{"id":48,"date":"2019-01-30T14:53:33","date_gmt":"2019-01-30T14:53:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/?post_type=chapter&#038;p=48"},"modified":"2025-03-07T22:09:58","modified_gmt":"2025-03-07T22:09:58","slug":"composition-of-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/composition-of-functions\/","title":{"raw":"1.5 Composition of Functions","rendered":"1.5 Composition of Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Create a new function by composition of functions.<\/li>\r\n \t<li>Evaluate composite functions.<\/li>\r\n \t<li>Decompose a composite function into its component functions.<\/li>\r\n<\/ul>\r\nOptional:\r\n<ul>\r\n \t<li>Find the domain of a composite function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165134094620\">Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.<\/p>\r\n<p id=\"fs-id1165134038788\">Using descriptive variables, we can notate these two functions. The function [latex]C\\left(T\\right)[\/latex] gives the cost [latex]C[\/latex] of heating a house for a given average daily temperature in [latex]T[\/latex] degrees Celsius. The function [latex]T\\left(d\\right)[\/latex] gives the average daily temperature on day [latex]d[\/latex] of the year. For any given day, [latex]\\text{Cost}=C\\left(T\\left(d\\right)\\right)[\/latex] means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature [latex]T\\left(d\\right).[\/latex] For example, we could evaluate [latex]T\\left(5\\right)[\/latex] to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the <span class=\"no-emphasis\">cost function<\/span> at that temperature. We would write [latex]C\\left(T\\left(5\\right)\\right).[\/latex]<\/p>\r\n<img class=\"size-full wp-image-3158 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142304\/336d97bc881544921ca4876a504752b89148f9fd.jpeg\" alt=\"Explanation of C(T(5)), which is the cost for the temperature and T(5) is the temperature on day 5.\" width=\"487\" height=\"140\" \/>\r\n<p id=\"fs-id1165137460512\">By combining these two relationships into one function, we have performed function composition, which is the focus of this section.<\/p>\r\n\r\n<div id=\"fs-id1165135690697\" class=\"bc-section section\">\r\n<h3>Create a Function by Composition of Functions<\/h3>\r\n<p id=\"fs-id1165137451258\">We can create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of <span class=\"no-emphasis\">combining functions<\/span> so that the output of one function becomes the input of another is known as a <span class=\"no-emphasis\">composition of functions<\/span><strong>.<\/strong> The resulting function is known as a <strong>composite function<\/strong>. We represent this combination by the following notation:<\/p>\r\n\r\n<div id=\"fs-id1165137605973\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137466004\">We read the left-hand side as [latex]\\text{\u201c}f[\/latex] composed with [latex]g[\/latex] at [latex]x,\\text{\u201d}[\/latex] and the right-hand side as [latex]\\text{\u201c}f[\/latex] of [latex]g[\/latex] of [latex]x.\\text{\u201d}[\/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\\circ [\/latex] is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number.\u00a0\u00a0It is important to realize that the product of functions [latex]fg[\/latex] is not the same as the function composition [latex]f\\left(g\\left(x\\right)\\right),[\/latex] because, in general, [latex]f\\left(x\\right)g\\left(x\\right)\\ne f\\left(g\\left(x\\right)\\right).[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function [latex]g[\/latex] takes the input [latex]x[\/latex] first and yields an output [latex]g\\left(x\\right).[\/latex] Then the function [latex]f[\/latex] takes [latex]g\\left(x\\right)[\/latex] as an input and yields an output [latex]f\\left(g\\left(x\\right)\\right).[\/latex]<\/span>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<img class=\"size-full wp-image-3176 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15153746\/c02500912de8e653c7e833a6aa2d53696a4c7630.jpeg\" alt=\"explanation of a composite function\" width=\"487\" height=\"171\" \/>\r\n<div id=\"fs-id1165135690697\" class=\"bc-section section\">\r\n<p id=\"fs-id1165137851405\">In general, [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] are different functions. In other words, in many cases [latex]f\\left(g\\left(x\\right)\\right)\\ne g\\left(f\\left(x\\right)\\right)[\/latex] for all [latex]x.[\/latex] We will also see that sometimes two functions can be composed only in one specific order.<\/p>\r\n<p id=\"fs-id1165135189964\">For example, if [latex]f\\left(x\\right)={x}^{2}[\/latex] and [latex]g\\left(x\\right)=x+2,[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id1165134047682\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(g\\left(x\\right)\\right)&amp;=f\\left(x+2\\right)\\hfill \\\\ \\text{ }&amp;={\\left(x+2\\right)}^{2}\\hfill \\\\ \\text{ }&amp;={x}^{2}+4x+4\\hfill \\end{align*}[\/latex]<\/div>\r\n<p id=\"fs-id1165137595004\">but<\/p>\r\n\r\n<div id=\"fs-id1165137461653\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}g\\left(f\\left(x\\right)\\right)&amp;=g\\left({x}^{2}\\right)\\hfill \\\\ \\text{ }&amp;={x}^{2}+2.\\hfill \\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137430833\">These expressions are not equal for all values of [latex]x,[\/latex] so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value [latex]x=-\\frac{1}{2}.[\/latex]<\/p>\r\n<p id=\"fs-id1165137531472\">Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.<\/p>\r\n\r\n<div id=\"fs-id1165134190746\">\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137783932\">When the output of one function is used as the input of another, we call the entire operation a <strong>composition of functions<\/strong>. For any input [latex]x[\/latex] and functions [latex]f[\/latex] and [latex]g,[\/latex] this action defines a composite function, which we write as [latex]f\\circ g[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1165137836657\" style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165135503928\">The domain of the composite function [latex]f\\circ g[\/latex] is all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_04_02\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137544021\">\r\n<div id=\"fs-id1165137482990\">\r\n<h3>Example 1:\u00a0 Determining Whether Composition of Functions is Commutative<\/h3>\r\n<p id=\"fs-id1165137600071\">Using the functions provided, find [latex]f\\left(g\\left(4\\right)\\right)[\/latex] and [latex]g\\left(f\\left(4\\right)\\right).[\/latex] Determine whether the composition of the functions is <span class=\"no-emphasis\">commutative<\/span>.<\/p>\r\n\r\n<div id=\"fs-id1165135530570\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+1\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }g\\left(x\\right)=3-x[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135195227\">[reveal-answer q=\"fs-id1165135195227\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135195227\"]\r\n<p id=\"fs-id1165137603003\">Let\u2019s begin by evaluating [latex]g\\left(4\\right)=3-4=-1.[\/latex]\u00a0 The output of [latex]g[\/latex] is the input to [latex]f[\/latex] so<\/p>\r\n\r\n<div id=\"fs-id1165135499437\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} f\\left(g\\left(4\\right)\\right)&amp;=f\\left(-1\\right)\\\\\\text{ }&amp;=2\\left(-1\\right)+1\\hfill \\\\ \\text{ }&amp;=-1\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137786392\">For the second composition\u00a0[latex]g\\left(f\\left(4\\right)\\right),[\/latex] we begin by evaluating [latex]f\\left(4\\right)=2\\left(4\\right)+1=9.[\/latex] The output of [latex]f[\/latex] will be the input to [latex]g[\/latex] so<\/p>\r\n\r\n<div id=\"fs-id1165135440456\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} g\\left(f\\left(4\\right)\\right)&amp;=g\\left(9\\right)\\\\\\text{}&amp;=3-9\\hfill \\\\ \\text{ }&amp;=-6\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135160089\">We find that [latex]g\\left(f\\left(4\\right)\\right)\\ne f\\left(g\\left(4\\right)\\right),[\/latex] so the operation of function composition is not commutative and order matters.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_04_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137603612\">\r\n<div id=\"fs-id1165137761836\">\r\n<h3>Example 2:\u00a0 Interpreting Composite Functions<\/h3>\r\n<p id=\"fs-id1165137695118\">The function [latex]c\\left(s\\right)[\/latex] gives the number of calories burned completing [latex]s[\/latex] sit-ups, and [latex]s\\left(t\\right)[\/latex] gives the number of sit-ups a person can complete in [latex]t[\/latex] minutes. Interpret [latex]c\\left(s\\left(3\\right)\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134170107\">[reveal-answer q=\"fs-id1165134170107\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134170107\"]\r\n<p id=\"fs-id1165137659141\">The inside expression in the composition is [latex]s\\left(3\\right).[\/latex] Because the input to the <em>s<\/em>-function is time, [latex]t=3[\/latex] represents 3 minutes, and [latex]s\\left(3\\right)[\/latex] is the number of sit-ups completed in 3 minutes.<\/p>\r\n<p id=\"fs-id1165137844294\">Using [latex]s\\left(3\\right)[\/latex] as the input to the function [latex]c\\left(s\\right)[\/latex] gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups).<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_04_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137551439\">\r\n<div id=\"fs-id1165135512587\">\r\n<h3>Example 3:\u00a0 Investigating the Order of Function Composition<\/h3>\r\n<p id=\"fs-id1165137643603\">Suppose [latex]f\\left(x\\right)[\/latex] gives the number of miles that can be driven in [latex]x[\/latex] hours and [latex]g\\left(y\\right)[\/latex] gives the number of gallons of gas used in driving [latex]y[\/latex] miles. Which of these expressions is meaningful: [latex]f\\left(g\\left(y\\right)\\right)[\/latex] or [latex]g\\left(f\\left(x\\right)\\right)?[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137451433\">[reveal-answer q=\"fs-id1165137451433\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137451433\"]\r\n<p id=\"fs-id1165135545914\">The function [latex]y=f\\left(x\\right)[\/latex] is a function whose output is the number of miles driven corresponding to the number of hours driven.<\/p>\r\n\r\n<div id=\"fs-id1165137648357\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\text{number of miles }=f\\text{ }\\left(\\text{number of hours}\\right)[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137749316\">The function [latex]g\\left(y\\right)[\/latex] is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:<\/p>\r\n\r\n<div id=\"fs-id1165137531080\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\text{number of gallons }=g\\text{ }\\left(\\text{number of miles}\\right)[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135190218\">The expression [latex]g\\left(y\\right)[\/latex] takes miles as the input and a number of gallons as the output. The function [latex]f\\left(x\\right)[\/latex] requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression [latex]f\\left(g\\left(y\\right)\\right)[\/latex] is meaningless.<\/p>\r\n<p id=\"fs-id1165137863600\">The expression [latex]f\\left(x\\right)[\/latex] takes hours as input and a number of miles driven as the output. The function [latex]g\\left(y\\right)[\/latex] requires a number of miles as the input. Using [latex]f\\left(x\\right)[\/latex] (miles driven) as an input value for [latex]g\\left(y\\right),[\/latex] where gallons of gas depends on miles driven, does make sense. The expression [latex]g\\left(f\\left(x\\right)\\right)[\/latex] makes sense, and will yield the number of gallons of gas used, [latex]g,[\/latex] driving a certain number of miles, [latex]f\\left(x\\right),[\/latex] in [latex]x[\/latex] hours.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137768571\" class=\"precalculus qa key-takeaways\">\r\n<h3>Q&amp;A<\/h3>\r\nAre there any situations where [latex]f\\left(g\\left(y\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex] would both be meaningful or useful expressions?\r\n<p id=\"fs-id1165137682006\"><em>Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134558042\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_01_04_01\">\r\n<div id=\"fs-id1165135560725\">\r\n<p id=\"fs-id1165135560726\">The gravitational force on a planet a distance [latex]r[\/latex] from the sun is given by the function [latex]G\\left(r\\right).[\/latex] The acceleration of a planet subjected to any force [latex]F[\/latex] is given by the function [latex]a\\left(F\\right).[\/latex] Form a meaningful composition of these two functions, and explain what it means.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137417357\">[reveal-answer q=\"fs-id1165137417357\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137417357\"]\r\n<p id=\"fs-id1165137417358\">A gravitational force is still a force, so [latex]a\\left(G\\left(r\\right)\\right)[\/latex] makes sense as the acceleration of a planet at a distance [latex]r[\/latex] from the sun (due to gravity), but [latex]G\\left(a\\left(F\\right)\\right)[\/latex] does not make sense.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137560475\" class=\"bc-section section\">\r\n<h3>Evaluating Composite Functions<\/h3>\r\n<p id=\"fs-id1165135168147\">Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function\u2019s output as the input for the outer function.<\/p>\r\n\r\n<div id=\"fs-id1165137760886\" class=\"bc-section section\">\r\n<h4>Evaluating Composite Functions Using Tables<\/h4>\r\n<p id=\"fs-id1165137725253\">When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.<\/p>\r\n\r\n<div id=\"Example_01_04_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137416770\">\r\n<div id=\"fs-id1165137465672\">\r\n<h3>Example 4:\u00a0 Using a Table to Evaluate a Composite Function<\/h3>\r\n<p id=\"fs-id1165135177664\">Using <a class=\"autogenerated-content\" href=\"#Table_01_04_01\">Table 1<\/a>, evaluate [latex]f\\left(g\\left(3\\right)\\right)[\/latex] and [latex]g\\left(f\\left(3\\right)\\right).[\/latex]<\/p>\r\n\r\n<table id=\"Table_01_04_01\" summary=\"Five rows and three columns. The first column is labeled, \u201cx\u201d, the second column is labeled, \u201cf(x)\u201d, and the third column is labeled, \u201cg(x)\u201d. We have the following values for f(x): f(1)=6, f(2)=8, f(3)=3, and f(4)=1. And for g(1)=3, g(2)=5, g(3)=2, and g(4)=7.\"><caption>Table 1<\/caption><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th class=\"border\" style=\"text-align: center;\">[latex]x[\/latex]<\/th>\r\n<th class=\"border\" style=\"text-align: center;\">[latex]f\\left(x\\right)[\/latex]<\/th>\r\n<th class=\"border\" style=\"text-align: center;\">[latex]g\\left(x\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">6<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165135582219\">[reveal-answer q=\"fs-id1165135582219\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135582219\"]\r\n<p id=\"fs-id1165135582221\">To evaluate [latex]f\\left(g\\left(3\\right)\\right),[\/latex] we start from the inside with the input value 3. We then evaluate the inside expression [latex]g\\left(3\\right)[\/latex] using the table that defines the function [latex]g:[\/latex] [latex]g\\left(3\\right)=2.[\/latex] We can then use that result as the input to the function [latex]f,[\/latex] so [latex]g\\left(3\\right)[\/latex] is replaced by 2 and we get [latex]f\\left(2\\right).[\/latex] Then, using the table that defines the function [latex]f,[\/latex] we find that [latex]f\\left(2\\right)=8.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165137415997\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}g\\left(3\\right)&amp;=2\\hfill \\\\ f\\left(g\\left(3\\right)\\right)&amp;=f\\left(2\\right)=8\\hfill \\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165134259282\">To evaluate [latex]g\\left(f\\left(3\\right)\\right),[\/latex] we first evaluate the inside expression [latex]f\\left(3\\right)[\/latex] using the first table: [latex]f\\left(3\\right)=3.[\/latex] Then, using the table for [latex]g\\text{, }[\/latex] we can evaluate<\/p>\r\n\r\n<div id=\"fs-id1165137841687\" class=\"unnumbered\" style=\"text-align: center;\">[latex]g\\left(f\\left(3\\right)\\right)=g\\left(3\\right)=2[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165134231482\"><a class=\"autogenerated-content\" href=\"#Table_01_04_02\">Table 2<\/a> shows the composite functions [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] as tables.<\/p>\r\n\r\n<table id=\"Table_01_04_02\" summary=\"Two rows and five columns. When x=3, g(3)=2, f(g(3))=8, f(3)=3, and g(f(3))=2.\"><caption>Table 2<\/caption><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]g\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]f\\left(g\\left(x\\right)\\right)[\/latex]<\/td>\r\n<td class=\"border\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]g\\left(f\\left(x\\right)\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137724943\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_01_04_02\">\r\n<div id=\"fs-id1165134297600\">\r\n<p id=\"fs-id1165134297603\">Using <a class=\"autogenerated-content\" href=\"#Table_01_04_01\">Table 1<\/a>, evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex] and [latex]g\\left(f\\left(4\\right)\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134258662\">[reveal-answer q=\"fs-id1165134258662\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134258662\"]\r\n<p id=\"fs-id1165137836967\">[latex]f\\left(g\\left(1\\right)\\right)=f\\left(3\\right)=3[\/latex] and [latex]g\\left(f\\left(4\\right)\\right)=g\\left(1\\right)=3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137756068\" class=\"bc-section section\">\r\n<h4>Evaluating Composite Functions Using Graphs<\/h4>\r\n<p id=\"fs-id1165137428192\">When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the [latex]x\\text{-}[\/latex] and [latex]y\\text{-}[\/latex]axes of the graphs.<\/p>\r\n\r\n<div id=\"fs-id1165137634364\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137660530\"><strong>Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs. <\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165133248553\" type=\"1\">\r\n \t<li>Locate the given input to the inner function on the [latex]x\\text{-}[\/latex]axis of its graph.<\/li>\r\n \t<li>Read off the output of the inner function from the [latex]y\\text{-}[\/latex]axis of its graph.<\/li>\r\n \t<li>Locate the inner function output on the [latex]x\\text{-}[\/latex]axis of the graph of the outer function.<\/li>\r\n \t<li>Read the output of the outer function from the [latex]y\\text{-}[\/latex]axis of its graph. This is the output of the composite function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_04_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137758193\">\r\n<div id=\"fs-id1165137758196\">\r\n<h3>Example 5:\u00a0 Using a Graph to Evaluate a Composite Function<\/h3>\r\n<p id=\"fs-id1165134226787\">Using <a class=\"autogenerated-content\" href=\"#Figure_01_04_002\">Figure 1<\/a>, evaluate [latex]f\\left(g\\left(1\\right)\\right).[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_3160\" align=\"aligncenter\" width=\"975\"]<img class=\"size-full wp-image-3160\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142515\/710401844b9cc0b203a1628994a0471665da4606.jpeg\" alt=\"Explanation of the composite function.\" width=\"975\" height=\"543\" \/> Figure 1[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135545946\">[reveal-answer q=\"fs-id1165135545946\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135545946\"]\r\n<p id=\"fs-id1165135545948\">To evaluate [latex]f\\left(g\\left(1\\right)\\right),[\/latex] we start with the inside evaluation. See <a class=\"autogenerated-content\" href=\"#Figure_01_04_004\">Figure 2<\/a>.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3162\" align=\"alignnone\" width=\"975\"]<img class=\"wp-image-3162 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142744\/0a3ffe6f2329045c480a8c60df63fe2df0466dcf.jpeg\" alt=\"Two graphs of a positive parabola (g(x)) and a negative parabola (f(x)). The following points are plotted: g(1)=3 and f(3)=6.\" width=\"975\" height=\"543\" \/> Figure 2[\/caption]\r\n<p id=\"fs-id1165137566106\">We evaluate [latex]g\\left(1\\right)[\/latex] using the graph of [latex]g\\left(x\\right),[\/latex] finding the input of 1 on the [latex]x\\text{-}[\/latex]axis and finding the output value of the graph at that input. Here, [latex]g\\left(1\\right)=3.[\/latex] We use this value as the input to the function [latex]f.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135432881\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(g\\left(1\\right)\\right)=f\\left(3\\right)[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137831420\">We can then evaluate the composite function by looking to the graph of [latex]f\\left(x\\right),[\/latex] finding the input of 3 on the [latex]x\\text{-}[\/latex]axis and reading the output value of the graph at this input. Here, [latex]f\\left(3\\right)=6,[\/latex] so [latex]f\\left(g\\left(1\\right)\\right)=6.[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\n<p id=\"fs-id1165137602379\"><a class=\"autogenerated-content\" href=\"#Figure_01_04_005\">Figure 3<\/a> shows how we can mark the graphs with arrows to trace the path from the input value to the output value.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3163\" align=\"aligncenter\" width=\"975\"]<img class=\"size-full wp-image-3163\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142833\/1cb774956db9f261e1d4be38aaffd0d296a652ed-1.jpeg\" alt=\"Two graphs of a positive and negative parabola.\" width=\"975\" height=\"520\" \/> Figure 3[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137723402\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_01_04_03\">\r\n<div id=\"fs-id1165137588093\">\r\n\r\nUsing <a class=\"autogenerated-content\" href=\"#Figure_01_04_002\">Figure 1<\/a>, evaluate [latex]g\\left(f\\left(2\\right)\\right).[\/latex]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137843133\">[reveal-answer q=\"fs-id1165137843133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137843133\"]\r\n<p id=\"fs-id1165137502186\">[latex]g\\left(f\\left(2\\right)\\right)=g\\left(5\\right)=3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137452389\" class=\"bc-section section\">\r\n<h4>Evaluating Composite Functions Using Formulas<\/h4>\r\n<p id=\"fs-id1165137432176\">When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression.<\/p>\r\n<p id=\"fs-id1165137567159\">While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition [latex]f\\left(g\\left(x\\right)\\right).[\/latex] To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like [latex]f\\left(t\\right)={t}^{2}-t,[\/latex] we substitute the value inside the parentheses into the formula wherever we see the input variable.<\/p>\r\n\r\n<div id=\"fs-id1165137584280\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137676972\"><strong>Given a formula for a composite function, evaluate the function. <\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137564125\" type=\"1\">\r\n \t<li>Evaluate the inside function using the input value or variable provided.<\/li>\r\n \t<li>Use the resulting output as the input to the outside function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_04_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134070851\">\r\n<div id=\"fs-id1165134070853\">\r\n<h3>Example 6:\u00a0 Evaluating a Composition of Functions Expressed as Formulas<\/h3>\r\n<p id=\"fs-id1165137447886\">Given [latex]f\\left(t\\right)={t}^{2}-t[\/latex] and [latex]h\\left(x\\right)=3x+2,[\/latex] evaluate<\/p>\r\n\r\n<ol>\r\n \t<li>[latex]f\\left(h\\left(1\\right)\\right).[\/latex]<\/li>\r\n \t<li>[latex]f\\left(h\\left(x\\right)\\right).[\/latex]<\/li>\r\n \t<li>[latex]h\\left(f\\left(t\\right)\\right).[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165135543322\">[reveal-answer q=\"fs-id1165135543322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135543322\"]\r\n<div><\/div>\r\n<div style=\"padding-left: 30px;\">\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">1.\u00a0 Because the inside expression is [latex]h\\left(1\\right),[\/latex] we start by evaluating [latex]h\\left(x\\right)[\/latex] at 1.<\/span>\r\n<p style=\"padding-left: 60px; text-align: center;\">[latex]\\begin{align*}h\\left(1\\right)&amp;=3\\left(1\\right)+2\\\\ &amp;=5\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">Then [latex]f\\left(h\\left(1\\right)\\right)=f\\left(5\\right),[\/latex] so we evaluate [latex]f\\left(t\\right)[\/latex] at an input of 5.<\/p>\r\n<p style=\"padding-left: 60px; text-align: center;\">[latex]\\begin{align*}f\\left(h\\left(1\\right)\\right)&amp;=f\\left(5\\right)\\\\ &amp;={5}^{2}-5\\\\ &amp;=20\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\n<p style=\"text-align: left;\">2.\u00a0 \u00a0The inside expression for\u00a0[latex]f\\left(h\\left(x\\right)\\right)[\/latex] is [latex]h\\left(x\\right)[\/latex] so we use its output, [latex]3x+2,[\/latex] as the input to the function [latex]f[\/latex].\u00a0 We then must evaluate [latex]f\\left(3x+2\\right).[\/latex]\u00a0 Begin by replacing each variable [latex]t,[\/latex] with [latex]3x+2[\/latex] in the function [latex]f[\/latex] to get<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(3x+2\\right)=\\left(3x+2\\right)^2-\\left(3x+2\\right).[\/latex]<\/p>\r\nThen simplify to get\r\n<p style=\"text-align: center;\">[latex]f\\left(h\\left(x\\right)\\right)=9x^2+6x+6x+4-3x-2=9x^2+9x+2.[\/latex][latex]\\\\[\/latex]<\/p>\r\n3.\u00a0The inside expression for\u00a0[latex]h\\left(f\\left(t\\right)\\right)[\/latex] is [latex]f\\left(t\\right),[\/latex] so we use its output, [latex]t^2-t,[\/latex] as the input to the function [latex]h[\/latex].\u00a0 We then must evaluate [latex]h\\left(t^2-t\\right).[\/latex]\u00a0 Begin by replacing each variable [latex]x,[\/latex] with [latex]t^2-t[\/latex] in the function [latex]h[\/latex] to get\r\n<p style=\"text-align: center;\">[latex]h\\left(t^2-t\\right)=3\\left(t^2-t\\right)+2.[\/latex]<\/p>\r\nThen simplify to get\r\n<p style=\"text-align: center;\">[latex]h\\left(f\\left(t\\right)\\right)=3t^2-3t+2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"unnumbered\"><\/div>\r\n<div class=\"unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137482990\">\r\n<div class=\"textbox examples\">\r\n<div id=\"fs-id1165137482990\">\r\n<h3>Example 7:\u00a0 Determining Whether Composition of Functions is Commutative<\/h3>\r\n<p id=\"fs-id1165137600071\">Using the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right).[\/latex] Determine whether the composition of the functions is <span class=\"no-emphasis\">commutative<\/span>.<\/p>\r\n\r\n<div id=\"fs-id1165135530570\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+1\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }g\\left(x\\right)=3-x[\/latex]<\/div>\r\n<div>[reveal-answer q=\"462565\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462565\"]\r\n<div>\r\n<p id=\"fs-id1165137603003\">Let\u2019s begin by substituting [latex]g\\left(x\\right)=3-x[\/latex] into [latex]f\\left(x\\right).[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135499437\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} f\\left(g\\left(x\\right)\\right)&amp;=f\\left(3-x\\right)\\\\\\text{ }&amp;=2\\left(3-x\\right)+1\\hfill \\\\ \\text{ }&amp;=6-2x+1\\hfill \\\\ \\text{ }&amp;=7-2x\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137786392\">Now we can substitute [latex]f\\left(x\\right)=2x+1[\/latex] into [latex]g\\left(x\\right).[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135440456\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} g\\left(f\\left(x\\right)\\right)&amp;=g\\left(2x+1\\right)\\\\\\text{}&amp;=3-\\left(2x+1\\right)\\hfill \\\\ \\text{ }&amp;=3-2x-1\\hfill \\\\ \\text{ }&amp;=-2x+2\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135160089\">We find that [latex]g\\left(f\\left(x\\right)\\right)\\ne f\\left(g\\left(x\\right)\\right),[\/latex] so the operation of function composition is not commutative.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137570158\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_01_04_04\">\r\n<div>\r\n<p id=\"fs-id1165134544990\">Given [latex]f\\left(t\\right)={t}^{2}-t[\/latex] and [latex]h\\left(x\\right)=3x+2,[\/latex] evaluate<\/p>\r\n\r\n<ol id=\"fs-id1165137758652\" type=\"a\">\r\n \t<li>[latex]h\\left(f\\left(2\\right)\\right)[\/latex]<\/li>\r\n \t<li>[latex]h\\left(f\\left(-2\\right)\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165137434572\">[reveal-answer q=\"fs-id1165137434572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137434572\"]\r\n<p id=\"fs-id1165134261859\">a. 8; b. 20<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135567444\" class=\"bc-section section\">\r\n<div id=\"fs-id1165135547254\">\r\n<h3>Decomposing a Composite Function into its Component Functions<\/h3>\r\nIn some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to\u00a0decompose a composite function, so we may choose the decomposition that appears to be most expedient.\u00a0 However, in calculus, you will be studying the chain rule in order to find the derivative of a composite function.\u00a0 While we can\u2019t begin to try to define the concept of a derivative at this point, we can help you begin to think about an \u201cinner\u201d and \u201couter\u201d function in a composition.\u00a0 When you study the chain rule, you will\u00a0 think of the outer function as [latex]f\\left(x\\right)[\/latex] and the inner function as [latex]g\\left(x\\right).[\/latex]\r\n\r\nLet\u2019s consider the function [latex]h\\left(x\\right)={\\left(x+10\\right)}^{5}.[\/latex]\u00a0 This reminds us of the function [latex]f\\left(x\\right)={x}^{5}[\/latex] where the input [latex]x[\/latex] has been replaced by [latex]x+10.[\/latex]\u00a0 We can see this as though we have a place holder where we would normally see the [latex]x.[\/latex]\u00a0 This would look like [latex]{\\left(\\text{__}\\right)}^{5},[\/latex] where we are thinking about the \u201couter\u201d function as being the function that raises the \u201cinside\u201d to the 5<sup>th<\/sup> power.\u00a0 We would let the outside function be [latex]f\\left(x\\right)={x}^{5}[\/latex] and the \u201cinside\" function be [latex]g\\left(x\\right)=x+10.[\/latex]\u00a0 Can you see that [latex]f\\left(g\\left(x\\right)\\right)={\\left(x+10\\right)}^{5}?[\/latex]\r\n\r\nNow let\u2019s try a function that isn\u2019t as obvious with outer and inner functions.\u00a0 Consider [latex]k\\left(x\\right)={2}^{3x+4}.[\/latex]\u00a0 This reminds us of the function [latex]f\\left(x\\right)={2}^{x}[\/latex] where the [latex]x[\/latex] has been replaced by [latex]3x+4.[\/latex]\u00a0 Let\u2019s let our outer function be [latex]f\\left(x\\right)={2}^{x},[\/latex] and our inner function be [latex]g\\left(x\\right)=3x+4.[\/latex]\u00a0 Now form the composite function [latex]f\\left(g\\left(x\\right)\\right).[\/latex]\u00a0 You get [latex]k\\left(x\\right)={2}^{3x+4}.[\/latex]\r\n\r\nSometimes the easiest way to see the \u201cinside\u201d and \u201coutside\u201d is to consider what a simpler function would look like if the input were simply x and not some more complex expression.\r\n\r\nLet\u2019s do one more.\u00a0 Consider [latex]r\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{2x+5}}.[\/latex]\u00a0 \u00a0We can see that this fits the form of a simpler function [latex]f\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{x}}[\/latex] where the [latex]x[\/latex] has been replaced by [latex]2x+5.[\/latex]\u00a0 Our outer function would be [latex]f\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{x}}[\/latex]\u00a0and our inner function would be [latex]g\\left(x\\right)=2x+5.[\/latex]\u00a0 If you form the composite function [latex]f\\left(g\\left(x\\right)\\right),[\/latex] you will get\u00a0[latex]r\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{2x+5}}.[\/latex]\r\n<div id=\"Example_01_04_10\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134319721\">\r\n<div id=\"fs-id1165134319723\">\r\n<h3>Example 8:\u00a0 Decomposing a Function<\/h3>\r\n<p id=\"fs-id1165137771903\">Write [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135194520\">[reveal-answer q=\"fs-id1165135194520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135194520\"]\r\n<p id=\"fs-id1165135194522\">We are looking for two functions, [latex]g[\/latex] and [latex]h,[\/latex] so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right).[\/latex] To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right).[\/latex] As one possibility, we might notice that the expression [latex]5-{x}^{2}[\/latex] is the inside of the square root. We could then decompose the function as<\/p>\r\n\r\n<div id=\"fs-id1165137413960\" class=\"unnumbered\" style=\"text-align: center;\">[latex]h\\left(x\\right)=5-{x}^{2}\\text{ and }g\\left(x\\right)=\\sqrt{x}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135250635\">We can check our answer by recomposing the functions.<\/p>\r\n\r\n<div id=\"fs-id1165137730234\" class=\"unnumbered\" style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g\\left(5-{x}^{2}\\right)=\\sqrt{5-{x}^{2}}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165131878633\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_01_04_08\">\r\n<div id=\"fs-id1165137501975\">\r\n<p id=\"fs-id1165137501976\">Write [latex]f\\left(x\\right)=\\frac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137411051\">[reveal-answer q=\"fs-id1165137411051\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137411051\"]\r\n<p id=\"fs-id1165137411053\">Possible answer:<\/p>\r\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\r\n[latex]h\\left(x\\right)=\\frac{4}{3-x}[\/latex]\r\n\r\n[latex]f=h\\circ g[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h3>Finding the Domain of a Composite Function (Optional)<\/h3>\r\n<p id=\"fs-id1165135519324\">As we discussed previously, the <span class=\"no-emphasis\">domain of a composite function<\/span> such as [latex]f\\circ g[\/latex] is dependent on the domain of [latex]g[\/latex] and the domain of [latex]f.[\/latex] It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as [latex]f\\circ g.[\/latex] Let us assume we know the domains of the functions [latex]f[\/latex] and [latex]g[\/latex] separately. If we write the composite function for an input [latex]x[\/latex] as [latex]f\\left(g\\left(x\\right)\\right),[\/latex] we can see right away that [latex]x[\/latex] must be a member of the domain of [latex]g[\/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\\left(x\\right)[\/latex] must be a member of the domain of [latex]f,[\/latex] otherwise the second function evaluation in [latex]f\\left(g\\left(x\\right)\\right)[\/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\\circ g[\/latex] consists of only those inputs in the domain of [latex]g[\/latex] that produce outputs from [latex]g[\/latex] belonging to the domain of [latex]f.[\/latex] Note that the domain of [latex]f[\/latex] composed with [latex]g[\/latex] is the set of all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex]<\/p>\r\n\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\nThe <strong>domain of a composite function\u00a0<\/strong>[latex]f\\left(g\\left(x\\right)\\right)[\/latex] is the set of those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137663970\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165135203267\"><strong>Given a function composition [latex]f\\left(g\\left(x\\right)\\right),[\/latex] determine its domain.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137714200\" type=\"1\">\r\n \t<li>Find the domain of [latex]g.[\/latex]<\/li>\r\n \t<li>Find the domain of [latex]f.[\/latex]<\/li>\r\n \t<li>Find those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex] That is, exclude those inputs [latex]x[\/latex] from the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is not in the domain of [latex]f.[\/latex] The resulting set is the domain of [latex]f\\circ g.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_01_04_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137646695\">\r\n<div id=\"fs-id1165137646697\">\r\n<h3>Example 9:\u00a0 Finding the Domain of a Composite Function<\/h3>\r\n<p id=\"fs-id1165135640630\">Find the domain of<\/p>\r\n\r\n<div id=\"fs-id1165135349196\" class=\"unnumbered\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }f\\left(x\\right)=\\frac{5}{x-1}\\text{ and }\\text{ }\\text{ }g\\left(x\\right)=\\frac{4}{3x-2}[\/latex]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137627920\">[reveal-answer q=\"fs-id1165137627920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137627920\"]\r\n<p id=\"fs-id1165137627922\">The domain of [latex]g\\left(x\\right)[\/latex] consists of all real numbers except [latex]x=\\frac{2}{3},[\/latex] since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[\/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\\left(x\\right)[\/latex] that value of [latex]x[\/latex] for which [latex]g\\left(x\\right)=1.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165137455472\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}\\frac{4}{3x-2}&amp;=1\\hfill \\\\ \\text{ }4&amp;=3x-2\\hfill \\\\ 6&amp;=3x\\hfill \\\\ x&amp;=2\\hfill \\end{align*}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137891240\">So the domain of [latex]f\\circ g[\/latex] is the set of all real numbers except [latex]\\frac{2}{3}[\/latex] and [latex]2.[\/latex] This means that[latex]\\\\[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165131959460\" class=\"unnumbered\" style=\"text-align: center;\">[latex]x\\ne \\frac{2}{3}\\text{ }\\text{ }\\text{or}\\text{ }\\text{ }x\\ne 2[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135152098\">We can write this in interval notation as [latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_01_04_09\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137444218\">\r\n<h3>Example 10:\u00a0 Finding the Domain of a Composite Function Involving Radicals<\/h3>\r\n<p id=\"fs-id1165135547426\">Find the domain of [latex]\\left(f\\circ g\\right)\\left(x\\right)[\/latex] where [latex]f\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{x+2}[\/latex] and [latex]g\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{3-x}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137812351\">[reveal-answer q=\"fs-id1165137812351\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137812351\"]\r\n<p id=\"fs-id1165137812353\">Because we cannot take the square root of a negative number, the domain of [latex]g[\/latex] is [latex]\\left(-\\infty ,3\\right].[\/latex] Now we check the domain of the composite function<\/p>\r\n\r\n<div id=\"fs-id1165137874760\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{\\sqrt[\\leftroot{1}\\uproot{2} ]{3-x}+2}[\/latex]<\/div>\r\n<div>[latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137871761\">For [latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{\\sqrt[\\leftroot{1}\\uproot{2} ]{3-x}+2},[\/latex] we know that [latex]\\sqrt{3-x}+2\\ge 0,[\/latex] since the radicand of a square root must be positive. Since square roots are positive, [latex]\\text{}\\sqrt{3-x}\\ge 0.[\/latex] Squaring both sides gives us [latex]3-x\\ge 0.[\/latex] Therefore, [latex]x\\le3[\/latex] which gives a domain of [latex]\\left(-\\infty ,3\\right][\/latex].<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\nThis example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\\circ g[\/latex] can contain values that are not in the domain of [latex]f,[\/latex] though they must be in the domain of [latex]g.[\/latex] Note that the domain of [latex]f[\/latex] is [latex]\\left(-2,\\text{ }\\infty\\right).[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135435597\" class=\"precalculus tryit\">\r\n<h3>Try it #6<\/h3>\r\n<div id=\"ti_01_04_07\">\r\n<div id=\"fs-id1165137937116\">\r\n<p id=\"fs-id1165137937117\">Find the domain of [latex]\\left(f\\circ g\\right)\\left(x\\right)[\/latex] where [latex]f\\left(x\\right)=\\frac{1}{x-2}[\/latex] and [latex]g\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{x+4}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137469124\">[reveal-answer q=\"fs-id1165137469124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137469124\"]\r\n<p id=\"fs-id1165137827717\">[latex]\\left[-4,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137425738\" class=\"bc-section section\">\r\n<div class=\"precalculus media\">\r\n<p id=\"fs-id1165137600228\">Access these online resources for additional instruction and practice with composite functions.<\/p>\r\n\r\n<ul id=\"fs-id1165137600231\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/compfunction\">Composite Functions<\/a><\/li>\r\n<\/ul>\r\nhttps:\/\/www.youtube.com\/watch?v=qxBmISCJSME\r\n<ul id=\"fs-id1165137600231\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/compfuncnot\">Composite Function Notation Application<\/a><\/li>\r\n<\/ul>\r\nhttps:\/\/youtu.be\/VI2kJp69jNg\r\n<ul id=\"fs-id1165137600231\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/compfuncgraph\">Composite Functions Using Graphs<\/a><\/li>\r\n<\/ul>\r\nhttps:\/\/youtu.be\/b-i7N0hE-Ys\r\n<ul id=\"fs-id1165137600231\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/decompfunction\">Decompose Functions<\/a><\/li>\r\n<\/ul>\r\nhttps:\/\/youtu.be\/gFSSk8jaAwA\r\n<ul id=\"fs-id1165137600231\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/compfuncvalue\">Composite Function Values<\/a><\/li>\r\n<\/ul>\r\nhttps:\/\/youtu.be\/y2kJI9XnyLY\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135388428\" class=\"key-equations\">\r\n<h3>Key Equation<\/h3>\r\n<table id=\"eip-id1165134118229\" summary=\"..\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 119px;\">Composite function<\/td>\r\n<td class=\"border\" style=\"width: 400.6px;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165137768015\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1165137754842\">\r\n \t<li>When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.<\/li>\r\n \t<li>The function produced by combining two functions is a composite function.<\/li>\r\n \t<li>The order of function composition must be considered when interpreting the meaning of composite functions.<\/li>\r\n \t<li>A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.<\/li>\r\n \t<li>A composite function can be evaluated from a table.<\/li>\r\n \t<li>A composite function can be evaluated from a graph.<\/li>\r\n \t<li>A composite function can be evaluated from a formula.<\/li>\r\n \t<li>Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.<\/li>\r\n \t<li>Functions can often be decomposed in more than one way.<\/li>\r\n \t<li>(Optional)The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl>\r\n \t<dt>commutative property<\/dt>\r\n \t<dd>the order of the operations being preformed does not matter if the commutative property holds<\/dd>\r\n \t<dt>composite function<\/dt>\r\n \t<dd id=\"fs-id1165137832426\">the new function formed by function composition, when the output of one function is used as the input of another<\/dd>\r\n \t<dd><\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Create a new function by composition of functions.<\/li>\n<li>Evaluate composite functions.<\/li>\n<li>Decompose a composite function into its component functions.<\/li>\n<\/ul>\n<p>Optional:<\/p>\n<ul>\n<li>Find the domain of a composite function.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165134094620\">Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.<\/p>\n<p id=\"fs-id1165134038788\">Using descriptive variables, we can notate these two functions. The function [latex]C\\left(T\\right)[\/latex] gives the cost [latex]C[\/latex] of heating a house for a given average daily temperature in [latex]T[\/latex] degrees Celsius. The function [latex]T\\left(d\\right)[\/latex] gives the average daily temperature on day [latex]d[\/latex] of the year. For any given day, [latex]\\text{Cost}=C\\left(T\\left(d\\right)\\right)[\/latex] means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature [latex]T\\left(d\\right).[\/latex] For example, we could evaluate [latex]T\\left(5\\right)[\/latex] to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the <span class=\"no-emphasis\">cost function<\/span> at that temperature. We would write [latex]C\\left(T\\left(5\\right)\\right).[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-3158 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142304\/336d97bc881544921ca4876a504752b89148f9fd.jpeg\" alt=\"Explanation of C(T(5)), which is the cost for the temperature and T(5) is the temperature on day 5.\" width=\"487\" height=\"140\" \/><\/p>\n<p id=\"fs-id1165137460512\">By combining these two relationships into one function, we have performed function composition, which is the focus of this section.<\/p>\n<div id=\"fs-id1165135690697\" class=\"bc-section section\">\n<h3>Create a Function by Composition of Functions<\/h3>\n<p id=\"fs-id1165137451258\">We can create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of <span class=\"no-emphasis\">combining functions<\/span> so that the output of one function becomes the input of another is known as a <span class=\"no-emphasis\">composition of functions<\/span><strong>.<\/strong> The resulting function is known as a <strong>composite function<\/strong>. We represent this combination by the following notation:<\/p>\n<div id=\"fs-id1165137605973\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137466004\">We read the left-hand side as [latex]\\text{\u201c}f[\/latex] composed with [latex]g[\/latex] at [latex]x,\\text{\u201d}[\/latex] and the right-hand side as [latex]\\text{\u201c}f[\/latex] of [latex]g[\/latex] of [latex]x.\\text{\u201d}[\/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\\circ[\/latex] is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number.\u00a0\u00a0It is important to realize that the product of functions [latex]fg[\/latex] is not the same as the function composition [latex]f\\left(g\\left(x\\right)\\right),[\/latex] because, in general, [latex]f\\left(x\\right)g\\left(x\\right)\\ne f\\left(g\\left(x\\right)\\right).[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function [latex]g[\/latex] takes the input [latex]x[\/latex] first and yields an output [latex]g\\left(x\\right).[\/latex] Then the function [latex]f[\/latex] takes [latex]g\\left(x\\right)[\/latex] as an input and yields an output [latex]f\\left(g\\left(x\\right)\\right).[\/latex]<\/span><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-3176 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15153746\/c02500912de8e653c7e833a6aa2d53696a4c7630.jpeg\" alt=\"explanation of a composite function\" width=\"487\" height=\"171\" \/><\/p>\n<div id=\"fs-id1165135690697\" class=\"bc-section section\">\n<p id=\"fs-id1165137851405\">In general, [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] are different functions. In other words, in many cases [latex]f\\left(g\\left(x\\right)\\right)\\ne g\\left(f\\left(x\\right)\\right)[\/latex] for all [latex]x.[\/latex] We will also see that sometimes two functions can be composed only in one specific order.<\/p>\n<p id=\"fs-id1165135189964\">For example, if [latex]f\\left(x\\right)={x}^{2}[\/latex] and [latex]g\\left(x\\right)=x+2,[\/latex] then<\/p>\n<div id=\"fs-id1165134047682\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(g\\left(x\\right)\\right)&=f\\left(x+2\\right)\\hfill \\\\ \\text{ }&={\\left(x+2\\right)}^{2}\\hfill \\\\ \\text{ }&={x}^{2}+4x+4\\hfill \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1165137595004\">but<\/p>\n<div id=\"fs-id1165137461653\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}g\\left(f\\left(x\\right)\\right)&=g\\left({x}^{2}\\right)\\hfill \\\\ \\text{ }&={x}^{2}+2.\\hfill \\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137430833\">These expressions are not equal for all values of [latex]x,[\/latex] so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value [latex]x=-\\frac{1}{2}.[\/latex]<\/p>\n<p id=\"fs-id1165137531472\">Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.<\/p>\n<div id=\"fs-id1165134190746\">\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137783932\">When the output of one function is used as the input of another, we call the entire operation a <strong>composition of functions<\/strong>. For any input [latex]x[\/latex] and functions [latex]f[\/latex] and [latex]g,[\/latex] this action defines a composite function, which we write as [latex]f\\circ g[\/latex] such that<\/p>\n<div id=\"fs-id1165137836657\" style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165135503928\">The domain of the composite function [latex]f\\circ g[\/latex] is all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"Example_01_04_02\" class=\"textbox examples\">\n<div id=\"fs-id1165137544021\">\n<div id=\"fs-id1165137482990\">\n<h3>Example 1:\u00a0 Determining Whether Composition of Functions is Commutative<\/h3>\n<p id=\"fs-id1165137600071\">Using the functions provided, find [latex]f\\left(g\\left(4\\right)\\right)[\/latex] and [latex]g\\left(f\\left(4\\right)\\right).[\/latex] Determine whether the composition of the functions is <span class=\"no-emphasis\">commutative<\/span>.<\/p>\n<div id=\"fs-id1165135530570\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+1\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }g\\left(x\\right)=3-x[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165135195227\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135195227\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135195227\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137603003\">Let\u2019s begin by evaluating [latex]g\\left(4\\right)=3-4=-1.[\/latex]\u00a0 The output of [latex]g[\/latex] is the input to [latex]f[\/latex] so<\/p>\n<div id=\"fs-id1165135499437\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} f\\left(g\\left(4\\right)\\right)&=f\\left(-1\\right)\\\\\\text{ }&=2\\left(-1\\right)+1\\hfill \\\\ \\text{ }&=-1\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137786392\">For the second composition\u00a0[latex]g\\left(f\\left(4\\right)\\right),[\/latex] we begin by evaluating [latex]f\\left(4\\right)=2\\left(4\\right)+1=9.[\/latex] The output of [latex]f[\/latex] will be the input to [latex]g[\/latex] so<\/p>\n<div id=\"fs-id1165135440456\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} g\\left(f\\left(4\\right)\\right)&=g\\left(9\\right)\\\\\\text{}&=3-9\\hfill \\\\ \\text{ }&=-6\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135160089\">We find that [latex]g\\left(f\\left(4\\right)\\right)\\ne f\\left(g\\left(4\\right)\\right),[\/latex] so the operation of function composition is not commutative and order matters.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_04_03\" class=\"textbox examples\">\n<div id=\"fs-id1165137603612\">\n<div id=\"fs-id1165137761836\">\n<h3>Example 2:\u00a0 Interpreting Composite Functions<\/h3>\n<p id=\"fs-id1165137695118\">The function [latex]c\\left(s\\right)[\/latex] gives the number of calories burned completing [latex]s[\/latex] sit-ups, and [latex]s\\left(t\\right)[\/latex] gives the number of sit-ups a person can complete in [latex]t[\/latex] minutes. Interpret [latex]c\\left(s\\left(3\\right)\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134170107\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134170107\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134170107\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137659141\">The inside expression in the composition is [latex]s\\left(3\\right).[\/latex] Because the input to the <em>s<\/em>-function is time, [latex]t=3[\/latex] represents 3 minutes, and [latex]s\\left(3\\right)[\/latex] is the number of sit-ups completed in 3 minutes.<\/p>\n<p id=\"fs-id1165137844294\">Using [latex]s\\left(3\\right)[\/latex] as the input to the function [latex]c\\left(s\\right)[\/latex] gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_04_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137551439\">\n<div id=\"fs-id1165135512587\">\n<h3>Example 3:\u00a0 Investigating the Order of Function Composition<\/h3>\n<p id=\"fs-id1165137643603\">Suppose [latex]f\\left(x\\right)[\/latex] gives the number of miles that can be driven in [latex]x[\/latex] hours and [latex]g\\left(y\\right)[\/latex] gives the number of gallons of gas used in driving [latex]y[\/latex] miles. Which of these expressions is meaningful: [latex]f\\left(g\\left(y\\right)\\right)[\/latex] or [latex]g\\left(f\\left(x\\right)\\right)?[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137451433\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137451433\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137451433\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135545914\">The function [latex]y=f\\left(x\\right)[\/latex] is a function whose output is the number of miles driven corresponding to the number of hours driven.<\/p>\n<div id=\"fs-id1165137648357\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\text{number of miles }=f\\text{ }\\left(\\text{number of hours}\\right)[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137749316\">The function [latex]g\\left(y\\right)[\/latex] is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:<\/p>\n<div id=\"fs-id1165137531080\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\text{number of gallons }=g\\text{ }\\left(\\text{number of miles}\\right)[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135190218\">The expression [latex]g\\left(y\\right)[\/latex] takes miles as the input and a number of gallons as the output. The function [latex]f\\left(x\\right)[\/latex] requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression [latex]f\\left(g\\left(y\\right)\\right)[\/latex] is meaningless.<\/p>\n<p id=\"fs-id1165137863600\">The expression [latex]f\\left(x\\right)[\/latex] takes hours as input and a number of miles driven as the output. The function [latex]g\\left(y\\right)[\/latex] requires a number of miles as the input. Using [latex]f\\left(x\\right)[\/latex] (miles driven) as an input value for [latex]g\\left(y\\right),[\/latex] where gallons of gas depends on miles driven, does make sense. The expression [latex]g\\left(f\\left(x\\right)\\right)[\/latex] makes sense, and will yield the number of gallons of gas used, [latex]g,[\/latex] driving a certain number of miles, [latex]f\\left(x\\right),[\/latex] in [latex]x[\/latex] hours.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137768571\" class=\"precalculus qa key-takeaways\">\n<h3>Q&amp;A<\/h3>\n<p>Are there any situations where [latex]f\\left(g\\left(y\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex] would both be meaningful or useful expressions?<\/p>\n<p id=\"fs-id1165137682006\"><em>Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1165134558042\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_01_04_01\">\n<div id=\"fs-id1165135560725\">\n<p id=\"fs-id1165135560726\">The gravitational force on a planet a distance [latex]r[\/latex] from the sun is given by the function [latex]G\\left(r\\right).[\/latex] The acceleration of a planet subjected to any force [latex]F[\/latex] is given by the function [latex]a\\left(F\\right).[\/latex] Form a meaningful composition of these two functions, and explain what it means.<\/p>\n<\/div>\n<div id=\"fs-id1165137417357\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137417357\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137417357\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137417358\">A gravitational force is still a force, so [latex]a\\left(G\\left(r\\right)\\right)[\/latex] makes sense as the acceleration of a planet at a distance [latex]r[\/latex] from the sun (due to gravity), but [latex]G\\left(a\\left(F\\right)\\right)[\/latex] does not make sense.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137560475\" class=\"bc-section section\">\n<h3>Evaluating Composite Functions<\/h3>\n<p id=\"fs-id1165135168147\">Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function\u2019s output as the input for the outer function.<\/p>\n<div id=\"fs-id1165137760886\" class=\"bc-section section\">\n<h4>Evaluating Composite Functions Using Tables<\/h4>\n<p id=\"fs-id1165137725253\">When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.<\/p>\n<div id=\"Example_01_04_05\" class=\"textbox examples\">\n<div id=\"fs-id1165137416770\">\n<div id=\"fs-id1165137465672\">\n<h3>Example 4:\u00a0 Using a Table to Evaluate a Composite Function<\/h3>\n<p id=\"fs-id1165135177664\">Using <a class=\"autogenerated-content\" href=\"#Table_01_04_01\">Table 1<\/a>, evaluate [latex]f\\left(g\\left(3\\right)\\right)[\/latex] and [latex]g\\left(f\\left(3\\right)\\right).[\/latex]<\/p>\n<table id=\"Table_01_04_01\" summary=\"Five rows and three columns. The first column is labeled, \u201cx\u201d, the second column is labeled, \u201cf(x)\u201d, and the third column is labeled, \u201cg(x)\u201d. We have the following values for f(x): f(1)=6, f(2)=8, f(3)=3, and f(4)=1. And for g(1)=3, g(2)=5, g(3)=2, and g(4)=7.\">\n<caption>Table 1<\/caption>\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th class=\"border\" style=\"text-align: center;\">[latex]x[\/latex]<\/th>\n<th class=\"border\" style=\"text-align: center;\">[latex]f\\left(x\\right)[\/latex]<\/th>\n<th class=\"border\" style=\"text-align: center;\">[latex]g\\left(x\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\n<td class=\"border\" style=\"text-align: center;\">6<\/td>\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\n<td class=\"border\" style=\"text-align: center;\">5<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\n<td class=\"border\" style=\"text-align: center;\">7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165135582219\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135582219\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135582219\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135582221\">To evaluate [latex]f\\left(g\\left(3\\right)\\right),[\/latex] we start from the inside with the input value 3. We then evaluate the inside expression [latex]g\\left(3\\right)[\/latex] using the table that defines the function [latex]g:[\/latex] [latex]g\\left(3\\right)=2.[\/latex] We can then use that result as the input to the function [latex]f,[\/latex] so [latex]g\\left(3\\right)[\/latex] is replaced by 2 and we get [latex]f\\left(2\\right).[\/latex] Then, using the table that defines the function [latex]f,[\/latex] we find that [latex]f\\left(2\\right)=8.[\/latex]<\/p>\n<div id=\"fs-id1165137415997\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}g\\left(3\\right)&=2\\hfill \\\\ f\\left(g\\left(3\\right)\\right)&=f\\left(2\\right)=8\\hfill \\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165134259282\">To evaluate [latex]g\\left(f\\left(3\\right)\\right),[\/latex] we first evaluate the inside expression [latex]f\\left(3\\right)[\/latex] using the first table: [latex]f\\left(3\\right)=3.[\/latex] Then, using the table for [latex]g\\text{, }[\/latex] we can evaluate<\/p>\n<div id=\"fs-id1165137841687\" class=\"unnumbered\" style=\"text-align: center;\">[latex]g\\left(f\\left(3\\right)\\right)=g\\left(3\\right)=2[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165134231482\"><a class=\"autogenerated-content\" href=\"#Table_01_04_02\">Table 2<\/a> shows the composite functions [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] as tables.<\/p>\n<table id=\"Table_01_04_02\" summary=\"Two rows and five columns. When x=3, g(3)=2, f(g(3))=8, f(3)=3, and g(f(3))=2.\">\n<caption>Table 2<\/caption>\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]g\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]f\\left(g\\left(x\\right)\\right)[\/latex]<\/td>\n<td class=\"border\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]g\\left(f\\left(x\\right)\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137724943\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_01_04_02\">\n<div id=\"fs-id1165134297600\">\n<p id=\"fs-id1165134297603\">Using <a class=\"autogenerated-content\" href=\"#Table_01_04_01\">Table 1<\/a>, evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex] and [latex]g\\left(f\\left(4\\right)\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165134258662\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134258662\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134258662\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137836967\">[latex]f\\left(g\\left(1\\right)\\right)=f\\left(3\\right)=3[\/latex] and [latex]g\\left(f\\left(4\\right)\\right)=g\\left(1\\right)=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137756068\" class=\"bc-section section\">\n<h4>Evaluating Composite Functions Using Graphs<\/h4>\n<p id=\"fs-id1165137428192\">When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the [latex]x\\text{-}[\/latex] and [latex]y\\text{-}[\/latex]axes of the graphs.<\/p>\n<div id=\"fs-id1165137634364\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137660530\"><strong>Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs. <\/strong><\/p>\n<ol id=\"fs-id1165133248553\" type=\"1\">\n<li>Locate the given input to the inner function on the [latex]x\\text{-}[\/latex]axis of its graph.<\/li>\n<li>Read off the output of the inner function from the [latex]y\\text{-}[\/latex]axis of its graph.<\/li>\n<li>Locate the inner function output on the [latex]x\\text{-}[\/latex]axis of the graph of the outer function.<\/li>\n<li>Read the output of the outer function from the [latex]y\\text{-}[\/latex]axis of its graph. This is the output of the composite function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_04_06\" class=\"textbox examples\">\n<div id=\"fs-id1165137758193\">\n<div id=\"fs-id1165137758196\">\n<h3>Example 5:\u00a0 Using a Graph to Evaluate a Composite Function<\/h3>\n<p id=\"fs-id1165134226787\">Using <a class=\"autogenerated-content\" href=\"#Figure_01_04_002\">Figure 1<\/a>, evaluate [latex]f\\left(g\\left(1\\right)\\right).[\/latex]<\/p>\n<div id=\"attachment_3160\" style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3160\" class=\"size-full wp-image-3160\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142515\/710401844b9cc0b203a1628994a0471665da4606.jpeg\" alt=\"Explanation of the composite function.\" width=\"975\" height=\"543\" \/><\/p>\n<p id=\"caption-attachment-3160\" class=\"wp-caption-text\">Figure 1<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135545946\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135545946\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135545946\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135545948\">To evaluate [latex]f\\left(g\\left(1\\right)\\right),[\/latex] we start with the inside evaluation. See <a class=\"autogenerated-content\" href=\"#Figure_01_04_004\">Figure 2<\/a>.<\/p>\n<div id=\"attachment_3162\" style=\"width: 985px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3162\" class=\"wp-image-3162 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142744\/0a3ffe6f2329045c480a8c60df63fe2df0466dcf.jpeg\" alt=\"Two graphs of a positive parabola (g(x)) and a negative parabola (f(x)). The following points are plotted: g(1)=3 and f(3)=6.\" width=\"975\" height=\"543\" \/><\/p>\n<p id=\"caption-attachment-3162\" class=\"wp-caption-text\">Figure 2<\/p>\n<\/div>\n<p id=\"fs-id1165137566106\">We evaluate [latex]g\\left(1\\right)[\/latex] using the graph of [latex]g\\left(x\\right),[\/latex] finding the input of 1 on the [latex]x\\text{-}[\/latex]axis and finding the output value of the graph at that input. Here, [latex]g\\left(1\\right)=3.[\/latex] We use this value as the input to the function [latex]f.[\/latex]<\/p>\n<div id=\"fs-id1165135432881\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(g\\left(1\\right)\\right)=f\\left(3\\right)[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137831420\">We can then evaluate the composite function by looking to the graph of [latex]f\\left(x\\right),[\/latex] finding the input of 3 on the [latex]x\\text{-}[\/latex]axis and reading the output value of the graph at this input. Here, [latex]f\\left(3\\right)=6,[\/latex] so [latex]f\\left(g\\left(1\\right)\\right)=6.[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p id=\"fs-id1165137602379\"><a class=\"autogenerated-content\" href=\"#Figure_01_04_005\">Figure 3<\/a> shows how we can mark the graphs with arrows to trace the path from the input value to the output value.<\/p>\n<div id=\"attachment_3163\" style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3163\" class=\"size-full wp-image-3163\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/01\/15142833\/1cb774956db9f261e1d4be38aaffd0d296a652ed-1.jpeg\" alt=\"Two graphs of a positive and negative parabola.\" width=\"975\" height=\"520\" \/><\/p>\n<p id=\"caption-attachment-3163\" class=\"wp-caption-text\">Figure 3<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137723402\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_01_04_03\">\n<div id=\"fs-id1165137588093\">\n<p>Using <a class=\"autogenerated-content\" href=\"#Figure_01_04_002\">Figure 1<\/a>, evaluate [latex]g\\left(f\\left(2\\right)\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137843133\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137843133\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137843133\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137502186\">[latex]g\\left(f\\left(2\\right)\\right)=g\\left(5\\right)=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137452389\" class=\"bc-section section\">\n<h4>Evaluating Composite Functions Using Formulas<\/h4>\n<p id=\"fs-id1165137432176\">When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression.<\/p>\n<p id=\"fs-id1165137567159\">While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition [latex]f\\left(g\\left(x\\right)\\right).[\/latex] To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like [latex]f\\left(t\\right)={t}^{2}-t,[\/latex] we substitute the value inside the parentheses into the formula wherever we see the input variable.<\/p>\n<div id=\"fs-id1165137584280\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137676972\"><strong>Given a formula for a composite function, evaluate the function. <\/strong><\/p>\n<ol id=\"fs-id1165137564125\" type=\"1\">\n<li>Evaluate the inside function using the input value or variable provided.<\/li>\n<li>Use the resulting output as the input to the outside function.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_04_07\" class=\"textbox examples\">\n<div id=\"fs-id1165134070851\">\n<div id=\"fs-id1165134070853\">\n<h3>Example 6:\u00a0 Evaluating a Composition of Functions Expressed as Formulas<\/h3>\n<p id=\"fs-id1165137447886\">Given [latex]f\\left(t\\right)={t}^{2}-t[\/latex] and [latex]h\\left(x\\right)=3x+2,[\/latex] evaluate<\/p>\n<ol>\n<li>[latex]f\\left(h\\left(1\\right)\\right).[\/latex]<\/li>\n<li>[latex]f\\left(h\\left(x\\right)\\right).[\/latex]<\/li>\n<li>[latex]h\\left(f\\left(t\\right)\\right).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165135543322\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135543322\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135543322\" class=\"hidden-answer\" style=\"display: none\">\n<div><\/div>\n<div style=\"padding-left: 30px;\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">1.\u00a0 Because the inside expression is [latex]h\\left(1\\right),[\/latex] we start by evaluating [latex]h\\left(x\\right)[\/latex] at 1.<\/span><\/p>\n<p style=\"padding-left: 60px; text-align: center;\">[latex]\\begin{align*}h\\left(1\\right)&=3\\left(1\\right)+2\\\\ &=5\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">Then [latex]f\\left(h\\left(1\\right)\\right)=f\\left(5\\right),[\/latex] so we evaluate [latex]f\\left(t\\right)[\/latex] at an input of 5.<\/p>\n<p style=\"padding-left: 60px; text-align: center;\">[latex]\\begin{align*}f\\left(h\\left(1\\right)\\right)&=f\\left(5\\right)\\\\ &={5}^{2}-5\\\\ &=20\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p style=\"text-align: left;\">2.\u00a0 \u00a0The inside expression for\u00a0[latex]f\\left(h\\left(x\\right)\\right)[\/latex] is [latex]h\\left(x\\right)[\/latex] so we use its output, [latex]3x+2,[\/latex] as the input to the function [latex]f[\/latex].\u00a0 We then must evaluate [latex]f\\left(3x+2\\right).[\/latex]\u00a0 Begin by replacing each variable [latex]t,[\/latex] with [latex]3x+2[\/latex] in the function [latex]f[\/latex] to get<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(3x+2\\right)=\\left(3x+2\\right)^2-\\left(3x+2\\right).[\/latex]<\/p>\n<p>Then simplify to get<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(h\\left(x\\right)\\right)=9x^2+6x+6x+4-3x-2=9x^2+9x+2.[\/latex][latex]\\\\[\/latex]<\/p>\n<p>3.\u00a0The inside expression for\u00a0[latex]h\\left(f\\left(t\\right)\\right)[\/latex] is [latex]f\\left(t\\right),[\/latex] so we use its output, [latex]t^2-t,[\/latex] as the input to the function [latex]h[\/latex].\u00a0 We then must evaluate [latex]h\\left(t^2-t\\right).[\/latex]\u00a0 Begin by replacing each variable [latex]x,[\/latex] with [latex]t^2-t[\/latex] in the function [latex]h[\/latex] to get<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(t^2-t\\right)=3\\left(t^2-t\\right)+2.[\/latex]<\/p>\n<p>Then simplify to get<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(f\\left(t\\right)\\right)=3t^2-3t+2[\/latex]<\/p>\n<\/div>\n<div class=\"unnumbered\"><\/div>\n<div class=\"unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137482990\">\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137482990\">\n<h3>Example 7:\u00a0 Determining Whether Composition of Functions is Commutative<\/h3>\n<p id=\"fs-id1165137600071\">Using the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right).[\/latex] Determine whether the composition of the functions is <span class=\"no-emphasis\">commutative<\/span>.<\/p>\n<div id=\"fs-id1165135530570\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+1\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }g\\left(x\\right)=3-x[\/latex]<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462565\">Show Solution<\/span><\/p>\n<div id=\"q462565\" class=\"hidden-answer\" style=\"display: none\">\n<div>\n<p id=\"fs-id1165137603003\">Let\u2019s begin by substituting [latex]g\\left(x\\right)=3-x[\/latex] into [latex]f\\left(x\\right).[\/latex]<\/p>\n<div id=\"fs-id1165135499437\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} f\\left(g\\left(x\\right)\\right)&=f\\left(3-x\\right)\\\\\\text{ }&=2\\left(3-x\\right)+1\\hfill \\\\ \\text{ }&=6-2x+1\\hfill \\\\ \\text{ }&=7-2x\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137786392\">Now we can substitute [latex]f\\left(x\\right)=2x+1[\/latex] into [latex]g\\left(x\\right).[\/latex]<\/p>\n<div id=\"fs-id1165135440456\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*} g\\left(f\\left(x\\right)\\right)&=g\\left(2x+1\\right)\\\\\\text{}&=3-\\left(2x+1\\right)\\hfill \\\\ \\text{ }&=3-2x-1\\hfill \\\\ \\text{ }&=-2x+2\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135160089\">We find that [latex]g\\left(f\\left(x\\right)\\right)\\ne f\\left(g\\left(x\\right)\\right),[\/latex] so the operation of function composition is not commutative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1165137570158\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_01_04_04\">\n<div>\n<p id=\"fs-id1165134544990\">Given [latex]f\\left(t\\right)={t}^{2}-t[\/latex] and [latex]h\\left(x\\right)=3x+2,[\/latex] evaluate<\/p>\n<ol id=\"fs-id1165137758652\" type=\"a\">\n<li>[latex]h\\left(f\\left(2\\right)\\right)[\/latex]<\/li>\n<li>[latex]h\\left(f\\left(-2\\right)\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165137434572\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137434572\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137434572\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134261859\">a. 8; b. 20<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135567444\" class=\"bc-section section\">\n<div id=\"fs-id1165135547254\">\n<h3>Decomposing a Composite Function into its Component Functions<\/h3>\n<p>In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to\u00a0decompose a composite function, so we may choose the decomposition that appears to be most expedient.\u00a0 However, in calculus, you will be studying the chain rule in order to find the derivative of a composite function.\u00a0 While we can\u2019t begin to try to define the concept of a derivative at this point, we can help you begin to think about an \u201cinner\u201d and \u201couter\u201d function in a composition.\u00a0 When you study the chain rule, you will\u00a0 think of the outer function as [latex]f\\left(x\\right)[\/latex] and the inner function as [latex]g\\left(x\\right).[\/latex]<\/p>\n<p>Let\u2019s consider the function [latex]h\\left(x\\right)={\\left(x+10\\right)}^{5}.[\/latex]\u00a0 This reminds us of the function [latex]f\\left(x\\right)={x}^{5}[\/latex] where the input [latex]x[\/latex] has been replaced by [latex]x+10.[\/latex]\u00a0 We can see this as though we have a place holder where we would normally see the [latex]x.[\/latex]\u00a0 This would look like [latex]{\\left(\\text{__}\\right)}^{5},[\/latex] where we are thinking about the \u201couter\u201d function as being the function that raises the \u201cinside\u201d to the 5<sup>th<\/sup> power.\u00a0 We would let the outside function be [latex]f\\left(x\\right)={x}^{5}[\/latex] and the \u201cinside&#8221; function be [latex]g\\left(x\\right)=x+10.[\/latex]\u00a0 Can you see that [latex]f\\left(g\\left(x\\right)\\right)={\\left(x+10\\right)}^{5}?[\/latex]<\/p>\n<p>Now let\u2019s try a function that isn\u2019t as obvious with outer and inner functions.\u00a0 Consider [latex]k\\left(x\\right)={2}^{3x+4}.[\/latex]\u00a0 This reminds us of the function [latex]f\\left(x\\right)={2}^{x}[\/latex] where the [latex]x[\/latex] has been replaced by [latex]3x+4.[\/latex]\u00a0 Let\u2019s let our outer function be [latex]f\\left(x\\right)={2}^{x},[\/latex] and our inner function be [latex]g\\left(x\\right)=3x+4.[\/latex]\u00a0 Now form the composite function [latex]f\\left(g\\left(x\\right)\\right).[\/latex]\u00a0 You get [latex]k\\left(x\\right)={2}^{3x+4}.[\/latex]<\/p>\n<p>Sometimes the easiest way to see the \u201cinside\u201d and \u201coutside\u201d is to consider what a simpler function would look like if the input were simply x and not some more complex expression.<\/p>\n<p>Let\u2019s do one more.\u00a0 Consider [latex]r\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{2x+5}}.[\/latex]\u00a0 \u00a0We can see that this fits the form of a simpler function [latex]f\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{x}}[\/latex] where the [latex]x[\/latex] has been replaced by [latex]2x+5.[\/latex]\u00a0 Our outer function would be [latex]f\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{x}}[\/latex]\u00a0and our inner function would be [latex]g\\left(x\\right)=2x+5.[\/latex]\u00a0 If you form the composite function [latex]f\\left(g\\left(x\\right)\\right),[\/latex] you will get\u00a0[latex]r\\left(x\\right)=\\frac{3}{\\sqrt[\\leftroot{1}\\uproot{2} ]{2x+5}}.[\/latex]<\/p>\n<div id=\"Example_01_04_10\" class=\"textbox examples\">\n<div id=\"fs-id1165134319721\">\n<div id=\"fs-id1165134319723\">\n<h3>Example 8:\u00a0 Decomposing a Function<\/h3>\n<p id=\"fs-id1165137771903\">Write [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.<\/p>\n<\/div>\n<div id=\"fs-id1165135194520\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135194520\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135194520\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135194522\">We are looking for two functions, [latex]g[\/latex] and [latex]h,[\/latex] so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right).[\/latex] To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right).[\/latex] As one possibility, we might notice that the expression [latex]5-{x}^{2}[\/latex] is the inside of the square root. We could then decompose the function as<\/p>\n<div id=\"fs-id1165137413960\" class=\"unnumbered\" style=\"text-align: center;\">[latex]h\\left(x\\right)=5-{x}^{2}\\text{ and }g\\left(x\\right)=\\sqrt{x}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135250635\">We can check our answer by recomposing the functions.<\/p>\n<div id=\"fs-id1165137730234\" class=\"unnumbered\" style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g\\left(5-{x}^{2}\\right)=\\sqrt{5-{x}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165131878633\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_01_04_08\">\n<div id=\"fs-id1165137501975\">\n<p id=\"fs-id1165137501976\">Write [latex]f\\left(x\\right)=\\frac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.<\/p>\n<\/div>\n<div id=\"fs-id1165137411051\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137411051\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137411051\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137411053\">Possible answer:<\/p>\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\n<p>[latex]h\\left(x\\right)=\\frac{4}{3-x}[\/latex]<\/p>\n<p>[latex]f=h\\circ g[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Finding the Domain of a Composite Function (Optional)<\/h3>\n<p id=\"fs-id1165135519324\">As we discussed previously, the <span class=\"no-emphasis\">domain of a composite function<\/span> such as [latex]f\\circ g[\/latex] is dependent on the domain of [latex]g[\/latex] and the domain of [latex]f.[\/latex] It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as [latex]f\\circ g.[\/latex] Let us assume we know the domains of the functions [latex]f[\/latex] and [latex]g[\/latex] separately. If we write the composite function for an input [latex]x[\/latex] as [latex]f\\left(g\\left(x\\right)\\right),[\/latex] we can see right away that [latex]x[\/latex] must be a member of the domain of [latex]g[\/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\\left(x\\right)[\/latex] must be a member of the domain of [latex]f,[\/latex] otherwise the second function evaluation in [latex]f\\left(g\\left(x\\right)\\right)[\/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\\circ g[\/latex] consists of only those inputs in the domain of [latex]g[\/latex] that produce outputs from [latex]g[\/latex] belonging to the domain of [latex]f.[\/latex] Note that the domain of [latex]f[\/latex] composed with [latex]g[\/latex] is the set of all [latex]x[\/latex] such that [latex]x[\/latex] is in the domain of [latex]g[\/latex] and [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex]<\/p>\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p>The <strong>domain of a composite function\u00a0<\/strong>[latex]f\\left(g\\left(x\\right)\\right)[\/latex] is the set of those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137663970\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135203267\"><strong>Given a function composition [latex]f\\left(g\\left(x\\right)\\right),[\/latex] determine its domain.<\/strong><\/p>\n<ol id=\"fs-id1165137714200\" type=\"1\">\n<li>Find the domain of [latex]g.[\/latex]<\/li>\n<li>Find the domain of [latex]f.[\/latex]<\/li>\n<li>Find those inputs [latex]x[\/latex] in the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is in the domain of [latex]f.[\/latex] That is, exclude those inputs [latex]x[\/latex] from the domain of [latex]g[\/latex] for which [latex]g\\left(x\\right)[\/latex] is not in the domain of [latex]f.[\/latex] The resulting set is the domain of [latex]f\\circ g.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_01_04_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137646695\">\n<div id=\"fs-id1165137646697\">\n<h3>Example 9:\u00a0 Finding the Domain of a Composite Function<\/h3>\n<p id=\"fs-id1165135640630\">Find the domain of<\/p>\n<div id=\"fs-id1165135349196\" class=\"unnumbered\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }f\\left(x\\right)=\\frac{5}{x-1}\\text{ and }\\text{ }\\text{ }g\\left(x\\right)=\\frac{4}{3x-2}[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1165137627920\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137627920\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137627920\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137627922\">The domain of [latex]g\\left(x\\right)[\/latex] consists of all real numbers except [latex]x=\\frac{2}{3},[\/latex] since that input value would cause us to divide by 0. Likewise, the domain of [latex]f[\/latex] consists of all real numbers except 1. So we need to exclude from the domain of [latex]g\\left(x\\right)[\/latex] that value of [latex]x[\/latex] for which [latex]g\\left(x\\right)=1.[\/latex]<\/p>\n<div id=\"fs-id1165137455472\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}\\frac{4}{3x-2}&=1\\hfill \\\\ \\text{ }4&=3x-2\\hfill \\\\ 6&=3x\\hfill \\\\ x&=2\\hfill \\end{align*}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137891240\">So the domain of [latex]f\\circ g[\/latex] is the set of all real numbers except [latex]\\frac{2}{3}[\/latex] and [latex]2.[\/latex] This means that[latex]\\\\[\/latex]<\/p>\n<div id=\"fs-id1165131959460\" class=\"unnumbered\" style=\"text-align: center;\">[latex]x\\ne \\frac{2}{3}\\text{ }\\text{ }\\text{or}\\text{ }\\text{ }x\\ne 2[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135152098\">We can write this in interval notation as [latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_01_04_09\" class=\"textbox examples\">\n<div id=\"fs-id1165137444218\">\n<h3>Example 10:\u00a0 Finding the Domain of a Composite Function Involving Radicals<\/h3>\n<p id=\"fs-id1165135547426\">Find the domain of [latex]\\left(f\\circ g\\right)\\left(x\\right)[\/latex] where [latex]f\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{x+2}[\/latex] and [latex]g\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{3-x}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137812351\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137812351\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137812351\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137812353\">Because we cannot take the square root of a negative number, the domain of [latex]g[\/latex] is [latex]\\left(-\\infty ,3\\right].[\/latex] Now we check the domain of the composite function<\/p>\n<div id=\"fs-id1165137874760\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{\\sqrt[\\leftroot{1}\\uproot{2} ]{3-x}+2}[\/latex]<\/div>\n<div>[latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137871761\">For [latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{\\sqrt[\\leftroot{1}\\uproot{2} ]{3-x}+2},[\/latex] we know that [latex]\\sqrt{3-x}+2\\ge 0,[\/latex] since the radicand of a square root must be positive. Since square roots are positive, [latex]\\text{}\\sqrt{3-x}\\ge 0.[\/latex] Squaring both sides gives us [latex]3-x\\ge 0.[\/latex] Therefore, [latex]x\\le3[\/latex] which gives a domain of [latex]\\left(-\\infty ,3\\right][\/latex].<\/p>\n<h3>Analysis<\/h3>\n<p>This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of [latex]f\\circ g[\/latex] can contain values that are not in the domain of [latex]f,[\/latex] though they must be in the domain of [latex]g.[\/latex] Note that the domain of [latex]f[\/latex] is [latex]\\left(-2,\\text{ }\\infty\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135435597\" class=\"precalculus tryit\">\n<h3>Try it #6<\/h3>\n<div id=\"ti_01_04_07\">\n<div id=\"fs-id1165137937116\">\n<p id=\"fs-id1165137937117\">Find the domain of [latex]\\left(f\\circ g\\right)\\left(x\\right)[\/latex] where [latex]f\\left(x\\right)=\\frac{1}{x-2}[\/latex] and [latex]g\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{x+4}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137469124\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137469124\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137469124\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137827717\">[latex]\\left[-4,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137425738\" class=\"bc-section section\">\n<div class=\"precalculus media\">\n<p id=\"fs-id1165137600228\">Access these online resources for additional instruction and practice with composite functions.<\/p>\n<ul id=\"fs-id1165137600231\">\n<li><a href=\"http:\/\/openstax.org\/l\/compfunction\">Composite Functions<\/a><\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Composite Functions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qxBmISCJSME?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ul id=\"fs-id1165137600231\">\n<li><a href=\"http:\/\/openstax.org\/l\/compfuncnot\">Composite Function Notation Application<\/a><\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Intro Composite Function Notation Application Problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VI2kJp69jNg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ul id=\"fs-id1165137600231\">\n<li><a href=\"http:\/\/openstax.org\/l\/compfuncgraph\">Composite Functions Using Graphs<\/a><\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Evaluate Composite Functions from Graphs\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/b-i7N0hE-Ys?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ul id=\"fs-id1165137600231\">\n<li><a href=\"http:\/\/openstax.org\/l\/decompfunction\">Decompose Functions<\/a><\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Decompose Functions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/gFSSk8jaAwA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<ul id=\"fs-id1165137600231\">\n<li><a href=\"http:\/\/openstax.org\/l\/compfuncvalue\">Composite Function Values<\/a><\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex 1:  Composite Function Values\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/y2kJI9XnyLY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135388428\" class=\"key-equations\">\n<h3>Key Equation<\/h3>\n<table id=\"eip-id1165134118229\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 119px;\">Composite function<\/td>\n<td class=\"border\" style=\"width: 400.6px;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137768015\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165137754842\">\n<li>When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.<\/li>\n<li>The function produced by combining two functions is a composite function.<\/li>\n<li>The order of function composition must be considered when interpreting the meaning of composite functions.<\/li>\n<li>A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.<\/li>\n<li>A composite function can be evaluated from a table.<\/li>\n<li>A composite function can be evaluated from a graph.<\/li>\n<li>A composite function can be evaluated from a formula.<\/li>\n<li>Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.<\/li>\n<li>Functions can often be decomposed in more than one way.<\/li>\n<li>(Optional)The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl>\n<dt>commutative property<\/dt>\n<dd>the order of the operations being preformed does not matter if the commutative property holds<\/dd>\n<dt>composite function<\/dt>\n<dd id=\"fs-id1165137832426\">the new function formed by function composition, when the output of one function is used as the input of another<\/dd>\n<dd><\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-48\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Composition of Functions. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:Nis4nvtB@10\/Composition-of-Functions\">https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:Nis4nvtB@10\/Composition-of-Functions<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 1. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Composition of Functions\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:Nis4nvtB@10\/Composition-of-Functions\",\"project\":\"Essential Precalcus, Part 1\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-48","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/48","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":27,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/48\/revisions"}],"predecessor-version":[{"id":3270,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/48\/revisions\/3270"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/48\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=48"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=48"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=48"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=48"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}