Solution Midterm Test
University of Toronto Scarborough Department of Computer & Mathematical Sciences MATA30: Calculus I - Solution Midterm Test 1. Let f (x) = 2e−x + 3. (a) [ 2 marks] Find the domain and range of f . solution : Since e−x is defined for all x ∈ R, then the domain of f (x) = 2e−x + 3 is R. 0 < ex < ∞ thus 0 < 1/ex = e−x < ∞ and 3 < 2e−x + 3 < ∞. Thus the range of f is the interval (3, ∞).
(b) [ 1 mark] Show that f is invertible. The graph of f is the graph of y = ex flipped about the y−axis, stretched vertically by a factor of 2 and then shifted upward by 3 units. Thus the graph of f is decreasing in its domain. Thus f is a one to one function and so f is invertible 10 9 8 7 6 5 4 3 2 1
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(c) [ 3 marks] Find f −1 , the inverse of f , and give its domain and range. solution : y = 2e−x + 3 y − 3 = 2e−x (y − 3) = e−x 2 ! " y−3 ln = −x 2 − ln
!
y−3 2
"
= x
thus the inverse of f is ! " x−3 −1 f (x) = − ln = −[ln (x − 3) − ln(2)] = ln(2) − ln (x − 3) 2 The domain of f −1 is dom(f −1) = range(f ) = (3, ∞). The range of f −1 is range(f −1 ) = dom(f ) = R. (d) [ 4 marks] Give the graphs of f and f −1 . 9
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2. (a) [ 3 marks] Find all x satisfying: 2 log4 (x + 1) = log4 (x + 7). 2
solution : 2 log4 (x + 1) = log4 (x + 7) ⇐⇒ 4log4 (x+1) = 4log4 (x+7) thus we have: (x + 1)2 = (x + 7) or x2 + 2x + 1 = x + 7. Simplify: x2 + x − 6 = (x − 2)(x + 3) = 0. Thus x = 2 or x = −3 However, log4 (x + 1) is not defined if x = −3, therefore the only solution is x = 2.
(b) [ 3 marks] Prove the identity: solution : LHS = = = = =
(sin(x) + cos(x))2 = 1 + sin(2x).
(sin(x) + cos(x))2 sin2 x + 2 sin x cos x + cos2 x 1 + 2 sin x cos x 1 + sin(2x) RHS
therefore (sin(x) + cos(x))2 = 1 + sin(2x).. # #√ $$ x2 +1 (c) [ 4 marks] Simplify the following: sec arctan 2 solution : & %√ √ # π π$ x2 + 1 x2 + 1 , then = tan θ where θ ∈ − , . Let θ = arctan 2 2 2 2 √ x2 + 1 Using the identity sec2 θ = 1 + tan2 θ and tan θ = we get: 2 #√ $2 x2 + 5 x2 + 1 x2 +1 = sec2 θ = 1 + tan2 θ = 1 + = 1 + 2 4 4 √ 2 x +5 Finally sec θ = . 2 & %√ x2 + 1 then since θ = arctan 2 %√ && x2 + 1 sec arctan ] = sec θ 2 %
hyp = = adj
√
x2 + 5 2
2
3. [5 marks each; 20 marks total] Evaluate each of the following limits, else explain why the limit does not exist. Justify your answer. Do not use L’Hˆospital’s rule. (a) lim (tan(ax) csc(bx)), where a '= 0, b '= 0. x→0
sin ax x→0 cos ax sin bx ! " ! " # $ ! " sin ax bx ax 1 = lim · · · x→0 ax sin bx bx cos ax
lim (tan(ax) csc(bx)) = lim
x→0
= 1·1·
(b)
e2x − 1 x→0 ex − 1 lim
e2x − 1 (ex − 1)(ex + 1) = lim x→0 ex − 1 x→0 ex − 1 lim
ex + 1 x→0 1
= lim = 2
a a 1 · = b 1 b
3
(c)
|1 − x| x→1 x2 − x ' x if x ≥ 0 |x| = −x if x < 0 lim
1 − x if x ≤ 1 −1 + x if x > 1
lim−
|1 − x| 1−x −(x − 1) −1 = lim− = lim− = lim− = −1 2 x→1 x − x x→1 x(x − 1) x→1 x(x − 1) x
lim+
|1 − x| x−1 1 = lim = lim =1 x2 − x x→1+ x(x − 1) x→1+ x
lim−
|1 − x| |1 − x| |1 − x| '= lim+ 2 therefore lim 2 does not exist. 2 x→1 x − x x − x x→1 x − x
x→1
x→1
x→1
(d)
|1 − x| =
'
# 1 $ lim (x + 1)2 arctan e x+1
x→−1
Since −
π π < arctan x < for all x, then: 2 2 # 1 $ π π − < arctan e x+1 < for all x '= −1 2 2
(∗)
Since (x + 1)2 > 0 for all x '= −1, then, multiplying all sides of (∗) with (x + 1)2 gives: −
# 1 $ π(x + 1)2 π(x + 1)2 < (x + 1)2 arctan e x+1 < for all x '= −1 2 2
Since lim − x→−1
theorem:
π(x + 1)2 = 0, therefore by the Squeeze x→−1 2 # 1 $ lim (x + 1)2 arctan e x+1 = 0
π(x + 1)2 = 0 and 2
lim
x→−1
——————————————————————————— Alternatively (A) lim (x + 1)2 = 0 x→−1
And
# 1 $ lim − arctan e x+1 = 0
x→−1
(x + 1 < 1, x + 1 → 0, so Thus
1 1 → −∞ and e x+1 → 0 ) x+1 # 1 $ 2 lim − (x + 1) arctan e x+1 = 0
x→−1
And (B) A
# 1 $ π lim + arctan e x+1 = x→−1 2 1 1 (x + 1 > 1, x + 1 → 0, so → ∞ and e x+1 → ∞ ) x+1 Thus # 1 $ π lim + (x + 1)2 arctan e x+1 = 0 · = 0 x→−1 2
From (A) and (B), Thus
# 1 $ # 1 $ lim − (x + 1)2 arctan e x+1 = lim + (x + 1)2 arctan e x+1 = 0
x→−1
x→−1
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4. [10 marks] Let f (x) =
√
x2 − 1 + 1 . x−1
(a) [2 marks] Give the domain of f . √ solution : x ∈ dom(f ) is x2 − 1 is defined and if x − 1 '= 0. So x '= 1 and x2 − 1 ≥ 0 2 x '= 1 and x ≥ 1 x '= 1 and |x| ≥ 1 x '= 1 and x ≤ −1 or x ≥ 1 Thus x ∈ (−∞, −1] ∪ (1, ∞)
(b) [8 marks] Find all the horizontal and vertical asymptotes of f . solution : For vertical asymptotes of f : x − 1 = 0 if x = 1. So we have a possible vertical asymptote at 1 Since f is not defined for x ∈ (−1, 1] we only consider the right limit or f as at 1: √ √ x2 − 1 + 1 lim+ = ∞ (since x − 1 → 0 but x − 1 > 0 and x2 − 1 + 1 → 1.) x→1 x−1 Thus x = 1 is the vertical asymptote of f . For horizontal asymptotes: ( √ √ x2 −1 x2 −1+1 2 + x −1+1 x2 x lim = lim = lim x−1 x→∞ x→∞ x→∞ x−1 1 − x1 x √ (x = x2 since x > 0) √
√ x2 −1+1 x lim x−1 x→−∞ x
x2 − 1 + 1 lim = x→−∞ x−1 √ (x = − x2 since x < 0)
= lim
x→−∞
1 x
x→∞
( − 1− 1
= lim
1 x2 − x1
+
1 x
Thus y = 1 and y = −1 are the horizontal asymptotes of f .
(
1 + x1 x2 − x1
1− 1
= −1
=1
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5. [10 marks] Given some numbers a, b and c, define f : R → R by 3−x if x ≤ a , √ x−1 if a < x < 5 , f (x) = x − b esin(πx) if 5 ≤ x ≤ 6 , cx2 if x > 6 , Find a, b and c so that f is continuous on R.
solution : 3 − x is continuous everywhere, thus continuous on (−∞, a) for all a. √
x − 1 is continuous on [1.∞) and so continuous on (a, 5) if a ≥ 1
x − b esin(πx)) is continuous everywhere since both x and esin(πx) are, thus it is also continuous on (5, 6). cx2 is a polynomial therefore it is continuous everywhere, thus continuous on (6, ∞). -For a function to be continuous at ”a” we need: lim f (x) = lim− f (x) = f (a) = a − 3
x→a+
x→a
lim f (x) = lim− 3 − x = 3 − a = f (a),
x→a−
x→a
lim+ f (x) = lim+
x→a
x→a
√
x−1=
√
a − 1, thus we need 3 − a =
√
a − 1 or (3 − a)2 = a − 1
a2 − 6x + 9 = a − 1 ⇐⇒ a2 − 7x + 10 = 0 ⇐⇒ (a − 5)(a − 2) = 0 ⇐⇒ a = 5 of a = 2. However, since a < 5 (second condition of definition of f ), then a = 2 therefore a=2 -For a function to be continuous at 5 we need: lim f (x) = lim− f (x) = f (5) = 5 − b esin(π5) = 5 − be0 = 5 − b
x→5+
x→5
lim f (x) = lim+ x − b esin(πx) = 5 − b esin(π5) = 5 − b, x→5 √ √ lim− f (x) = lim− x − 1 = 5 − 1 = 2, thus we need 5 − b = 2 or b = 3 therefore
x→5+ x→5
x→5
b=3
6
For a function to be continuous at 6 we need: lim f (x) = lim− f (x) = f (6) = 6 − b esin(π6) = 6 − b = 6 − 3 = 3
x→6+
x→6
lim f (x) = lim− x − b esin(πx) = 6 − b = f (6),
x→6−
x→6
lim+ f (x) = lim+ cx2 = 36c, thus we need 3 = 36c or
x→6
x→6
c=1/12 Since b = 3, then a = 2 and c =
f (x) =
is continuous on R.
1 and 12 3−x √ x−1
if x ≤ 2 , if 2 < x < 5 ,
x − 3 esin(πx) if 5 ≤ x ≤ 6 , x2 if x > 6 , 12
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6. [5 marks] Fully justify that arctan x = 10x − ex has a positive root. solution : Since arctan x, 10x and ex are all continuous functions on R, then their linear combination f (x) = arctan x − 10x + ex is also a continuous function on R. π Since f (0) = arctan (0)−10(0)+e0 = 1 and f (1) = arctan (1)−10+e1 = −10+e < 0, 4 or f (1) < 0 < f (0) then by the IVT there is some c ∈ (0, 1) such that f (c) = 0 or arctan x = 10x − ex has a positive root.
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7. (a) [5 marks] Let f (x) = solution : (a)
√
x − 2. Use the definition of the derivative to find f % (x).
f (x + h) − f (x) = lim lim h→0 h→0 h = lim
√ √
h→0
x+h−2− h x+h−2− h
√ √
x−2 x−2
√ √ x+h−2+ x−2 √ ·√ x+h−2+ x−2
x + h − 2 − (x − 2) √ √ h→0 h( x + h − 2 + x − 2)
= lim
h = lim √ √ h→0 ( x + h − 2 + x − 2)h = lim √ h→0
=
1 √ x+h−2+ x−2
1 2 x−2 √
(b) [7 marks] There are many tangent lines to the graph of f . Find the equation(s) of the tangent line(s) that pass through the point (5, 2). (Note that the point (5, 2) is NOT on the graph of f .) solution : Suppose the tangent line touches the graph of f at (a, f (a)) = √ 1 (a, a − 2). Then the slope of the tangent at that point is m = f % (a) = √ . 2 a−2 Since (5, 2) is also a point on the line , then:
which gives
√ 2− a−2 1 = m= √ 5−a 2 a−2 5−a (5 − a) a+1 (a + 1)2 2 a + 2a + 1 2 a − 14a + 33 (a − 11)(a − 3)
= = = = = = =
√ √ (2√− a − 2)2 a − 2 4√a − 2 − 2(a − 2) 4 a−2 16(a − 2) 16a − 32 0 0
So a = 11, a = 3. 1 y−2 1 1 (A) If a = 11, then m = √ giving the equation of = √ = = 6 x−5 2 11 − 2 2 9 the tangent: y = x/6 + 7/6 1 y−2 1 (B) If a = 3, then m = √ giving the equation of the tangent: = = 2 x−5 2 3−2 y = x/2 − 1/2
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8. [3 marks] Consider the function f : R → R whose graph is given below.
Let A be the set of points {x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 , x10 , x11 } ∈ R.
i. Find all points in A where f is differentiable. Your answer: x2 , x5 , x7 , x8 ii. Find all points in A where f is continuous but not differentiable. Your answer: x4 , x9 iii. Find all points in A where f % is zero. Your answer: x2 , x5 , x8
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9. For the following given functions use differentiation rules to find their required derivatives. Do not simplify. (2 + x)2 (a) [5 marks] f (x) = . Find f % (x) ln(2 + x) solution : 2(2 + x) ln(2 + x) − (2 + x)2 · % f (x) = (ln(2 + x))2 =
1 2+x
= 2(2 + x) ln(2 + x) − (2 + x) (ln(2 + x))2
(b) [5 marks] g(x) = sin(tan(sec(x2 ))). Find g %(x). g % (x) = cos(tan(sec(x2 ))) · sec2 (sec(x2 )) · sec(x2 ) tan(x2 ) · 2x
11
x2
(c) [5 marks] h(x) = 58 solution :
where a is a positive constant real number. Find h% (x).
x2
2
h% (x) = (ln 5) · (58 ) · (ln 8)(8x ) · 2x dy (d) [5 marks] If y cos x = 2x2 + 4y 2, find . dx solution : y cos x % y cos x − y sin x y % cos x − 8yy % y % (cos x − 8y) y%
2x2 + 4y 2 4x + 8yy % 4x + y sin x 4x + y sin x 4x + y sin x = cos x − 8y
= = = =
(b) [ 1 mark] Show that f is invertible. The graph of f is the graph of y = ex flipped about the y−axis, stretched vertically by a factor of 2 and then shifted upward by 3 units. Thus the graph of f is decreasing in its domain. Thus f is a one to one function and so f is invertible 10 9 8 7 6 5 4 3 2 1
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(c) [ 3 marks] Find f −1 , the inverse of f , and give its domain and range. solution : y = 2e−x + 3 y − 3 = 2e−x (y − 3) = e−x 2 ! " y−3 ln = −x 2 − ln
!
y−3 2
"
= x
thus the inverse of f is ! " x−3 −1 f (x) = − ln = −[ln (x − 3) − ln(2)] = ln(2) − ln (x − 3) 2 The domain of f −1 is dom(f −1) = range(f ) = (3, ∞). The range of f −1 is range(f −1 ) = dom(f ) = R. (d) [ 4 marks] Give the graphs of f and f −1 . 9
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2. (a) [ 3 marks] Find all x satisfying: 2 log4 (x + 1) = log4 (x + 7). 2
solution : 2 log4 (x + 1) = log4 (x + 7) ⇐⇒ 4log4 (x+1) = 4log4 (x+7) thus we have: (x + 1)2 = (x + 7) or x2 + 2x + 1 = x + 7. Simplify: x2 + x − 6 = (x − 2)(x + 3) = 0. Thus x = 2 or x = −3 However, log4 (x + 1) is not defined if x = −3, therefore the only solution is x = 2.
(b) [ 3 marks] Prove the identity: solution : LHS = = = = =
(sin(x) + cos(x))2 = 1 + sin(2x).
(sin(x) + cos(x))2 sin2 x + 2 sin x cos x + cos2 x 1 + 2 sin x cos x 1 + sin(2x) RHS
therefore (sin(x) + cos(x))2 = 1 + sin(2x).. # #√ $$ x2 +1 (c) [ 4 marks] Simplify the following: sec arctan 2 solution : & %√ √ # π π$ x2 + 1 x2 + 1 , then = tan θ where θ ∈ − , . Let θ = arctan 2 2 2 2 √ x2 + 1 Using the identity sec2 θ = 1 + tan2 θ and tan θ = we get: 2 #√ $2 x2 + 5 x2 + 1 x2 +1 = sec2 θ = 1 + tan2 θ = 1 + = 1 + 2 4 4 √ 2 x +5 Finally sec θ = . 2 & %√ x2 + 1 then since θ = arctan 2 %√ && x2 + 1 sec arctan ] = sec θ 2 %
hyp = = adj
√
x2 + 5 2
2
3. [5 marks each; 20 marks total] Evaluate each of the following limits, else explain why the limit does not exist. Justify your answer. Do not use L’Hˆospital’s rule. (a) lim (tan(ax) csc(bx)), where a '= 0, b '= 0. x→0
sin ax x→0 cos ax sin bx ! " ! " # $ ! " sin ax bx ax 1 = lim · · · x→0 ax sin bx bx cos ax
lim (tan(ax) csc(bx)) = lim
x→0
= 1·1·
(b)
e2x − 1 x→0 ex − 1 lim
e2x − 1 (ex − 1)(ex + 1) = lim x→0 ex − 1 x→0 ex − 1 lim
ex + 1 x→0 1
= lim = 2
a a 1 · = b 1 b
3
(c)
|1 − x| x→1 x2 − x ' x if x ≥ 0 |x| = −x if x < 0 lim
1 − x if x ≤ 1 −1 + x if x > 1
lim−
|1 − x| 1−x −(x − 1) −1 = lim− = lim− = lim− = −1 2 x→1 x − x x→1 x(x − 1) x→1 x(x − 1) x
lim+
|1 − x| x−1 1 = lim = lim =1 x2 − x x→1+ x(x − 1) x→1+ x
lim−
|1 − x| |1 − x| |1 − x| '= lim+ 2 therefore lim 2 does not exist. 2 x→1 x − x x − x x→1 x − x
x→1
x→1
x→1
(d)
|1 − x| =
'
# 1 $ lim (x + 1)2 arctan e x+1
x→−1
Since −
π π < arctan x < for all x, then: 2 2 # 1 $ π π − < arctan e x+1 < for all x '= −1 2 2
(∗)
Since (x + 1)2 > 0 for all x '= −1, then, multiplying all sides of (∗) with (x + 1)2 gives: −
# 1 $ π(x + 1)2 π(x + 1)2 < (x + 1)2 arctan e x+1 < for all x '= −1 2 2
Since lim − x→−1
theorem:
π(x + 1)2 = 0, therefore by the Squeeze x→−1 2 # 1 $ lim (x + 1)2 arctan e x+1 = 0
π(x + 1)2 = 0 and 2
lim
x→−1
——————————————————————————— Alternatively (A) lim (x + 1)2 = 0 x→−1
And
# 1 $ lim − arctan e x+1 = 0
x→−1
(x + 1 < 1, x + 1 → 0, so Thus
1 1 → −∞ and e x+1 → 0 ) x+1 # 1 $ 2 lim − (x + 1) arctan e x+1 = 0
x→−1
And (B) A
# 1 $ π lim + arctan e x+1 = x→−1 2 1 1 (x + 1 > 1, x + 1 → 0, so → ∞ and e x+1 → ∞ ) x+1 Thus # 1 $ π lim + (x + 1)2 arctan e x+1 = 0 · = 0 x→−1 2
From (A) and (B), Thus
# 1 $ # 1 $ lim − (x + 1)2 arctan e x+1 = lim + (x + 1)2 arctan e x+1 = 0
x→−1
x→−1
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4. [10 marks] Let f (x) =
√
x2 − 1 + 1 . x−1
(a) [2 marks] Give the domain of f . √ solution : x ∈ dom(f ) is x2 − 1 is defined and if x − 1 '= 0. So x '= 1 and x2 − 1 ≥ 0 2 x '= 1 and x ≥ 1 x '= 1 and |x| ≥ 1 x '= 1 and x ≤ −1 or x ≥ 1 Thus x ∈ (−∞, −1] ∪ (1, ∞)
(b) [8 marks] Find all the horizontal and vertical asymptotes of f . solution : For vertical asymptotes of f : x − 1 = 0 if x = 1. So we have a possible vertical asymptote at 1 Since f is not defined for x ∈ (−1, 1] we only consider the right limit or f as at 1: √ √ x2 − 1 + 1 lim+ = ∞ (since x − 1 → 0 but x − 1 > 0 and x2 − 1 + 1 → 1.) x→1 x−1 Thus x = 1 is the vertical asymptote of f . For horizontal asymptotes: ( √ √ x2 −1 x2 −1+1 2 + x −1+1 x2 x lim = lim = lim x−1 x→∞ x→∞ x→∞ x−1 1 − x1 x √ (x = x2 since x > 0) √
√ x2 −1+1 x lim x−1 x→−∞ x
x2 − 1 + 1 lim = x→−∞ x−1 √ (x = − x2 since x < 0)
= lim
x→−∞
1 x
x→∞
( − 1− 1
= lim
1 x2 − x1
+
1 x
Thus y = 1 and y = −1 are the horizontal asymptotes of f .
(
1 + x1 x2 − x1
1− 1
= −1
=1
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5. [10 marks] Given some numbers a, b and c, define f : R → R by 3−x if x ≤ a , √ x−1 if a < x < 5 , f (x) = x − b esin(πx) if 5 ≤ x ≤ 6 , cx2 if x > 6 , Find a, b and c so that f is continuous on R.
solution : 3 − x is continuous everywhere, thus continuous on (−∞, a) for all a. √
x − 1 is continuous on [1.∞) and so continuous on (a, 5) if a ≥ 1
x − b esin(πx)) is continuous everywhere since both x and esin(πx) are, thus it is also continuous on (5, 6). cx2 is a polynomial therefore it is continuous everywhere, thus continuous on (6, ∞). -For a function to be continuous at ”a” we need: lim f (x) = lim− f (x) = f (a) = a − 3
x→a+
x→a
lim f (x) = lim− 3 − x = 3 − a = f (a),
x→a−
x→a
lim+ f (x) = lim+
x→a
x→a
√
x−1=
√
a − 1, thus we need 3 − a =
√
a − 1 or (3 − a)2 = a − 1
a2 − 6x + 9 = a − 1 ⇐⇒ a2 − 7x + 10 = 0 ⇐⇒ (a − 5)(a − 2) = 0 ⇐⇒ a = 5 of a = 2. However, since a < 5 (second condition of definition of f ), then a = 2 therefore a=2 -For a function to be continuous at 5 we need: lim f (x) = lim− f (x) = f (5) = 5 − b esin(π5) = 5 − be0 = 5 − b
x→5+
x→5
lim f (x) = lim+ x − b esin(πx) = 5 − b esin(π5) = 5 − b, x→5 √ √ lim− f (x) = lim− x − 1 = 5 − 1 = 2, thus we need 5 − b = 2 or b = 3 therefore
x→5+ x→5
x→5
b=3
6
For a function to be continuous at 6 we need: lim f (x) = lim− f (x) = f (6) = 6 − b esin(π6) = 6 − b = 6 − 3 = 3
x→6+
x→6
lim f (x) = lim− x − b esin(πx) = 6 − b = f (6),
x→6−
x→6
lim+ f (x) = lim+ cx2 = 36c, thus we need 3 = 36c or
x→6
x→6
c=1/12 Since b = 3, then a = 2 and c =
f (x) =
is continuous on R.
1 and 12 3−x √ x−1
if x ≤ 2 , if 2 < x < 5 ,
x − 3 esin(πx) if 5 ≤ x ≤ 6 , x2 if x > 6 , 12
7
6. [5 marks] Fully justify that arctan x = 10x − ex has a positive root. solution : Since arctan x, 10x and ex are all continuous functions on R, then their linear combination f (x) = arctan x − 10x + ex is also a continuous function on R. π Since f (0) = arctan (0)−10(0)+e0 = 1 and f (1) = arctan (1)−10+e1 = −10+e < 0, 4 or f (1) < 0 < f (0) then by the IVT there is some c ∈ (0, 1) such that f (c) = 0 or arctan x = 10x − ex has a positive root.
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7. (a) [5 marks] Let f (x) = solution : (a)
√
x − 2. Use the definition of the derivative to find f % (x).
f (x + h) − f (x) = lim lim h→0 h→0 h = lim
√ √
h→0
x+h−2− h x+h−2− h
√ √
x−2 x−2
√ √ x+h−2+ x−2 √ ·√ x+h−2+ x−2
x + h − 2 − (x − 2) √ √ h→0 h( x + h − 2 + x − 2)
= lim
h = lim √ √ h→0 ( x + h − 2 + x − 2)h = lim √ h→0
=
1 √ x+h−2+ x−2
1 2 x−2 √
(b) [7 marks] There are many tangent lines to the graph of f . Find the equation(s) of the tangent line(s) that pass through the point (5, 2). (Note that the point (5, 2) is NOT on the graph of f .) solution : Suppose the tangent line touches the graph of f at (a, f (a)) = √ 1 (a, a − 2). Then the slope of the tangent at that point is m = f % (a) = √ . 2 a−2 Since (5, 2) is also a point on the line , then:
which gives
√ 2− a−2 1 = m= √ 5−a 2 a−2 5−a (5 − a) a+1 (a + 1)2 2 a + 2a + 1 2 a − 14a + 33 (a − 11)(a − 3)
= = = = = = =
√ √ (2√− a − 2)2 a − 2 4√a − 2 − 2(a − 2) 4 a−2 16(a − 2) 16a − 32 0 0
So a = 11, a = 3. 1 y−2 1 1 (A) If a = 11, then m = √ giving the equation of = √ = = 6 x−5 2 11 − 2 2 9 the tangent: y = x/6 + 7/6 1 y−2 1 (B) If a = 3, then m = √ giving the equation of the tangent: = = 2 x−5 2 3−2 y = x/2 − 1/2
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8. [3 marks] Consider the function f : R → R whose graph is given below.
Let A be the set of points {x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 , x10 , x11 } ∈ R.
i. Find all points in A where f is differentiable. Your answer: x2 , x5 , x7 , x8 ii. Find all points in A where f is continuous but not differentiable. Your answer: x4 , x9 iii. Find all points in A where f % is zero. Your answer: x2 , x5 , x8
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9. For the following given functions use differentiation rules to find their required derivatives. Do not simplify. (2 + x)2 (a) [5 marks] f (x) = . Find f % (x) ln(2 + x) solution : 2(2 + x) ln(2 + x) − (2 + x)2 · % f (x) = (ln(2 + x))2 =
1 2+x
= 2(2 + x) ln(2 + x) − (2 + x) (ln(2 + x))2
(b) [5 marks] g(x) = sin(tan(sec(x2 ))). Find g %(x). g % (x) = cos(tan(sec(x2 ))) · sec2 (sec(x2 )) · sec(x2 ) tan(x2 ) · 2x
11
x2
(c) [5 marks] h(x) = 58 solution :
where a is a positive constant real number. Find h% (x).
x2
2
h% (x) = (ln 5) · (58 ) · (ln 8)(8x ) · 2x dy (d) [5 marks] If y cos x = 2x2 + 4y 2, find . dx solution : y cos x % y cos x − y sin x y % cos x − 8yy % y % (cos x − 8y) y%
2x2 + 4y 2 4x + 8yy % 4x + y sin x 4x + y sin x 4x + y sin x = cos x − 8y
= = = =
