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Basic Concepts of Functions PDF

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Summary

This document covers basic concepts of functions, including definitions, examples, and evaluating functions. It also includes operations on functions. The handout explains how to determine if an equation represents a function using the vertical line test.

Full Transcript

SH1902 Basic Concepts of Functions I. Definition of Functions Definition 1.1 A function is a correspondence between two (2) sets of elements such that with each element in the first set, there corresponds only one (1) element...

SH1902 Basic Concepts of Functions I. Definition of Functions Definition 1.1 A function is a correspondence between two (2) sets of elements such that with each element in the first set, there corresponds only one (1) element in the second set. The first set is called the domain, and the set of all corresponding elements in the second set is called the range. A function associates each element in the domain with only one (1) element in the range. This association is called a mapping. An input must only have one (1) output. However, two (2) inputs can have the same output. Definition 1.2 The independent variable in a function is the variable that may take any value that is within the function’s domain. The dependent variable in a function is the variable which is affected by the changes in the independent variable with respect to the rule set by the function. Example 1.2 The equations 𝑦𝑦 = 𝑥𝑥 2 + 2𝑥𝑥 and 10𝑦𝑦 = 𝑥𝑥 3 are functions. In the first equation, for any real number 𝑥𝑥, 𝑥𝑥 2 is a unique real number and so is 2𝑥𝑥. Hence, 𝑥𝑥 2 + 2𝑥𝑥 is a unique real number. In the second equation, if we will solve for 𝑦𝑦, we will get 𝑥𝑥 3 𝑦𝑦 = 10 and since for any real number 𝑥𝑥, 𝑥𝑥 3 is a unique real number, then the second equation is a function. However, the equation 𝑥𝑥 2 + 𝑦𝑦 2 = 16 is not a function because if you solve 𝑦𝑦 in terms of 𝑥𝑥, 𝑦𝑦 2 = 16 − 𝑥𝑥 2 𝑦𝑦 = ± 16 − 𝑥𝑥 2 You get two (2) different values: √16 − 𝑥𝑥 2 and −√16 − 𝑥𝑥 2. Drawing a vertical line on the graph of an equation is an easy way to determine whether the equation is a function or not. 𝑦𝑦 = 𝑥𝑥 2 + 2𝑥𝑥 10𝑦𝑦 = 𝑥𝑥 3 𝑥𝑥 2 + 𝑦𝑦 2 = 16 https://www.desmos.com/calculator 01 Handout 1 *Property of STI  [email protected] Page 1 of 3 SH1902 The Vertical Line Test An equation defines a function if each vertical line in a rectangular coordinate system passes through at most one (1) point on the graph of the equation. If any vertical line passes through two (2) or more points on the graph of an equation, then the equation does not define a function. II. Evaluating Functions To evaluate a function, simply replace its variable with a given number or expression. We call this process substitution. Example 2.1 Evaluate 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 + 1 when 𝑥𝑥 = 3 Solution: 𝑓𝑓(3) = 2(3) + 1 = 6 + 1 = 7 Example 2.2 Evaluate 𝑔𝑔(𝑥𝑥) = 3𝑥𝑥 2 − 4 when 𝑥𝑥 = −2 Solution: 𝑔𝑔(−2) = 3(−2)2 − 4 = 3(4) − 4 = 12 − 4 =8 III. Operations on Functions Let 𝑓𝑓 and 𝑔𝑔 be functions. The four (4) basic operations on functions are: Addition: (𝑓𝑓 + 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) + 𝑔𝑔(𝑥𝑥) Subtraction: (𝑓𝑓 − 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) − 𝑔𝑔(𝑥𝑥) Multiplication: (𝑓𝑓 ∙ 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) ∙ 𝑔𝑔(𝑥𝑥) 𝑓𝑓 𝑓𝑓(𝑥𝑥) Division: 𝑔𝑔 (𝑥𝑥) = given that 𝑔𝑔(𝑥𝑥) ≠ 0 𝑔𝑔(𝑥𝑥) Example 3.1: Given the following functions: 1 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 − 3 𝑔𝑔(𝑥𝑥) = ℎ(𝑥𝑥) = 𝑥𝑥 − 2 𝑥𝑥 − 3 Find the following: a. (𝑓𝑓 + ℎ)(𝑥𝑥) b. (ℎ − 𝑓𝑓)(𝑥𝑥) c. (𝑓𝑓 ∙ 𝑔𝑔)(𝑥𝑥) 𝑓𝑓 d. ℎ (𝑥𝑥) Solution: a. (𝑓𝑓 + ℎ)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) + ℎ(𝑥𝑥) = (𝑥𝑥 − 3) + (𝑥𝑥 − 2) 01 Handout 1 *Property of STI  [email protected] Page 2 of 3 SH1902 = 𝑥𝑥 − 3 + 𝑥𝑥 − 2 = 2𝑥𝑥 − 5 b. (ℎ − 𝑓𝑓)(𝑥𝑥) = ℎ(𝑥𝑥) − 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 2) − (𝑥𝑥 − 3) = 𝑥𝑥 − 2 − 𝑥𝑥 + 3 = 𝑥𝑥 − 𝑥𝑥 − 2 + 3 =1 c. (𝑓𝑓 ∙ 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) ∙ 𝑔𝑔(𝑥𝑥) 1 = (𝑥𝑥 − 3) 𝑥𝑥 − 3 𝑥𝑥 − 3 = 𝑥𝑥 − 3 =1 𝑓𝑓 𝑓𝑓(𝑥𝑥) d. ℎ (𝑥𝑥) = ℎ(𝑥𝑥) 𝑥𝑥 − 3 = 𝑥𝑥 − 2 Composition of two (2) functions 𝑓𝑓 and 𝑔𝑔, denoted with the symbol “∘”, is a special operation wherein one (1) function is applied to the result of another. The operation (𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥) is like evaluating 𝑔𝑔(𝑥𝑥) at 𝑓𝑓(𝑥𝑥), that is, (𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑔𝑔 𝑓𝑓(𝑥𝑥). Example 3.2 Consider 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 + 1 and 𝑔𝑔(𝑥𝑥) = 3𝑥𝑥 − 1. Find a. (𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥) b. (𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥) Solution: a. (𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑔𝑔 𝑓𝑓(𝑥𝑥) = 𝑔𝑔(2𝑥𝑥 + 1) = 3(2𝑥𝑥 + 1) − 1 = 6𝑥𝑥 + 3 − 1 = 6𝑥𝑥 + 2 b. (𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓 𝑔𝑔(𝑥𝑥) = 𝑓𝑓(3𝑥𝑥 − 1) = 2(3𝑥𝑥 − 1) + 1 = 6𝑥𝑥 − 2 + 1 = 6𝑥𝑥 − 1 References Introduction to functions. (n.d.). In Math On Web. Retrieved from http://mathonweb.com/help_ebook/html/functions_6.htm Hunter, J. (2017). Sets and functions. Retrieved from https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch1.pdf 01 Handout 1 *Property of STI  [email protected] Page 3 of 3

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