06.1 Functions Concept Overview
FUNCTIONS | CONCEPT OVERVIEW The TOPIC of FUNCTIONS is not directly provided for us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook.
CONCEPT INTRO: An equation is considered a function if for any βπ₯β in a specified domain, the equation will yield exactly one defined, distinct value for βπ¦β. Generally speaking, a function can be written in the form: π(π₯) = π¦ FUNCTION NOTATION: Using the notation π(π₯) is nothing more than an elaborate way of writing the βπ¦β in a an equation, allowing us to identify unique and distinct notation for each individual result of the defined function. Using function notation allows us to represent the value of the function at a given βπ₯β within a defined domain in a clean compact way. Made with
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It also allows us to distinctly define derivatives and other manipulations of the given function at a particular point in the domain. Function notation does not mean that a function can only be represented as π(π₯), any letter can be used in the same form such as: π(π₯), π(π₯), β(π₯) β¦ etc Of course this list is not all-inclusive. The domain of the function also doesnβt need to be represented exclusively by x, this is just the standard in how we are most commonly taught. It is important to recall that this is not some letter times βπ₯β, this is just an elaborate way of writing βπ¦β. EVALUATING A FUNCTION: Evaluating a function is simply done by substituting everywhere you see an βπ₯β on the right side with the value that is within the parenthesis on the left side. As an illustration, say we have the following function: π(π₯) = π₯ 2 β 3π₯
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And are asked to evaluate this function at x = 3, this will give us: π(3) = (3)2 β 3(3) When π(π‘) = π(βπ‘), the function is a visual mirror of itself on a Cartesian plane such that either side of the vertical axis will show the same symmetry. This is known as EVEN SYMMMETRY. A function is EVEN, if for each x in the domain of π, π(βπ₯) = π(π₯). Even functions are characterized as having reflective symmetry across the y-axis. A function is ODD, if for each x in the domain of π, π(βπ₯) = βπ(π₯). Odd functions are characterized as having reflective symmetry across the x-axis. FINDING THE ROOTS OF A FUNCTION: The TOPIC of ROOTS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must take note to memorize this topic and understand its application independent of the NCEES Supplied Reference Handbook.
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A root of a given function is simply a number for which the function results in zero. In other words, where: π(π₯) = 0 One of the more important concepts of a function is that of the domain and range. The domain of a function is the set of all values that can be plugged into a function and have the function exist, resulting in a real number for a value. This means that for the domain, we need to avoid division by zero, square roots of negative numbers, logarithms of zero, negative numbers, etc. These will not result in a distinct, real number value. With a domain defined, we can determine the range of the function. The range of a function is simply the set of all possible resulting values given the domain. FUNCTION COMPOSITION: The TOPIC of COMPOSITIONS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Itβs important however and must be dialed in to memory with an ability to fully understand its application independent of the NCEES Supplied Reference Handbook.
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Given two functions π(π₯) and π(π₯), the composition of the two can be written as: (π β π) = π(π(π₯)) Compositions come across as complicated, but are evaluated simply by plugging the second function listed into the first function listed. The most important thing is the order in which the functions are listed. Missing out and interchanging this order will result in an incorrect result, and believe us when we say this, that result will be presented as an answer option come your exam day.
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FUNCTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Given the function π(π₯) = 9π₯ 2 and π(π₯) = 4π₯ 3 + 2 the composition (π β π) is most close to: A. 36(4π₯ 6 + 4π₯ 3 + 1) B. π₯ 6 + 4π₯ 3 + 1 C. 4π₯ 6 + π₯ 3 + 1 D. 9(π₯ 6 + π₯ 3 + 1)
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SOLUTION: The TOPIC of COMPOSITIONS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Itβs important however and must be dialed in to memory with an ability to fully understand its application independent of the NCEES Supplied Reference Handbook.
Recall that the composition of π(π₯) and g(x) is written as: (π β π) = π(π(π₯)) Therefore, given π(π₯) = 9π₯ 2 and π(π₯) = 4π₯ 3 + 2, we simply plug π(π₯) in to the function π(π₯) such that: π(π(π₯)) = π(4π₯ 3 + 2) = 9(4π₯ 3 + 2)2 Expanding the right side, we get: 9[(4π₯ 3 + 2)(4π₯ 3 + 2)] Expanding on it a little more: 9[16π₯ 6 + 16π₯ 3 + 4]
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Which gives us: 144π₯ 6 + 144π₯ 3 + 36 We can simplify to conclude that the composition (π β π) is: (π β π) = 36(4π₯ 6 + 4π₯ 3 + 1) The correct answer choice is A. ππ(πππ + πππ + π)
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CONCEPT INTRO: An equation is considered a function if for any βπ₯β in a specified domain, the equation will yield exactly one defined, distinct value for βπ¦β. Generally speaking, a function can be written in the form: π(π₯) = π¦ FUNCTION NOTATION: Using the notation π(π₯) is nothing more than an elaborate way of writing the βπ¦β in a an equation, allowing us to identify unique and distinct notation for each individual result of the defined function. Using function notation allows us to represent the value of the function at a given βπ₯β within a defined domain in a clean compact way. Made with
by Prepineer | Prepineer.com
It also allows us to distinctly define derivatives and other manipulations of the given function at a particular point in the domain. Function notation does not mean that a function can only be represented as π(π₯), any letter can be used in the same form such as: π(π₯), π(π₯), β(π₯) β¦ etc Of course this list is not all-inclusive. The domain of the function also doesnβt need to be represented exclusively by x, this is just the standard in how we are most commonly taught. It is important to recall that this is not some letter times βπ₯β, this is just an elaborate way of writing βπ¦β. EVALUATING A FUNCTION: Evaluating a function is simply done by substituting everywhere you see an βπ₯β on the right side with the value that is within the parenthesis on the left side. As an illustration, say we have the following function: π(π₯) = π₯ 2 β 3π₯
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And are asked to evaluate this function at x = 3, this will give us: π(3) = (3)2 β 3(3) When π(π‘) = π(βπ‘), the function is a visual mirror of itself on a Cartesian plane such that either side of the vertical axis will show the same symmetry. This is known as EVEN SYMMMETRY. A function is EVEN, if for each x in the domain of π, π(βπ₯) = π(π₯). Even functions are characterized as having reflective symmetry across the y-axis. A function is ODD, if for each x in the domain of π, π(βπ₯) = βπ(π₯). Odd functions are characterized as having reflective symmetry across the x-axis. FINDING THE ROOTS OF A FUNCTION: The TOPIC of ROOTS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must take note to memorize this topic and understand its application independent of the NCEES Supplied Reference Handbook.
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by Prepineer | Prepineer.com
A root of a given function is simply a number for which the function results in zero. In other words, where: π(π₯) = 0 One of the more important concepts of a function is that of the domain and range. The domain of a function is the set of all values that can be plugged into a function and have the function exist, resulting in a real number for a value. This means that for the domain, we need to avoid division by zero, square roots of negative numbers, logarithms of zero, negative numbers, etc. These will not result in a distinct, real number value. With a domain defined, we can determine the range of the function. The range of a function is simply the set of all possible resulting values given the domain. FUNCTION COMPOSITION: The TOPIC of COMPOSITIONS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Itβs important however and must be dialed in to memory with an ability to fully understand its application independent of the NCEES Supplied Reference Handbook.
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Given two functions π(π₯) and π(π₯), the composition of the two can be written as: (π β π) = π(π(π₯)) Compositions come across as complicated, but are evaluated simply by plugging the second function listed into the first function listed. The most important thing is the order in which the functions are listed. Missing out and interchanging this order will result in an incorrect result, and believe us when we say this, that result will be presented as an answer option come your exam day.
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by Prepineer | Prepineer.com
FUNCTIONS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.
Given the function π(π₯) = 9π₯ 2 and π(π₯) = 4π₯ 3 + 2 the composition (π β π) is most close to: A. 36(4π₯ 6 + 4π₯ 3 + 1) B. π₯ 6 + 4π₯ 3 + 1 C. 4π₯ 6 + π₯ 3 + 1 D. 9(π₯ 6 + π₯ 3 + 1)
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SOLUTION: The TOPIC of COMPOSITIONS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Itβs important however and must be dialed in to memory with an ability to fully understand its application independent of the NCEES Supplied Reference Handbook.
Recall that the composition of π(π₯) and g(x) is written as: (π β π) = π(π(π₯)) Therefore, given π(π₯) = 9π₯ 2 and π(π₯) = 4π₯ 3 + 2, we simply plug π(π₯) in to the function π(π₯) such that: π(π(π₯)) = π(4π₯ 3 + 2) = 9(4π₯ 3 + 2)2 Expanding the right side, we get: 9[(4π₯ 3 + 2)(4π₯ 3 + 2)] Expanding on it a little more: 9[16π₯ 6 + 16π₯ 3 + 4]
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Which gives us: 144π₯ 6 + 144π₯ 3 + 36 We can simplify to conclude that the composition (π β π) is: (π β π) = 36(4π₯ 6 + 4π₯ 3 + 1) The correct answer choice is A. ππ(πππ + πππ + π)
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