FUNCTION
61
Note that in the above example, we used the points (5, 70) and (10, 55) to estimate f 0 (5). This is known as right slope estimation. Similarly, we can estimate f 0 (5) by using a left slope estimation,i.e. f 0 (5) ≈
f (5) − f (0) = −6 5−0
An improved estimation consists of taking the average of the left slope and the right slope, that is, −3 − 6 f 0 (5) ≈ = −4.5. 2 Exercise 27 Estimate f 0 (20) in the previous exercise. Solution. Since 20 is a right endpoint then we will use a left slope estimation to obtain f 0 (20) ≈
f (20) − f (15) = −1.2 20 − 15
Now, if a formula for f is given then by applying the definition of f 0 (x) as the limit of the difference quotient we can find a formula of f as shown in the following two examples. Example 63 (Derivative of a Constant Function) Suppose that f (x) = k for all x. Find a formula for f 0 (x). Solution. (x) f 0 (x) = limh→0 f (x+h)−f h = limh→0 k−k h =0
Thus, f 0 (x) = 0. Example 64 (Derivative of a Linear Function) Find the derivative of the linear function f (x) = mx + b. Solution. f 0 (x) = = =
(x) limh→0 f (x+h)−f h m(x+h)+b−(mx+b) limh→0 h limh→0 mh h =m
Thus, f 0 (x) = m. Recommended Problems (pp. 104 - 6): 5, 6, 7, 9, 11, 13, 15, 19, 21.
62
2.3
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
Leibniz Notation for The Derivative
When dealing with mathematical models that involve derivatives it is convenient dy to denote the prime notation of the derivative of a function y = f (x) by dx . That is, dy = f 0 (x) dx This notation is called Leibniz notation (due to W.G. Leibniz). For example, dy we can write dx = 2x for y 0 = 2x. When using Leibniz notation to denote the value of the derivative at a point a we will write ¯ dy ¯¯ dx ¯x=a Thus, to evaluate
dy dx
= 2x at x = 2 we would write ¯ dy ¯¯ = 2x|x=2 = 2(2) = 4. dx ¯x=2
dy appears as a fraction but it is not. It is just an alRemark. Even though dx ternative notation for the derivative. A concept called differential will provide meaning to symbols like dy and dx.
One of the advantages of Leibniz notation is the recognition of the units of the derivative. For example, if the position function s(t) is expressed in meters and the time t in seconds then the units of the velocity function ds dt is meters/sec. In general, the units of the derivative are the units of the dependent variable divided by the units of the independent variable. Example 65 The cost, C ( in dollars) to produce x gallons of ice cream can be expressed as C =¯ f (x). What are the units of measurements and the meaning of the statement dC ¯ dx x=200 = 1.4? Solution. dC dx is measured in dollars per gallon. The notation ¯ dC ¯¯ = 1.4 dx ¯x=200 means that if 200 gallons of ice cream have already been produced then the cost of producing the next gallon will be roughly 1.4 dollars. Exercise 28 The derivative of the velocity function v is called acceleration and is denoted by a. Suppose that v is measured in meters/seconds, what are the units of a?
2.4. THE SECOND DERIVATIVE
63
Solution. The units of a are meters/seconds/seconds = meters/seconds2 . Finally, one can use the derivative at a point to approximate values of the function at nearby points. For example, if we know the values of f (a) and f 0 (a) then for a nearby point b the value of f (b) is found by the formula f (b) ≈ f 0 (a)(b − a) + f (a). Exercise 29 Climbing health care costs have been a source of concern for some time. Use the data in the table below to estimate the average per capita expenditure in 1991 and 2010 assuming that the costs climb at the same rate since 1990. Year Per capita expenditure ($)
1970 349
1975 591
1980 1055
1985 1596
1990 2714
Solution. = $223.60 Between 1985 and 1990 the rate of increase in the costs is 2714−1596 5 per year. Since we are assuming that the costs continue to increase at the same rate then C(1991) ≈ C(1990) + C 0 (1990)(1991 − 1990) = 2714 + 223.60 = $2937.60 and C(2010) = 2714 + 223.60(10) = $7, 186. Recommended Problems (pp. 111 - 3): 1, 3, 6, 8, 9, 12, 13, 14, 17, 23.
2.4
The Second Derivative
Let f (x) be a differentiable function. If the limit f 0 (x + h) − f 0 (x) h→0 h lim
exists then we say that the function f 0 (x) is differentiable and we denote its derivative by f 00 (x) or using Leibniz notation µ ¶ d2 y d dy = . dx2 dx dx We call f 00 (x) the second derivative of f (x). Now, recall that if f 0 (x) > 0 ( resp. f 0 (x) < 0) over an interval I then the function f (x) is increasing (resp. decreasing on I). So if f 00 (x) > 0 on I then f 0 (x) is increasing on I. This occurs only when the graph of f is bending up
64
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
on the interval I. In this case, we say that f is concave up on I. Similarly, if f 00 (x) < 0 on I then f 0 (x) is decreasing and this happens when the graph of f is bending down. In this case, we say that f is concave down. Remark. Note that when a curve is concave up then the tangent lines lie below the curve whereas when it is concave down then the tangent lines lie above the curve. Example 66 Give the signs of f 0 and f 00 for the following function Fig. 11. 8 6 4 2 0
0.5
1
1.5 x
2
2.5
3
Solution. Since f is always increasing then f 0 is always positive. Since the graph is concave up then f 00 is always positive. Example 67 Find where the graph of f (x) = x3 +3x+1 is concave up and where it is concave down. Solution. Finding the first and second derivatives of f we obtain f 0 (x) = 3x2 + 3 and f 00 (x) = 6x. Thus, the graph of f is concave up for x > 0 and concave down for x < 0. As an application to the second derivative, we consider the motion of an object determined by the position function s(t). Recall that the velocity of the object is defined to be the first derivative of s(t), i.e. ds dt and the absolute value of v(t) is the speed. When the object speeds up we say that he/she accelerates and when the object slows down we say that he/she decelerates. We define the acceleration of an object as the derivative of the velocity function and consequently as the second derivative of the position function ds d2 s . a(t) = 2 = dt dt v(t) = s0 (t) =
2.5. MARGINAL COST AND REVENUE
65
Example 68 A particle is moving along a straight line. If its distance, s, to the right of a fixed point is given by Fig. 12, estimate: (a) When the particle is moving to the right and when it is moving to the left. (b) When the particle has positive acceleration and when it has negative acceleration. Fig. 12.
3 2.5 2 1.5 1 0.5 0
0.5
1
1.5 x
2
2.5
3
Solution. (a) When s is increasing then the particle moves to the right. This occurs when 0 < t < 23 and for t > 2. On the other hand, the particle moves to the left when s is decreasing. This happens when 23 < t < 2. (b) Positive acceleration occurs when the graph is concave up. This occurs when t > 43 . The particle has negative acceleration when the curve is concave down, i.e for t < 43 . Recommended Problems (pp. 115 - 7): 2, 3, 8, 9, 11, 12, 13, 15, 17, 18, 23.
2.5
Marginal Cost and Revenue
Marginal analysis is an area of economics concerned with estimating the effect on quantities such as cost, revenue, and profit when the level of production is changed by a unit amount. For example, if C(q) is the cost of producing q units of a certain commodity, then the marginal cost, MC(q), is the additional cost of producing one more unit and is given by the difference M C(q) = C(q + 1) − C(q). Using the estimation C 0 (q) ≈ we find that
C(q + 1) − C(q) = C(q + 1) − C(q) (q + 1) − q M C(q) ≈ C 0 (q)
and for this reason, we will compute the marginal cost by the derivative C 0 (q). Similarly, if R(q) is the revenue obtained from producing q units of a commodity,
66
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
then the marginal revenue, MR(q), is the additional revenue obtained from producing one more unit, and we compute M R(q) by the derivative R0 (q). Example 69 Let C(q) represent the cost, R(q) the revenue, and P (q) the total profit, in dollars, of producing q units. (a) If C 0 (50) = 75 and R0 (50) = 84, approximately how much profit is earned by the 51st item? (b) If C 0 (90) = 71 and R0 (90) = 68, approximately how much profit is earned by the 91st item? Solution. (a) P 0 (50) = R0 (50) − C 0 (50) = 84 − 75 = 9. (b) P 0 (90) = R0 (90) − C 0 (90) = 68 − 71 = −3. A loss by 3 dollars. We will now consider the question of maximizing profit. To solve this problem, we want to maximize the profit function P (q) = R(q) − C(q). We will see later that the profit function attains its maximum for the level of production q for which P 0 (q) = 0. That is, when M R(q) = M C(q). Example 70 A manufacturer estimates that when q units of a particular commodity are produced each month, the total cost (in dollars) will be C(q) =
1 2 q + 4q + 200 8
and all units are sold at a price p = 49 − q dollars per unit. Determine the price that corresponds to the maximum profit. Solution. The revenue function is given by R(q) = pq = q(49 − q) = 49q − q 2 . Thus, M R(q) = 49 − 2q and M C(q) = 41 q + 4. The profit is maximized when M C(q) = M R(q). That is, when 14 q + 4 = 49 − 2q. Solving this equation for q we find q = 20 units. In this case, the price is p(20) = 49 − 20 = $29. Exercise 30 If the revenue and cost functions are given by the figure below, sketch graphs of M C and M R.
2.5. MARGINAL COST AND REVENUE
67
C
R
100 Solution.
MC
MR
100 Recommended Problems (pp. 122 - 4): 1, 3, 5, 7, 8, 9, 13.
68
CHAPTER 2. THE DERIVATIVE OF A FUNCTION
Chapter 3
Rules of Differentiation In this chapter we derive differentiation formulas for common functions such as exponential, logarithmic, polynomial, and rational functions. Also, we discuss general rules such as the product, quotient, and chain rules which allow us to differentiate combinations of the above list of functions.
3.1
Derivative Formulas for Power and Polynomials
Finding the derivative function by using the limit of the difference quotient is sometimes difficult for functions with complicated expressions. Fortunately, there is an indirect way for computing derivatives that does not compute limits but instead uses formulas which we will derive in this section and in the coming sections. We first derive a couple of formulas of differentiation. Theorem 1 If f is differentiable and k is a constant then the new function kf (x) is differentiable with derivative given by [kf (x)]0 = kf 0 (x). Proof. (x) (x)) [kf 0 (x) = limh→0 kf (x+h)−kf = limh→0 k(f (x+h)−f h h f (x+h)−f (x) 0 = kf (x) = k limh→0 h
where we used the fact that a constant can be taking across the limit sign by the properties of limits. Theorem 2 If f (x) and g(x) are two differentiable functions then the functions f + g and 69
70
CHAPTER 3. RULES OF DIFFERENTIATION
f − g are also differentiable with derivatives [f (x) ± g(x)]0 = f 0 (x) ± g 0 (x) Proof. Again by using the definition of the derivative and the fact that the limit of a sum/difference is the sum/difference of limits we find [f (x) + g(x)]0
= = = limh→0
(f (x+h)+g(x+h))−(f (x)+g(x)) h (f (x+h)−f (x))+(g(x+h)−g(x)) h f (x+h)−f (x) + limh→0 g(x+h)−g(x) = f 0 (x) h h
limh→0 limh→0
+ g 0 (x)
The same proof is valid for the difference formula. Next, we state without proof a rule for finding the derivative of a power function of the form f (x) = xn . Theorem 3 (Power Rule) For any real number n, the derivative of the function y = xn is given by the formula dy = nxn−1 dx Example 71 Use the power rule to differentiate the following: 4
(a) y = x 3
(b) y =
1 √ 3 x
(c) y = xπ .
Solution. 1 (a) Using the power rule with n = 43 to obtain y 0 = 43 x 3 . 1 4 (b) Since y = x− 3 then using the power rule with n = − 13 to obtain y 0 = − 13 x− 3 . (c) Using the power rule with n = π to obtain y 0 = πxπ−1 . Combining the results discussed above, we can find the derivative of functions that are combinations of power functions of the form axn . In particular, the derivative of a polynomial function f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 is given by the formula f 0 (x) = nan xn−1 + (n − 1)an−1 xn−2 + · · · + a1 . Example 72 √ Find the derivative of the function y = 3x7 −
x5 5
+ π.
Solution. √ The derivative is f 0 (x) = 7 3x6 − x4 . Example 73 √ Find the second derivative of y = 5 3 x −
10 x4
+
1 √ . 2 x
3.2. DERIVATIVE FORMULAS FOR EXPONENTIAL AND LOGARITHMIC FUNCTIONS71 Solution. 1 1 Note that the given function can be written in the form y = 5x 3 −10x−4 + 12 x− 2 . Thus, the first derivative is y0 =
1 3 5 −2 x 3 + 40x−5 − x− 2 . 3 4
The second derivative is y 00 = −
10 − 5 3 5 x 3 − 200x−6 + x− 2 . 9 8
Recommended Problems (pp. 141 - 2): 1, 3, 6, 11, 18, 29, 32, 34, 36, 38, 41, 42.
3.2
Derivative Formulas for Exponential and Logarithmic Functions
We start this section by looking at the limit eh − 1 . h→0 h lim
The chart below suggests that the limit is 1. h eh −1 h
-0.01 0.995
-0.001 0.9995
-0.0001 0.99995
0 undefined
0.0001 1.0000
0.001 1.0005
0.01 1.005
Now, let’s try and find the derivative of the function f (x) = ex at any number x. By the definition of the derivative and the limit above we see that f 0 (x) = = = =
(x) limh→0 f (x+h)−f h ex+h −ex limh→0 h x h limh→0 e (eh −1) h ex limh→0 e h−1 = ex
This means that ex is its own derivative: d x (e ) = ex . dx Example 74 Find the derivative of f (x) = ex+2 . Solution. Rewriting f to get f (x) = e2 · ex . Thus, f 0 (x) = e2 (ex )0 = e2 · ex = ex+2 . • Derivative of eu
72
CHAPTER 3. RULES OF DIFFERENTIATION
Now, suppose that the x in ex is replaced by a differentiable function of x, say u(x). We would like to find the derivative of eu with respect to x. That is, what d is dx (eu )? d u dx (e )
= = = = = =
eu(x+h) −eu(x) h u(x+h) −eu(x) u(x+h)−u(x) · limh→0 eu(x+h)−u(x) h eu(x+h) −eu(x) limh→0 u(x+h)−u(x) limh→0 u(x+h)−u(x) h u(x)+w −eu(x) limw→0 e limh→0 u(x+h)−u(x) ww h eu limw→0 e w−1 limh→0 u(x+h)−u(x) h d eu dx (u)
limh→0
where w = u(x + h) − u(x). Note that since u is differentiable then it is continuous so that limh→0 (u(x + h) − u(x)) = 0. Also, we have used the limit discussed at the beginning of this section. • Derivative of ax We can now use the above discussion to find the derivative of the function f (x) = ax . First, note that f (x) = ax = ex ln a since in general aloga x = x. Thus, by the previous result we see that d x d x ln a d (a ) = (e ) = ex ln a (x ln a) = ln aex ln a = ax ln a. dx dx dx Example 75 Find the derivative of each of the following functions: (a) f (x) = 3x
(b) y = 2 · 3x + 5 · e3x−4 .
Solution. (a) f 0 (x) = 3x ln 3. (b) y 0 = 2(3x )0 + 5(e3x−4 )0 = 2 · 3x ln 3 + 5(3)e3x−5 = 2 · 3x ln 3 + 15 · e3x−4 . • Derivative of ln x We end this section by finding the derivative of f (x) = ln x. Since eln x = x then taking the derivative of both sides to obtain eln x (ln x)0 = 1 or (ln x)0 = eln1 x = x1 . Using Leibnitz notation we have d 1 (ln x) = dx x Example 76 Find the equation of the tangent line to the graph of f (x) = ln x at x = 2. Solution. Since f 0 (x) = x1 then the slope of the tangent line is f 0 (2) = 21 . Thus, y = x2 + b. Since the line crosses the point (2 ln 2) then ln 2 = 22 + b or b = ln 2 − 1. Hence,
3.3. DERIVATIVES OF COMPOSITE FUNCTIONS: THE CHAIN RULE73 y=
x 2
+ ln 2 − 1.
Recommended Problems (pp. 28, 31, 32.
3.3
145 - 6): 1, 5, 11, 15, 16, 21, 25,
Derivatives of Composite Functions: The Chain Rule
In this section we want to find the derivative of a composite function f (g(x)) where f (x) and g(x) are two differentiable functions: d dx [f (g(x))]
= = = = =
f (g(x+h))−f (g(x)) h (g(x)) g(x+h)−g(x) limh→0 f (g(x+h))−f · g(x+h)−g(x) h f (g(x+h))−f (g(x)) limh→0 g(x+h)−g(x) · limh→0 g(x+h)−g(x) h (g(x)) · limh→0 g(x+h)−g(x) limw→0 f (w+g(x))−f w h 0 0
limh→0
f (g(x))g (x)
where w = g(x + h) − g(x) and limh→0 (g(x + h) − g(x)) = 0 since g(x) is continuous(since it is differentiable). This result is known as the chain rule. Thus, the derivative of f (g(x)) is the derivative of f (x) evaluated at g(x) times the derivative of g(x). Example 77 Let’s find the derivative of y = (4x2 + 1)7 . First note that y = f (g(x)) where f (x) = x7 and g(x) = 4x2 + 1. Thus, f 0 (x) = 7x6 , f 0 (g(x)) = 7(4x2 + 1)6 and g 0 (x) = 8x. So according to the chain rule, y 0 = 7(4x2 +1)6 (8x) = 56x(4x2 +1)6 . As a result of this rule we can have the following differentiation formulas: d u du (e ) = eu , dx dx
d n du (u ) = nun−1 , dx dx
d u0 (ln u) = dx u
where u is a function of x. Exercise 31 2 Find the derivative of the function f (x) = (3x2 − 5)3 + ln (x10 + 1) − 5e−x . Solution. Using the chain rule we have 2
f 0 (x) = [(3x2 − 5)3 ]0 + [ln (x10 + 1)]0 − 5[e−x ]0 10 0 −x2 = 3(3x2 − 5)2 (3x2 − 5)0 + (xx10+1) (−x2 )0 +1 − 5e 10x9 2 −x2 = 9(3x − 5) + x10 +1 + 10xe
74
CHAPTER 3. RULES OF DIFFERENTIATION
Exercise 32 Suppose that $1,000 is deposited into a bank account that pays 8% annual interest, compounded continuously. Find a formula f (t) for the balance t years after the initial deposit. Find f (10) and f 0 (10) and explain what your answers mean in terms of money. Solution. The formula is given by
f (t) = 1, 000e0.08t .
The balance after 10 years is f (10) = 1, 000e0.08(10) ≈ $2, 225.54 Next, we compute f 0 (10). Since f 0 (t) = 1, 000(0.08)e0.08t = 80e0.08t then f 0 (10) = 80e0.8 ≈ $178.04 This means that after 10 years, the balance is growing at the rate of about $178 per year. Recommended Problems (pp. 149 - 150): 3, 15, 23, 29, 30, 32, 33, 35, 36, 38.
3.4
The Product and Quotient Rules
At this point we don’t have the tools to find the derivative of either the function 2 2 f (x) = x3 ex or the function g(x) = xex . Looking closely at the function f (x) 2 we notice that this function is the product of two functions, namely, x3 and ex . On the other hand, the function g(x) is the ratio of two functions. Thus, we hope to have a rule for differentiating a product of two functions and one for differentiating the ratio of two functions. We start by finding the derivative of the product u(x)v(x), where u and v are differentiable function: 0
(u(x)v(x))
= limh→0 u(x+h)v(x+h)−u(x)v(x) h = limh→0 u(x+h)(v(x+h)−v(x))+v(x)(u(x+h)−u(x)) h = limh→0 u(x + h) limh→0 v(x+h)−v(x) + v(x) limh→0 u(x+h)−u(x) h h 0 0 = u(x)v (x) + u (x)v(x)
Note that since u is differentiable so it is continuous and therefore lim u(x + h) = u(x).
h→0
The formula d d d (u(x)v(x)) = u(x) (v(x)) + (u(x))v(x) dx dx dx is called the product rule.
(3.1)
3.4. THE PRODUCT AND QUOTIENT RULES
75
Example 78 2 2 Let u(x) = x3 and v(x) = ex . Then u0 (x) = 3x2 and v 0 (x) = 2xex . Thus, by the product rule we have 2
2
2
2
f 0 (x) = x3 (2x)ex + 3x2 ex = 2x4 ex + 3x2 ex . The quotient rule is obtained from the product rule as follows: Let f (x) = u(x) 0 0 v(x) . Then u(x) = f (x)v(x). Using the product rule, we find u (x) = f (x)v (x)+ 0 0 f (x)v(x). Solving for f (x) to obtain u0 (x) − f (x)v 0 (x) . v(x)
f 0 (x) = Now replace f (x) by
u(x) v(x)
³
to obtain u(x) v(x)
´0
u(x)
= = =
Example 79 Consider the function g(x) = quotient rule we have
x2 ex .
f 0 (x)
u0 (x)− v(x) v 0 (x) v(x) u0 (x)v(x)−u(x)v 0 (x) v(x)
v(x) u0 (x)v(x)−u(x)v 0 (x) (v(x))2
Let u(x) = x2 and v(x) = ex . Then by the
= =
(x2 )0 ex −x2 (ex )0 (ex )2 2xex −x2 ex e2x
Recommended Problems (pp. 152 - 3): 3, 5, 12, 22, 25, 27, 33, 34, 35, 38.
76
CHAPTER 3. RULES OF DIFFERENTIATION
Chapter 4
Applications of the Derivative In this chapter, we will apply the notion of derivative in solving optimization problems. By that we mean, finding the maximum and the minimum (if they exist) of a given function.
4.1
Using First and Second Derivatives
We start this section by reviewing what the first and second derivatives of a function tell us about its graph: • • • •
If If If If
f 0 (x) > 0 on an open interval I then f (x) is increasing on I. f 0 (x) < 0 on an open interval I then f (x) is decreasing on I. f 00 (x) > 0 on an open interval I then f (x) is concave up on I. f 00 (x) < 0 on an open interval I then f (x) is concave down on I.
Exercise 33 Consider the function f (x) = x3 − 9x2 − 48x + 52. (a) Find the intervals where the function is increasing and decreasing. (b) Find the intervals where is the function is concave up and concave down. Solution. (a) Finding the first derivative we obtain f 0 (x) = 3x2 −18x−48 = 3(x−8)(x+2). Constructing the chart of signs below
77
78
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
(-)(-)
(-)(+) -2
(+)(+) 8
(+)
(-)
(+)
we see that f (x) is increasing on (−∞, −2) ∪ (8, ∞) and decreasing on (−2, 8). (b) Finding the second derivative, we obtain f 00 (x) = 6x − 18 = b(x − 3). So, f (x) is concave up on (3, ∞) and concave down on (−∞, 3). • Local Maxima and Minima Points of interest on the graph of a function are those points that are the highest on the curve, or the lowest, in a specific interval. Such points are called local extrema. The highest point, say f (a), is called a local maximum and satisfies f (x) ≤ f (a) for all x in an interval I. A local minimum is a point f (a) such that f (a) ≤ f (x) for all x in an interval I. Exercise 34 Find the local maxima and the local minima of the given function. 500 –10
10 x 20
30
–500 –1000
Solution. The local maxima occur at x = −2 and x = 14 whereas the local minimum occurs at x = 8. Next we will discuss two procedures for finding local extrema. We notice from the previous exercise that local extrema occur at points p where the derivative is either zero or undefined. We call (p, f (p)) a critical point and f (p) a critical value. Also, the graph in the previous exercise suggests the following test for finding local extrema: First-Derivative Test Suppose that a continuous function f has a critical point at p. • If f 0 changes sign from negative to positive at p, then f has a local minimum
4.1. USING FIRST AND SECOND DERIVATIVES
79
at p. • If f 0 changes sign from positive to negative at p, then f has a local maximum at p. Exercise 35 (a) Find the local extrema of the function f (x) = x3 − 9x2 − 48x + 52. (b) Find the local extrema of the function f (x) = sin x + ex , x ≥ 0. Solution. (a) Using the chart of signs of f 0 discussed in the first exercise of this section, we find that f (x) has a local maximum at x = −2 and a local minimum at x = 8. (b) Finding the derivative to obtain f 0 (x) = cos x + ex . But for x ≥ 0, 1 ≤ ex . Since −1 ≤ cos x ≤ 1 then adding the two inequalities we see that 0 ≤ cos x+ex . This implies that f 0 does not change sign for x ≥ 0. Therefore, there are no local maxima or minima. We have seen that local extrema are always critical points. The converse of this statement is not true in general. That is, there are critical points that are not local extrema of a function. An example of this situation is discussed in the next exercise. Exercise 36 Show that f (x) = x3 has a critical point at x = 0 but 0 is neither a local maximum nor a local minimum. Solution. Finding the derivative to obtain f 0 (x) = 3x2 . Setting this to 0 we find the critical point x = 0. Since f 0 (x) does not change sign at 0 then 0 is neither a local maximum nor a local minimum. Looking again closely to the graph of the second problem of this section we can use concavity in finding the local extrema: Second-Derivative Test Let f be a continuous function such that f 0 (p) = 0. • if f 00 (p) > 0 then f has a local minimum at p. • if f 00 (p) < 0 then f has a local maximum at p. • if f 00 (p) = 0 then the test fails. In this case, it is recommended that you use the first derivative test. Exercise 37 Use the second derivative test to find the local extrema of the function f (x) = x3 − 9x2 − 48x + 52. Solution. The second derivative of f (x) is given by f 00 (x) = 6(x − 3). The critical numbers are −2 and 8. Since f (−2) = −30 < 0 then x = −2 is a local maximum. Since f (8) = 30 > 0 then x = 8 is a local minimum.
80
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
Exercise 38 Find the local extrema of the function f (x) = x4 . Solution. Let’s try and find the local extrema by using the second derivative test. Since f 0 (x) = 4x3 then x = 0 is the only critcal number. Since f 00 (x) = 12x2 then f 00 (0) = 0. So the second derivative test is inconclusive. Now, using the first derivative test, we see that f 0 (x) changes sign from negative to positive at x = 0. so that x = 0 is a local minimum. Recommended Problems (pp. 170 - 1): 1, 3, 5, 7, 9, 12, 15, 17, 19, 21.
4.2
Concavity and Points of Inflection
We have seen that a local extremum is a point where the first derivative changes sign. In this section we will discuss points where the second derivative changes sign. That is, the points where the graph of the function changes concavity. We call such points points of inflection. How do you find the points of inflection? Well, since f 00 changes sign on the two sides of an inflection point then it makes sense to say that points of inflection occur at points where either the second derivative is 0 or undefined. Exercise 39 Find the point(s) of inflection of the function f (x) = xe−x . Solution. Using the product rule to obtain f 0 (x) = e−x − xe−x . Using the product rule for the seond time we find f 00 (x) = e−x (x − 2). Thus, a candidate for a point of inflection is x = 2. Since f 00 (x) > 0 for x > 2 and f 00 (x) < 0 for x < 2 then x = 2 is a point of inflection. Remark. We have seen that not every value of x where the derivative is zero or undefined is a local maximum or minimum. The same thing applies for points of inflection. That is, it is not always true that if the second derivative is 0 or undefined then automatically you have a point of inflection. It is critical that f 00 changes sign at such a point in order to have a point of inflection. Exercise 40 Consider the function f (x) = x4 . Show that f 00 (0) = 0 but 0 is not a point of inflection. Solution. The second derivative is given by the formula f 00 (x) = 12x2 . Clearly, f 00 (0) = 0. Since f 00 (x) ≥ 0, that is, f 00 (x) does not change sign then 0 is not a point of inflection.
4.2. CONCAVITY AND POINTS OF INFLECTION
81
Exercise 41 Graph the derivative of the function f (x) = x + sin x. Determine where f is increasing most rapidly, and least rapidly.
Solution. The derivative of f (x) is given by the expression f 0 (x) = cos x + 1 ≥ 0 so that f (x0 is always increasing. Now, f (x) increases most rapidly at the maximum values of f 0 (x) and increases least rapidly at the minimum values of f 0 (x). Graphing the function f 0 (x) we find Plot of 1+cos(x) 2
±3π ±2π −π
π
2π
3π
Thus, f increases most rapidly at x = 2nπ and least rapidly at x = (2n + 1)π where n is an integer.
Exercise 42 Graph a function with the following properties: • f has a critical point at x = 4 and an inflection point at x = 8. • f 0 < 0 for x < 4 and f 0 > 0 for x > 4. • f 00 > 0 for x < 8 and f 00 < 0 for x > 8.
Solution.
82
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
4
8
Recommended Problems (pp. 175 - 6): 1, 3, 7, 9, 15, 16, 26, 27, 28, 29.
4.3
Global Maxima and Minima
In this section we will look for the largest or the smallest values of a function on its domain. Such points are called global extrema. If f (a) is the largest value then it satisfies the inequality f (x) ≤ f (a) for all x in the domain of f. We call f (a) the global or absolute maximum value of f and the point (a, f (a)) the global maximum point. Similarly, if f (a) is the smallest value of f (x) then f (a) ≤ f (x) for all x in the domain of f. We call f (a) the absolute or global minimum value of f and the point (a, f (a)) the global minimum. The process of finding the global extrema is called optimization. Problems that involve finding the global extrema are called optimization problems. How do we find the global extrema? • If the function is continuous on a closed interval then the global extrema occur at either the critical points or the endpoints of the interval. Example 80 Find the global extrema of the function f (x) = x3 − 9x2 − 48x + 52 on the closed interval [−5, 12]. Solution. Finding the derivative of f (x) we get f 0 (x) = 3x2 −18x−48. Solving the equation f 0 (x) = 0 that is, x2 − 6x − 16 = 0 we find the critical points at x = 8 and
4.3. GLOBAL MAXIMA AND MINIMA x = −2. Now, evaluating the function find f (−5) f (−2) f (8) f (12)
83
at these points and at the endpoints we = = = =
−58 104 −396 −92
It follows that (−2, 104) is the global maximum point and (8, −396) is the global minimum point. • If a function is continuous on an open interval or on all real numbers then it is recommended to find the global extrema by graphing the function. Example 81 Find the global extrema of the function f (x) = 100(e−0.02x − e−0.1x ) for x ≥ 0. Solution. Let’s sketch the graph of this function. The standard process of graphing consists of the following steps: Step 1. Find the critical numbers. Setting f 0 (x) = 0 to obtain 100(−0.02e−0.02x + 0.1e−0.1x ) = 0 0.02e−0.02x = 0.1e−0.1x e−0.02x 0.1 = e−0.1x 0.02 0.08x e = 5 0.08x = ln 5 ln 5 x = 0.08
= 20.12
Step 2. We construct the following chart:
x f’(x) f(x)
+ %
20.12 0 53.50
&
Step 3. Find the second derivative to obtain f 00 (x) = 100(0.0004e−0.02x −0.01e−0.1x ). Setting this to zero and solving for x as in Step 2 we find x ≈ 40.25. Now we construct the table
x f”(x) f(x) Step 4.Graph
∩
40.25 0 f(40.25)
+ ∪
84
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE 50 40 30 20 10 0
20
40
x
60
80
100
Thus, from the graph we see that (20.12, 53.50) is a global maximum. The function has a global minimum at x = 0. Recommended Problems (pp. 21, .
4.4
179 - 182): 2, 3, 5, 9, 11, 13, 19,
Applications of Optimization to Marginality
Management of most businesses always aim to maximizing profit. In this section we will use the derivative to optimizie profit and revenue functions. • Optimizing Profit Recall that the profit resulting from producing and selling q items is defined by P (q) = R(q) − C(q) where C(q) is the total cost of producing a quantity q and R(q) is the total revenue from selling a quantity q of some good. To maximize or minimize profit over a closed interval, we optimize the profit function P. We know that global extrema occur at the critical numbers of P or at the endpoints of the interval. Thus, the process of optimization requires finding the critical numbers which are the zeros of the marginal profit function P 0 (q) = R0 (q) − C 0 (q) = 0 where R0 (q) is the marginal revenue function and C 0 (q) is the marginal cost function. Thus, the global maximum or the global minimum of P occurs when M R(q) = M C(q) or at the endpoints of the interval. Example 82 Find the quantity q which maximizes profit given the total revenue and cost
4.4. APPLICATIONS OF OPTIMIZATION TO MARGINALITY
85
functions R(q) = 5q − 0.003q 2 C(q) = 300 + 1.1q. where 0 ≤ q ≤ 800 units. What production level gives the minimum profit? Solution. The profit function is given by P (q) = R(q) − C(q) = −0.003q 2 + 3.9q − 300. The critical numbers of P are the solutions to the equation P 0 (q) = 0. That is, 3.9 − 0.006q = 0 or q = 650 units. Since P (0) = −$300, P (800) = $900 and P (650) = $967.50 then the maximum profit occurs when q = 800 units and the minimum profit(or loss) occurs when q = 0, i.e. when there is no production. Exercise 43 The total revenue and total cost curves for a product are given below
C(q) S
R(q)
q
q 1
q 3
q 2
q 4
(a) Sketch the curves for the marginal revenue and marginal cost on the same axes. Show on this graph the quantities where marginal revenue equals marginal cost. What is the significance of these two quantities? At which quantity is profit maximum? (b) Graph the profit function π(q). Solution. Since R is a straight line with positive slope then its derivative is a positive constant. That is, the graph of the margnal revenue is a horizontal line at some value a > 0. Since C is always increasing then its derivative M C is always positive. For 0 < q < q3 the curve is concave down so that M C is decreasing. For
86
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
q > q3 the graph of C is concave up and so M C is increasing. Thus, the graphs of C and R are shown below.
MC MR
q
1
q
q
3
2
According to the graph, marginal revenue equals marginal costs at the values q = q1 and q = q2 . So maximum profit occurs either at q1 , q2 or at the endpoints. Notice that the production levels q1 and q2 correspond to the two points where the tangent line to C is parallel to the tangent line to R. Now, for 0 < q < q1 we have M R < M C so that P 0 = M R − M C < 0 and this shows that P is decreasing. For q1 < q < q2 , M R > M C so that P 0 >) and hence P is increasing. So P changes from decreasing to increasing at q1 which means that P has a minimum at q1 . Now, for q > q2 we have that M R < M C so that P 0 < 0 and P is decreasing. Thus, P changes from increasing to decreasing at q2 so q2 is a local maximum for P. So maximum profit occurs either at the endpoint or at q2 . Since profit is negative for q < q1 and q > q2 then the profit is maximum for q = q2 . (b)
q -C(0)
• Optimizing Revenue Example 83
q
1
4
q
3
q
2
4.5. AVERAGE COST
87
The demand equation for a product is p = 45−0.01q. Write the revenue function as a function of q and find the quantity that maximizes revenue. What price corresponds to this quantity? What is the total revenue at this price? Solution. The revenue function is given by R(q) = pq = 45q − 0.01q 2 . This is a parabola that opens down so its vertex is the global maximum. The maximum then occurs at the critical number of R(q). That is, at the solution of R0 (q) = 0 or 45 − 0.02q = 0. Solving for q we find q = 2250 units. The maximum revenue is R(2250) = $50, 625. The unit price in this case is p = 45−0.01(2250) = $22.50 Recommended Problems (pp. 187 - 8): 1, 2, 3, 5, 7, 8, 9, 16, 17.
4.5
Average Cost
We have seen that an important principle in economics is the problem of maximizing profit. A second general principle involves the relationship between the marginal cost and the average cost a(q) =
C(q) . q
Example 84 The cost of producing q items is C(q) = 2500 + 12q dollars. (a) What is the marginal cost of producing the 100th item? (b) What is the average cost of producing 100 items? Solution. (a) The marginal cost is given by M C(q) = 12. This means that after producing the 99 items, it costs an additional $12 to produce the 100th item. 2500+12(100) (b) a(100) = C(100) = $37 per item. 100 = 100 Since a(q) = C(q) = C(q)−0 then a(q) is the slope of the line passing through q q−0 the points (q, C(q)) and the origin (0, 0). The important question in this section is the question of minimizing the average cost function a(q). Let’s try to find the derivative of a(q). Using the quotient rule of differentiation we obtain a0 (q) =
C 0 (q)q − C(q) C 0 (q) − a(q) = . q2 q
Thus, a0 (q) = 0 when C 0 (q) = a(q). So critical numbers of a(q) satisfy the relationship C 0 (q) = a(q). In economics theory the global minimum occurs at a critical number. Graphically, the minimum average cost occurs at the point on the graph of C(q) where the line passing through the origin is tangent to the graph of C(q). Thus, if q0 is a critical number of a then for q < q0 the
88
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
marginal cost is less than the average cost. This means, increasing production will decrease the average cost. If, on the other hand, q > q0 then the marginal cost is greater than the average cost. This means that increasing production will increase the average cost.
Example 85 A total cost function, in thousands of dollars, is given by C(q) = q 3 − 6q 2 + 15q, where q is in thousands and 0 ≤ q ≤ 5. (a) Graph C(q). Estimate the quantity at which average cost is minimized. (b) Graph the average cost function. Use it to estimate the minimum average cost. (c) Determine analytically the exact value of q at which average cost is minimized.
Solution. (a) A graph of C(q) is given below. The average cost is minimized at the point where the line going through the origin is tangent to the graph of C(q). This occur at approximately q = 3. 50 40 30 20 10 0
1
2
x
3
4
5
(b) The average cost function is given by a(q) = C(q) = q 2 − 6q + 15. The q graph of this function is given below. Notice that the minimum occurs at approximately q = 3.
4.6. ELASTICITY OF DEMAND
89
14 12 10 8 6 0
1
2
x
3
4
5
(c) The minimum average cost occurs when C 0 (q) = a(q). That is, 3q 2 − 12q + 15 = q 2 − 6q + 15. This gives 2q 2 − 6q = 0. Solving for q we find q = 0 or q = 3. Since the average cost is not defined when q = 0 then the average cost is minimum at q = 3. Recommended Problems (pp. 192 - 3): 1, 3, 5, 9, 11, 13.
4.6
Elasticity of Demand
An important quantity in economics theory is the price elasticity of demand which measures the responsiveness of demand to a given change in price and is found using the formula ¯ ¯ ¯ change in quantity demanded ¯ E = ¯ percentage ¯ percentage ¯change ¯ in price ¯ dq ¯ q ¯ ¯ dp = ¯ ¯ ¯ p ¯ ¯ p dq ¯ = ¯ q · dp ¯ Changing the price of an item by 1% causes a change of E% in the quantity sold. If E > 1 then this means that an increase (or decrease) of 1% in price causes demand to drop (increase) by more than one percent. In this case, we say that the demand is elastic. If 0 ≤ E < 1 then an increase (decrease) of 1% in price causes demand to drop (increase) by less than one percent and in this case we say that the demand is inelastic. Note that we are assuming that increasing the price usually decreases demand dp and decreasing the price will increase demand so that dq q and p have opposite sign, that is their ratio is always negative.
Example 86 Raising the price of hotel rooms from $75 to $80 per night reduces weekly sales from 100 rooms to 90 rooms.
90
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
(a) What is the elasticity of demand for rooms at a price of $75? (b) Should the owner raise the price? Solution. (a) The percent change in price is ∆q q
∆p 80−75 = 0.067 = 6.7% p = 75 90−100 = −0.1 = −10%. Thus, 100
and the percent
change in demand is = the elasticity of 0.1 demand is E = 0.067 = 1.5. (b) The weekly revenue per room at the price of $75 is 100 · 75 = $7500 whereas at the price of $80 the weekly revenue is 90 · 80 = $7200. A price increase results in loss of revenue, so the price should not be raised. Example 87 The demand for a product is q = 2000 − 5p where q is units sold at a price of p dollars. Find the elasticity if the price is $10, and interpret your answer in terms of demand. Solution. Using Leibniz Notation we find
¯
dq ¯ dp ¯
p=10
= −5 and for p = 10 the corresponding
quantity is q = 2000 − 50 = 1950 so that the elasticity is ¯ ¯ ¯ p dq ¯ 10 · 5 ¯= ¯ = 0.03. E=¯ q dp ¯ 1950 The demand is inelastic at the given price; a 1% increase in price will result in a decrease of 0.03% in demand. Finally, we would like to know the price that maximizes revenue. That is, the price that brings the greatest revenue. Recall that the revenue function is dq p dq given by R = pq so that dR dp = q + p dp = q(1 + q dp )
dq dq If E > 1 then pq dp < −1 so that 1 + pq dp −1 so that 1 + pq dp > 0 and consequently dR dp > 0. This means that increasing price will increase revenue. Finally, note that if dR dp = 0 then E = 1. That is, E = 1 at the critical points of the revenue function.
R E=1 E1 R(p) p
4.6. ELASTICITY OF DEMAND
91
Recommended Problems (pp. 196 - 8): 1, 2, 3, 7, 8, 10, 12, 13, 17, 18.
92
CHAPTER 4. APPLICATIONS OF THE DERIVATIVE
Chapter 5
The Definite Integral Integration is the reverse process of differentiation. That is, given a function f (x) we would like to find a function F (x) such that F 0 (x) = f (x). In particular, given f 0 (x) we want to find the original function f (x). In this chapter, we introduce the concept of definite integral. This concept is used in many applications such as finding the area of a region, distance traveled, etc.
5.1
Measuring The Distance Traveled
We have seen that the velocity of an object moving along the curve s(t) is obtained by taking the average rate of change on smaller and samller intervals , that is finding the derivative of s, i..e v(t) = s0 (t). In this and the following sections we want to go the opposite direction. That is, given the velocity function v(t) we want to find the position function s(t). That is we want to estimate the total change of the distance using the average rate of change. To be more precise, suppose that we want to estimate the distance s traveled by a car after 10 seconds of departure. Assume for example, that we are given the velocity of the car every two seconds as shown in the table below Time (sec) Velocity (ft/sec)
0 20
2 30
4 38
6 44
8 48
10 50
Since we don’t know the instantaneous velocity of the car then what we can do is to estimate the distance traveled. For the first two seconds, the velocity is at least 20 miles per second so that the distance traveled is at least 20 × 2 = 40 feet. Likewise, at least 30 × 2 = 60 feet has been traveled the next two seconds and so on. Thus, we obtain an underestimate to the exact distance traveled 20 × 2 + 30 × 2 + 38 × 2 + 44 × 2 + 48 × 2 = 360 f eet. However, we can reason differently and get an overestimate to the total distance traveled as follows: For the first two seconds the car’s velocity is at most 30 feet so that the car travels at most 30 × 2 = 60 feet. In the next two seconds, it 93
94
CHAPTER 5. THE DEFINITE INTEGRAL
travels 38 × 2 = 76 feet and so on. So an overestimate of the total distance traveled is 30 × 2 + 38 × 2 + 44 × 2 + 48 × 2 + 50 × 2 = 420 f eet Hence, 360 f eet ≤ T otal distance traveled ≤ 420 f eet. Notice that the difference between the upper and lower estimates is 60 feet. The graph below shows both the lower estimate and the upper estimate. The graph of the velocity is obtained by plotting the points given in the above table and then connect them with a smooth curve. The area of the lower rectangles represent the lower estimate and the larger rectangles represent the upper estimate.
50 30 20 0
2
4
6
8
10
To visualize the difference between the upper and lower estimates, look at the above figure, and imagine that all the unshaded rectangles are pushed to the right and stacked on top of each other. This gives a rectangle of width 2 and height 30 so its area is the difference between the estimates. Exercise 44 Suppose that the velocity of the car is given every second instead as shown in the table below. Find the lower and upper estimates of the total distance traveled. What is the difference between the lower and upper estimates? Do you think that knowing the velocity at every second is a better estimate than knowing the velocity every two seconds? Time (sec) Velocity (ft/sec)
0 20
1 26
2 30
3 35
4 38
5 42
6 44
7 46
8 48
9 49
10 50
Solution. The lower estimate is (20)(1)+(26)(1)+(30)(1)+(35)(1)+(38)(1)+(42)(1)+(44)(1)+(46)(1)+(48)(1)+(49)(1) = 378 f eet
5.1. MEASURING THE DISTANCE TRAVELED
95
and the upper estimate is (26)(1)+(30)(1)+(35)(1)+(38)(1)+(42)(1)+(44)(1)+(46)(1)+(48)(1)+(49)(1)+(50)(1) = 408 f eet Hence, 378 f eet ≤ T otal distance traveled ≤ 408 f eet. So the difference between the upper and lower estimates is 408 − 378 = 30 f eet. This shows that by increasing the parition points we get better and better estimate. Remark. Once the upper estimate and the lower estimate are found then one can get an even better estimate by taking the average of the two estimates. The use of the average rate of change of the distance leads to finding the total distance traveled. This same method can be used to find the total change from the rate of change of other quantities. Example 88 The following table gives world oil consumptions, in billions of barrels per year. Estimate the total oil consumption during this 20-year period. Year Oil (barrels/yr)
1980 22.3
1985 23.0
1990 23.9
1995 24.9
2000 27.0
Solution. We underestimate the total oil consumption as follows: 22.3 × 5 + 23.0 × 5 + 23.9 × 5 + 24.9 × 5 = 470.5 billion barrels The overestimate is 23.0 × 5 + 23.9 × 5 + 24.9 × 5 + 27.0 × 5 = 494 billion barrels A good single estimate of the total oil consumption is the average of the above estimates. That is T otal oil consumption ≈
470.5 + 494 = 482.25 billion barrels 2
• Finding the Exact Distance Traveled Suppose that we want to find the total distance traveled over the time interval a ≤ t ≤ b. We take measurements of the velocity v(t) at equally spaced times, a = t0 , t1 , t2 , · · · , tn = b. This means that we devide the interval [a, b] into n equal pieces each of length ∆t = b−a n . We first use the left-end point of each interval [ti−1 , ti ] and construct the left-hand sum L(v, n) = v(t0 )∆t + v(t1 )∆t + · · · + v(tn−1 )∆t.
96
CHAPTER 5. THE DEFINITE INTEGRAL
Geometrically, this sum represents the sum of areas of rectangles constructed by taking the height to be the value of the function at the left-endpoint of each subinterval. 4 3 2 1
0 1
1.2
1.4
x
1.6
1.8
2
Secondly, we use the right-end point of each interval [ti−1 , ti ] and construct the right-hand sum R(v, n) = v(t1 )∆t + v(t2 )∆t + · · · + v(tn )∆t. Geometrically, this sum represents the sum of areas of rectangles constructed by taking the height to be the value of the function at the right-endpoint of each subinterval. 4 3 2 1
0 1
1.2
1.4
x
1.6
1.8
2
Notice that for an increasing function the left-hand sum is an underestimate whereas the right-hand sum is an overestimate. This role is reversed for a decreasing function. Now, the exact distance traveled lies between the two estimates. As we have seen earlier, by making the time interval smaller and smaller we can make the difference between the two estimates as small as we like. This is equivalent to letting n → ∞. If the function v(t) is continuous then the following two limits
5.2. THE DEFINITE INTEGRAL
97
are equal to the exact distance traveled from t = a to t = b. T otal distance traveled = lim L(v, n) = lim R(v, n). n→∞
n→∞
Geometrically, each of the above limit represents the area under the graph of v(t) bounded by the lines t = a, t = b and the horizontal axis. 16 14 12 10 8 6 4 2 0 1
1.5
2
2.5 x
3
3.5
4
Recommended Problems (pp. 223 - 5): 1, 3, 7, 9, 10, 13, 18.
5.2
The Definite Integral
We discussed in the previous section how to find the exact distance traveled by taking the limit of both the left-hand sum and the right-hand sum as the number of subintervals increases without bound. In this section, we will construct these sums for any continuous function on a closed interval [a, b]. We start by dividing the interval [a,b] into n subintervals each of length ∆x =
b−a . n
Let a = x0 , x1 , · · · , xn−1 , xn = b be the endpoints of the subdivisions. We construct the left-hand sum or the left Riemann sum L(f, n) = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x =
n−1 X
f (xi )∆x
i=0
and the right-hand sum or the right Riemann sum R(f, n) = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x =
n X
f (xi )∆x
i=1
It is shown in advanced calculus that for a continuous function on a closed interval [a, b] that as n → ∞ both the left-hand sum and the right-hand sum
98
CHAPTER 5. THE DEFINITE INTEGRAL
Rb exist and are equal. We denote the common value by the notation a f (x)dx. Thus, Rb f (x)dx = limn→∞ L(f, n) a = limn→∞ R(f, n) Rb We call a f (x)dx the definite integral of f from x = a to x = b. We call a the lower limit and b the upper limit. The function f is called the integrand. Example 89 (a) On a sketch of y = ln x, represent the left Riemann sum with n = 2 approxR2 imating 1 ln xdx. Write out the terms in the sum, but do not evaluate. (b) On a another sketch of y = ln x, represent the right Riemann sum with n = 2 R2 approximating 1 ln xdx. Write out the terms in the sum, but do not evaluate. (c) Which sum is an underestimate? Which sum is an overestimate? Solution. (a) The left Riemann sum is the sum L(ln x, 2) = ln 1(0.5) + ln (1.5)(0.5) = 0.5 ln (1.5). The sum is represented by the rectangle shaded in the figure to the left
1
1.5
2
Left Riemann Sum
1
1.5
2
Right Riemann Sum
(b) The right Riemann sum is the sum R(ln x, 2) = ln (1.5)(0.5) + ln 2(0.5) = 0.5 ln (2)(1.5) = 0.5 ln 3. The sum is represented by the rectangles shaded in the figure to the right. R2 (c) L(ln x, 2) < 1 ln xdx < R(ln x, 2). In the next section, we will see that a definite integral represents an area under the graph of a function. Recommended Problems (pp. 231 - 2): 1, 3, 5, 7, 9, 11, 15, 19, 23.
5.3. THE DEFINITE INTEGRAL AS AREA
5.3
99
The Definite Integral as Area
In this section, we will see how definite integrals are used to find areas. Case 1:f (x) ≥ 0 Looking closely to either the left Riemann sum or the right Riemann sum we see that if f (x) ≥ 0 then a term of the form f (x)∆x represents the area of a rectangle. As n increases without bound, that is, the width ∆t of the rectangles approaches zero, the rectangles fit the curve of the graph more exactly, and the sum of their areas gets closer and closer to the area under the graph, bounded by the vertical lines x = a and x = b and the x-axis. Thus, Z
b
f (x)dx = Area under graph of f between a and b. a
Example 90 R1 √ Consider the integral −1 1 − x2 dx. (a) Interpret the integral as an area, and find its exact value. (b) Estimate the integral using a calculator. Solution. (a) Note that the equation of a circle centered at the origin and with radius √ 1 is √ given by x2 + y 2 = 1. Solving for y we find y = ± 1 − x2 . The function √ y = 1 − x2 corresponds to the upper semicircle and the function y = − 1 − x2 corresponds to the lower semicircle.
100
CHAPTER 5. THE DEFINITE INTEGRAL 1 0.8 0.6 0.4 0.2
–1
0
–0.6
0.20.40.60.8 1 x
It follows that the given integral represents the area of the upper semicircle and therefore is equal to π2 . That is, Z
p π 1 − x2 dx = . 2 −1 1
(b) Using a TI-83 calculator we find p f nInt( 1 − x2 , x, −1, 1) ≈ 1.571. Case 2:f (x) ≤ 0 In this case, since each product of the form f (x)∆x is less than or equal to zero then the area gets counted negatively. That is, the absolute value of the integral gives the area above the curve between x = a and x = b. Example 91 Find the area above the graph of y = x2 − 1 from x = −1 to x = 1. Solution. The graph of y = x2 − 1 is given by –1
–0.6 –0.2 0 –0.2 –0.4 y –0.6 –0.8 –1
x 0.20.40.60.8 1
5.4. INTERPRETATIONS OF THE DEFINITE INTEGRAL
101
R1 The area is given by | −1 (x2 − 1)dx| ≈ | − 1.33| = 1.33. Case 3:f changes sign In this case, the integral is the sum of the areas above the x-axis, counted positively, and the areas below the x-axis, counted negatively. If the integral is positive then the region above the x-axis has larger area than the region below the x-axis. If the integral is negative then the region below the x-axis has a larger area then the region above the x-axis. Example 92 Find the area between the graph of y = x3 and the x-axis from x = −1 to x = 1. Solution. The area is shown below 1 0.5
–1
–0.6 –0.2
0.20.40.60.8 1 x
–0.5 –1
It follows that the area is given by ¯Z 0 ¯ Z ¯ ¯ 3 ¯ x dx¯¯ + ¯ −1
1
x3 dx = 0.5
0
Recommended Problems (pp. 235 - 7): 1, 3, 5, 7, 11, 13, 17, 19, 22, 25. 27, 30, 31, 32.
5.4
Interpretations of the Definite Integral
We start this section by showing that the definite integral of a rate of change gives the total change of the function. We define the total change of a function F (t) from t = a to t = b to be the difference F (b) − F (a). Suppose that F (t) is continuous in [a,b] and differentiable in (a, b). Divide the interval [a, b] into n equal subintervals each of length ∆t = b−a n . Let a = t0 , t1 , · · · , tn = b be the partition points of the subdivision. Then on the interval [t0 , t1 ] the change in F can be estimated by the formula F 0 (t0 ) ≈
F (t0 + ∆t) − F (t0 ) ∆t
102 or that is
CHAPTER 5. THE DEFINITE INTEGRAL
F (t0 + ∆t) − F (t0 ) ≈ F 0 (t0 )∆t F (t1 ) − F (t0 ) ≈ F 0 (t0 )∆t
On the interval [t1 , t2 ] we get the estimation F (t2 ) − F (t1 ) = F 0 (t1 )∆t Continuing in this fashion we find that on the interval [tn−1 , tn ] we have F (tn−1 ) − F (tn ) ≈ F 0 (tn−1 )∆t. Adding all these approximations we find that F (tn ) − F (t0 ) ≈
n−1 X
F 0 (ti )∆t
i=0
Letting n → ∞ we see that Z
b
F (b) − F (a) =
F 0 (t)dt.
a
Example 93 The amount of waste a company produces, W, in metric tons per week, is approximated by W = 3.75e−0.008t , where t is in weeks since January 1, 2000. Waste removal for the company costs $15/ton. How much does the company pay for waste removal during the year 2000? Solution. The R 52 amount of tons produced during the year 2000 is just the definite integral W (t)dt. Using a calculator we find that 0 Z T otal waste during the year =
52
3.75e−0.008t dt ≈ 159 tons
0
The cost to remove this quantity is 159 × 15 = $2385. Rb Remark. When using a f (x)dx in applications then its units is the product of the units of f (x) with the units of x. Recommended Problems (pp. 240 - 3): 1, 3, 5, 8, 9, 15, 18.
5.5
The Fundamental Theorem of Calculus
The following result is considered among the most important result in calculus.
5.5. THE FUNDAMENTAL THEOREM OF CALCULUS
103
The Fundamental Theorem of Calculus If f (x) is a continuous function on [a,b] and F 0 (x) = f (x) then Z
b
f (x)dx = F (b) − F (a) a
We call the function F (x) an antiderivative of f (x). Proof. Partition the interval [a,b] into n subintervals each of length ∆x = b−a n and let {a = x0 , x1 , · · · , xn = b} be the partition points. Applying the Mean Value Theorem on the interval [x0 , x1 ] we can find a number x0 < c1 < x1 such that F (x1 ) − F (x0 ) = F 0 (c1 )∆x. Continuing this process on the remaining intervals we find = .. .
F 0 (c1 )∆x
F (xn ) − F (xn−1 ) =
F 0 (cn )∆x
F (x2 ) − F (x1 )
Adding these equalities we find n X
F (xn ) − F (x0 ) =
f (ci )∆x
i=1
Letting n → ∞ to obtain Z F (b) − F (a) =
b
f (x)dx a
Example 94 R2 Use FTC to compute 1 2xdx. Use a calculator to find the answer to the integral and compare. Solution. Since the derivative of x2 is 2x then F (x) = x2 . Thus, by the FTC we have Z 2 2xdx = F (2) − F (1) = 4 − 1 = 3. 1
Using a calculator we find
R2 1
2xdx = 3.
Example 95 Let F (t) represent a bacteria population which is 5 million at time t = 0. After t hours, the population is growing at an instantaneous rate of 2t million bacteria per hour. Estimate the total increase in the bacteria population during the first hour, and the population at t = 1.
104
CHAPTER 5. THE DEFINITE INTEGRAL
Solution. Since total change is the definite integral of F 0 (t) = 2t from t = 0 to t = 1 then Z 1 Change in population = F (1) − F (0) = 2t dt ≈ 1.44 million bacteria 0
Since F (0) = 5 then Z
1
F (1) = F (0) +
2t dt ≈ 5 + 1.44 = 6.44 million.
0
If C(q) is the total cost to produce a quantity q of a certain comodity then we can use the Fundamental Theorem of Calculus and compute the total cost of producing b units as follows Z b C(b) − C(0) = C 0 (q)dq 0
or
Z
b
C(b) = C(0) + We call the quantity
Rb 0
C 0 (q)dq
0
C 0 (q)dq the total variable cost.
Example 96 The marginal cost function for a company is given by C 0 (q) = q 2 − 16q + 70 dollars/unit, where q is the quantity produced. If C(0) = 500, find the total cost of producing 20 units. What is the fixed cost and what is the total variable cost for this quantity? Solution. We find C(20) as follows: Z C(20) = C(0) +
20
Z C 0 (q)dq = 500 +
0
20
(q 2 − 16q + 70)dq.
0
where C(0) = 500 is the fixed cost. Using a calculator we find total variable cost to be Z 20 (q 2 − 16q + 70)dq ≈ 866.7 0
Thus, the total cost of producing 20 units is C(20) ≈ 500 + 866.7 = 1366.7 Recommended Problems (pp. 245 - 6): 2, 3, 5, 7, 9, 11, 16, 17, 20, 25.
5.6. THE DEFINITE INTEGRAL AS AN AVERAGE
5.6
105
The Definite Integral as an Average
We know that the average of n given numbers is just the sum divided by n. What is the average in the continuous case. That is, what is the average of a continuous function on a closed interval [a, b]? Partition the interval into n equal subintervals each of length ∆t = b−a n and let a = t0 , t1 , t2 , · · · , tn be the division points. Then Average of f (t) on [a, b] ≈ But n =
b−a ∆t
f (t0 ) + f (t1 ) + · · · + f (tn−1 ) n
so that
Average of f (t) on [a, b] ≈ =
1 b−a (f (t0 )
+ f (t1 ) + · · · + f (tn−1 ))∆t Pn−1 1 i=0 f (ti )∆t b−a
Letting n → ∞ we see that 1 Average of f (t) on [a, b] = b−a
Z
b
f (x)dx. a
Example 97 A bar of metal is cooling from 1000◦ C to room temperature, 20◦ C. The temperature, H, of the bar t minutes after it starts cooling is given by H = 20 + 980e−0.1t . Find the average temperature over the first hour. Solution. The average temperature is given by Average temperature f or the f irst hour =
1 60
Z
60
(20 + 980e−0.1t )dt ≈ 183◦ C.
0
Example 98 Suppose that C(t) represent the daily cost of heating your house, in dollars per day, where t isRtime in days and t =R0 corresponds to January 1, 2002. Interpret 90 90 1 C(t)dt. the quantities 0 C(t)dt and 90−0 0 Solution. R 90 The integral 0 C(t)dt represents the total cost in dollars to heat your house for the first 90 days of 2002. The second expression represents the average cost per day to heat your house during the first 90 days of 2002. Remark. From the definition of the average value we can write Z b (average value of f ) × (b − a) = f (x)dx a
106
CHAPTER 5. THE DEFINITE INTEGRAL
Geometrically, this says that the area of the rectangle of dimensions (average value of f )× (b − a) is equal to the area under the graph of f (x) from x = a to x = b. Recommended Problems (pp. 258 - 9): 1, 4, 5, 9, 11, 14, 18.
Chapter 6
Antiderivatives and Indefinite Integrals In this chapter we discuss the procedure of computing definite integrals using the Fundamental Theorem of Calculus. This requires finding the so-called an antiderivative of the integrand. We will discuss how to construct antiderivatives analytically, numerically, and graphically.
6.1
Finding Antiderivatives Graphically and Numerically
In this section we study the process of graphing a function f (x) given the graph of its derivative f 0 (x). Recall first that if a function f (x) is given then the new function f 0 (x) is called the derivative of f (x). For example, if f (x) = x2 then f 0 (x) = 2x. This process is referred to as differentiation. Now, if instead f 0 (x) is known then the process of finding f (x) is called integration or antidifferentiation. In this case, we call f (x) an antiderivative of f 0 (x). In general, if f and F are two functions such that F 0 = f then we say that F (x) is an antiderivative of f (x). For example, x2 is an antiderivative of 2x since (x2 )0 = 2x. Note that, there are infinitely many antiderivatives of 2x, namely, x2 + C where C is a constant. We call x2 + C the general antiderivative ( or the indefinite integral) of x2 and we represent this symbolically by Z 2xdx = x2 + C. Example 99 The graph of f 0 (x) is given.
107
108
CHAPTER 6. ANTIDERIVATIVES AND INDEFINITE INTEGRALS Graph of f’(x) 2 1
–2
–1
1 x
2
–1 –2
Sketch a graph of the function f (x) satisfying f (0) = 1. Solution. Note that since f 0 (x) is always increasing then the graph of f (x) is concave up since f 00 (x) > 0. Since f 0 (x) < 0 for x < 0 then f (x) is decreasing there. Similarly, since f 0 (x) > 0 for x > 0 then f (x) is increasing there. Since f 0 (0) = 0 and f (x) is decreasing and then increasing we conclude that x = 0 is a minimum. A graph of f (x) is given below. Graph of f(x) 3 2 y 1 –2
–1
0 –1
1 x
2
1 x
2
–2 –3
Example 100 2 The graph of f 0 (x) = e−x is given below. Graph of f’(x) 2 1.8 1.6 1.4 1.2 y 1 0.8 0.6 0.4 0.2 –2
–1
0
Sketch the graph of f (x) satisfying f (0) = 0. Solution.
6.1. FINDING ANTIDERIVATIVES GRAPHICALLY AND NUMERICALLY109 Since f 0 (x) is always positive then the graph of f (x) is always increasing. Now, for x < 0, f 0 (x) is increasing so that f 00 (x) > 0 and therefore f (x) is concave up. For x > 0 the function f 0 (x) is decreasing and so f 00 (x) < 0. That is, f (x) is concave down there. Thus, x = 0 is a point of inflection. Finally, since limx→±∞ f 0 (x) = 0 then the graph of f (x) levels off at both ends (See picture below) Graph of f(x) 4 y
–3
–2
2
–1 0
1 x 2
3
–2 –4
Next, we will estimate numerically the value of an antiderivative at a given point. For this purpose we remind the reader of the Fundamental Theorem of Rb Calculus (abbreviated by FTC): If F 0 (x) = f (x) then a f (x)dx = F (b) − F (a). In particular, we have Z
b
f 0 (x)dx = f (b) − f (a).
a
Rb Now, recall that for f (x) ≥ 0 the definite integral a f (x)dx represents the area under the graph of f (x) between the lines x = a and x = b. If the region is Rb below the x-axis then a f (x)dx is the negative of the area of that region. Example 101 Given below is the graph of f 0 (x). Suppose that f (−1) = −2. Find f (0), f (1), and f (3). 2
Graph of f’(x)
1.5 1 0.5
–1
0
1
x
2
3
110
CHAPTER 6. ANTIDERIVATIVES AND INDEFINITE INTEGRALS
Solution. By the FTC we have Z
b
f (b) = f (−1) +
f 0 (x)dx.
−1
Thus,
R0 f (0) = f (−1) + −1 f 0 (x)dx = −2 + 12 (1 · 2) = −1 R1 f (1) = f (0) + 0 f 0 (x)dx = −1 + 1 · 2 = 1 R3 0 f (3) = f (1) + 1 f (x)dx = 1 + 12 (2 · 2) = 3 Rt where we compute a f 0 (x)dx by determining the area between f 0 and the horizontal axis for a ≤ x ≤ t. Recommended Problems (pp. 294 - 6): 1, 3, 5, 9, 11, 17, 21, 22, 23, 24, 25.
6.2
Analytical Construction of Antiderivatives
In this section we will find analytical expressions of antiderivatives. As discussed in the previous section, there are infinitely many antiderivatives of a given function f (x). They all differ by a constant and the family of antiderivatives is represented by F (x) + C. The notation of the general antiderivative is called an indefinite integral and is written Z f (x)dx = F (x) + C. R The symbol is the symbol of integration, f (x) is called the integrand and C is called the constant of integration. Keep in mind the relationship between f (x) and F (x) which is given by F 0 (x) = f (x). Warning: The indefinite integral is a short-hand notation for a family of functions F (x) + C with the property F 0 (x) = f (x) for all x. It is not to be confused Rb with the definite integral a f (x)dx which is a real number. Example R102 Show that 0dx = C. Solution. Since the derivative of a constant function is always zero then Z 0dx = C.
6.2. ANALYTICAL CONSTRUCTION OF ANTIDERIVATIVES
111
Example R103 Show that kdx = kx + C where k is a constant. Solution. Since the derivative of kx is just k then Z kdx = kx + C. Example 104 R Show that xn dx =
xn+1 n+1
+ C for n 6= −1.
Solution. By the power rule, if F (x) =
xn+1 n+1
then F 0 (x) = xn . Thus,
Z xn dx =
xn+1 + C. n+1
Note that this formula is valid only if n 6= −1 for if n = −1 we would have which doesn’t make sense. The case n = −1 is treated in the next example. Example 105 Show that
Z
x0 0
dx = ln |x| + C. x
Solution. Suppose first that x > 0 so that ln |x| = ln x. Then (ln |x|)0 = (ln x)0 = x1 . Now, if −1 x < 0 then ln |x| = ln (−x) and by the chain rule (ln |x|)0 = (ln (−x))0 = −x = x1 . 1 0 Thus, in both cases (ln |x|) = x . Example 106 R Show that for a 6= 0, eax dx =
eax a
+ C.
Solution. ax If a is a nonzero constant and F (x) = ea then F 0 (x) = eax . This shows that Z eax eax dx = + C. a • Properties of Indefinite Integrals Z Z Z [f (x) ± g(x)]dx = f (x)dx ± g(x)dx. To see why this property is true, let F (x) be an antiderivative of f (x) and G(x) d [F (x) ± be an antiderivative of g(x). The result follows from the fact that dx G(x)] = f (x) ± g(x). Z Z cf (x)dx = c
f (x)dx.
112
CHAPTER 6. ANTIDERIVATIVES AND INDEFINITE INTEGRALS
R To see this, suppose that F (x) is an antiderivative of f (x). Then f (x)dx = d F (x) + C. R But dx (cF (x)) = cf0(x) so that cF (x) is an antiderivative of cf (x), that is, cf (x)dx = cF (x) + C . This implies Z Z Z Z 0 0 0 cf (x)dx = cF (x)+C = c( f (x)dx−C)+C = c f (x)dx−cC+C = c f (x)dx. 0 Note that the R constant −cC + C is ignored since a constant of integration will result from f (x)dx.
Once we have found an antiderivative of f (x), computing definite integrals is easy by the Fundamental Theorem of Calculus. ExampleR 107 2 Compute 1 3x2 dx. Solution. Since F (x) = x3 is an antiderivative of f (x) = 3x2 , then by FTC Z
2 1
3x2 dx = x3 |21 = 23 − 13 = 7.
Recommended Problems: pp. 281 - 2: 6, 8, 10, 13, 15, 22, 24, 32, 36, 38, 40, 41, 48. pp. 289 - 90: 1, 3, 5, 11, 13, 15, 17, 24, 25, 27, 28.
