{"id":1795,"date":"2023-10-12T00:32:12","date_gmt":"2023-10-12T00:32:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-compositions-of-functions\/"},"modified":"2025-10-24T00:19:44","modified_gmt":"2025-10-24T00:19:44","slug":"introduction-compositions-of-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-compositions-of-functions\/","title":{"raw":"Composition","rendered":"Composition"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li class=\"li2\"><span class=\"s1\">Combine functions using algebraic operations.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Create a new function by composition of functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Evaluate composite functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Find the domain of a composite function.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Decompose a composite function into its component functions.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nSuppose 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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195613\/CNX_Precalc_Figure_01_04_0062.jpg\" 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\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\nUsing 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 <strong>cost function<\/strong> at that temperature. We would write [latex]C\\left(T\\left(5\\right)\\right)[\/latex].\r\n\r\nBy combining these two relationships into one function, we have performed function composition, which is the focus of this section.\r\n<h2>Compositions of Functions<\/h2>\r\n<p id=\"fs-id1165137446477\">Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.<\/p>\r\n<p id=\"fs-id1165135533159\">Suppose we need to add two columns of numbers that represent a husband and wife\u2019s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year\u2019s incomes and then collecting all the data in a new column. If [latex]w\\left(y\\right)[\/latex] is the wife\u2019s income and [latex]h\\left(y\\right)[\/latex] is the husband\u2019s income in year [latex]y[\/latex], and we want [latex]T[\/latex] to represent the total income, then we can define a new function.<\/p>\r\n\r\n<div id=\"fs-id1165137641862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T\\left(y\\right)=h\\left(y\\right)+w\\left(y\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165135347766\">If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write<\/p>\r\n\r\n<div id=\"fs-id1165137665765\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=h+w[\/latex]<\/div>\r\n<p id=\"fs-id1165132957142\">Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.<\/p>\r\n<p id=\"fs-id1165137424116\">For two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] with real number outputs, we define new functions [latex]f+g,f-g,f\\cdot{g}[\/latex], and [latex]\\dfrac{f}{g}[\/latex] by the relations<\/p>\r\n\r\n<div id=\"fs-id1165137543139\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{}\\left(f+g\\right)\\left(x\\right)&amp;=f\\left(x\\right)+g\\left(x\\right) \\\\[2mm] \\left(f-g\\right)\\left(x\\right)&amp;=f\\left(x\\right)-g\\left(x\\right) \\\\[2mm] \\text{ }\\left(f\\cdot{g}\\right)\\left(x\\right)&amp;=f\\left(x\\right)\\cdot{g}\\left(x\\right) \\\\[2mm] \\text{ }\\left(\\dfrac{f}{g}\\right)\\left(x\\right)&amp;=\\dfrac{f\\left(x\\right)}{g\\left(x\\right)}\\\\&amp; \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"Example_01_04_01\" class=\"example\">\r\n<div id=\"fs-id1165137585443\" class=\"exercise\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Performing Algebraic Operations on Functions<\/h3>\r\nFind and simplify the functions [latex]\\left(g-f\\right)\\left(x\\right)[\/latex] and [latex]\\left(\\dfrac{g}{f}\\right)\\left(x\\right)[\/latex], given [latex]f\\left(x\\right)=x - 1[\/latex] and [latex]g\\left(x\\right)={x}^{2}-1[\/latex]. Give the domain of your result. Are they the same function?\r\n[reveal-answer q=\"117163\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"117163\"]\r\n<p id=\"fs-id1165137466263\">Begin by writing the general form, and then substitute the given functions.<\/p>\r\n\r\n<div id=\"fs-id1165135701567\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\left(g-f\\right)\\left(x\\right)&amp;=g\\left(x\\right)-f\\left(x\\right) \\\\[2mm] \\left(g-f\\right)\\left(x\\right)&amp;={x}^{2}-1-\\left(x - 1\\right)\\\\[2mm] \\text{ }&amp;={x}^{2}-x \\\\[2mm] \\text{ }&amp;=x\\left(x - 1\\right) \\\\[2mm]\\end{align}\\hspace{20mm}[\/latex] [latex]\\begin{align}\\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&amp;=\\frac{g\\left(x\\right)}{f\\left(x\\right)} \\\\[2mm] \\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&amp;=\\frac{{x}^{2}-1}{x - 1}\\\\[2mm] \\text{ }&amp;=\\frac{\\left(x+1\\right)\\left(x - 1\\right)}{x - 1}\\text{ where }x\\ne 1 \\\\[2mm] \\text{ }&amp;=x+1 \\end{align}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe domain [latex](g-f)(x)[\/latex] is all real numbers and the domain of [latex]\\left(\\dfrac{g}{f}\\right)(x)[\/latex] is [latex]x\\ne1[\/latex]. The functions are not the same.\r\n<p id=\"fs-id1165135351612\"><em>Note<\/em>: For [latex]\\left(\\dfrac{g}{f}\\right)\\left(x\\right)[\/latex], the condition [latex]x\\ne 1[\/latex] is necessary because when [latex]x=1[\/latex], [latex]f(x)=1-1=0[\/latex], which makes the quotient function undefined (dividing by zero).<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind and simplify the functions [latex]\\left(f\\cdot{g}\\right)\\left(x\\right)[\/latex] and [latex]\\left(f-g\\right)\\left(x\\right)[\/latex].\r\n<div id=\"fs-id1165137434994\" class=\"equation unnumbered\">[latex]f\\left(x\\right)=x - 1\\text{ and }g\\left(x\\right)={x}^{2}-1[\/latex]<\/div>\r\n<p id=\"fs-id1165137434911\">Are they the same function?\r\n[reveal-answer q=\"721147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"721147\"][latex]\\left(f\\cdot{g}\\right)\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)=\\left(x - 1\\right)\\left({x}^{2}-1\\right)={x}^{3}-{x}^{2}-x+1[\/latex]<\/p>\r\n[latex]\\left(f-g\\right)\\left(x\\right)=f\\left(x\\right)-g\\left(x\\right)=\\left(x - 1\\right)-\\left({x}^{2}-1\\right)=x-{x}^{2}[\/latex]\r\n\r\nNo, the functions are not the same.[\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=111999&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nLet's explore what happens to the graphs of functions that are combined using algebraic operations. Use an online graphing tool to graph the following functions:\r\n<ol>\r\n \t<li>[latex]f(x) = x^2+3x-4[\/latex]<\/li>\r\n \t<li>[latex]g(x) = \\dfrac{1}{x-1}[\/latex]<\/li>\r\n<\/ol>\r\nNow, enter [latex]h(x) = f(x)+g(x)[\/latex] into the next line.\r\n\r\nEvaluate [latex]h(1)[\/latex], why do you think you get this result?\r\n\r\n<\/div>\r\n<h3>Create a New Function Using a Composition<\/h3>\r\n<\/div>\r\n<\/div>\r\nPerforming algebraic operations on functions combines them into a new function, but we can also 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 combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/p>\r\nWe read the left-hand side as [latex]\"f[\/latex] composed with [latex]g[\/latex] at [latex]x,\"[\/latex] and the right-hand side as [latex]\"f[\/latex] of [latex]g[\/latex] of [latex]x.\"[\/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. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases [latex]f\\left(g\\left(x\\right)\\right)\\ne f\\left(x\\right)g\\left(x\\right)[\/latex].\r\n\r\nIt 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].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195616\/CNX_Precalc_Figure_01_04_0012.jpg\" alt=\"Explanation of the composite function. g(x), the output of g is the input of f. X is the input of g.\" width=\"487\" height=\"171\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\nIn 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].\r\n\r\nFor example if [latex]f\\left(x\\right)={x}^{2}[\/latex] and [latex]g\\left(x\\right)=x+2[\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&amp;=f\\left(x+2\\right) \\\\[2mm] &amp;={\\left(x+2\\right)}^{2} \\\\[2mm] &amp;={x}^{2}+4x+4\\hfill \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">but<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&amp;=g\\left({x}^{2}\\right) \\\\[2mm] \\text{ }&amp;={x}^{2}+2\\hfill \\end{align}[\/latex]<\/p>\r\nThese 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].\r\n\r\nNote 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. Additionally, in applied settings,\u00a0function composition usually only makes sense in one specific order.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Composition of Functions<\/h3>\r\nWhen the output of one function is used as the input of another, we call the entire operation a composition of functions. 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\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/p>\r\nThe 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].\r\n\r\nIt 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].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Define a Composition of Functions<\/h3>\r\nUsing the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex].\r\n[latex]f\\left(x\\right)=2x+1\\\\g\\left(x\\right)=3-x[\/latex]\r\n[reveal-answer q=\"822785\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"822785\"]\r\n\r\nTo determine [latex]f\\left(g\\left(x\\right)\\right)[\/latex], we substitute [latex]g\\left(x\\right)[\/latex] into [latex]f\\left(x\\right)[\/latex].\r\n\r\n[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&amp;=2\\left(3-x\\right)+1 \\\\[2mm] &amp;=6 - 2x+1 \\\\[2mm] &amp;=7 - 2x \\end{align}[\/latex]\r\n\r\nNow we can determine\u00a0[latex]g\\left(f\\left(x\\right)\\right)[\/latex] by substituting [latex]f\\left(x\\right)[\/latex] into [latex]g\\left(x\\right)[\/latex].\r\n\r\n[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&amp;=3-\\left(2x+1\\right) \\\\[2mm] &amp;=3 - 2x - 1 \\\\[2mm] &amp;=-2x+2 \\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32907&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nNow we will explore the graph of a composition of functions. Use an online graphing tool to graph the following functions:\r\n<ol>\r\n \t<li>[latex]f(x) = x^2+3x-4[\/latex]<\/li>\r\n \t<li>[latex]g(x) = \\dfrac{1}{x-1}[\/latex]<\/li>\r\n<\/ol>\r\nNow define a new function:\r\n<p style=\"padding-left: 30px;\">[latex]h(x) = f(f(x))[\/latex]<\/p>\r\nNow zoom in (use the plus sign in the upper right-hand corner of the graph) on the point [latex](0.23607,-3.236)[\/latex], notice that both graphs pass through this point. Continue this pattern for a few more iterations: define a new function that contains another composition of [latex]f(x)[\/latex] with itself. For example: [latex]p(x) = f(f(f(x)))[\/latex], and so on. Continue to use the zoom as you iterate.\r\n\r\nThe point\u00a0[latex](0.23607,-3.236)[\/latex] is called a fixed point of the function [latex]f(x)=x^2+3x-4[\/latex]. Fixed points are used in mathematical applications such as the page rank algorithm that Google uses to generate internet search results. Read more about <a href=\"https:\/\/en.wikipedia.org\/wiki\/Fixed_point_(mathematics)\">fixed points here<\/a>, and <a href=\"https:\/\/en.wikipedia.org\/wiki\/PageRank\">page rank here<\/a>.\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Interpreting Composite Functions<\/h3>\r\nThe 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].\r\n[reveal-answer q=\"542344\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"542344\"]\r\n\r\nThe inside expression in the composition is [latex]s\\left(3\\right)[\/latex]. Because the input to the s-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.\r\n\r\nUsing [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).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Investigating the Order of Function Composition<\/h3>\r\nSuppose [latex]f\\left(x\\right)[\/latex] gives miles that can be driven in [latex]x[\/latex] hours and [latex]g\\left(y\\right)[\/latex] gives the 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]\r\n[reveal-answer q=\"736073\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736073\"]\r\n\r\nThe 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.\r\n\r\n[latex]\\text{number of miles }=f\\left(\\text{number of hours}\\right)[\/latex]\r\n\r\nThe 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:\r\n\r\n[latex]\\text{number of gallons }=g\\left(\\text{number of miles}\\right)[\/latex]\r\n\r\nThe 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.\r\n\r\nThe 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.\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=57278&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>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?<\/strong>\r\n\r\n<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>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe gravitational force on a planet a distance r 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.\r\n[reveal-answer q=\"349230\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"349230\"]\r\n\r\nA 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 r from the Sun (due to gravity), but [latex]G\\left(a\\left(F\\right)\\right)[\/latex] does not make sense.[\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3674&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Evaluate a Composition of Functions<\/h2>\r\nOnce 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.\r\n<h3>Evaluating Composite Functions Using Tables<\/h3>\r\nWhen 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.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a Table to Evaluate a Composite Function<\/h3>\r\nUsing the table below,\u00a0evaluate [latex]f\\left(g\\left(3\\right)\\right)[\/latex] and [latex]g\\left(f\\left(3\\right)\\right)[\/latex].\r\n<table summary=\"Five rows and three columns. The first column is labeled,\"><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]f\\left(x\\right)[\/latex]<\/th>\r\n<th>[latex]g\\left(x\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>1<\/td>\r\n<td>6<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>8<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4<\/td>\r\n<td>1<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"195147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"195147\"]\r\n\r\nTo 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;g\\left(3\\right)=2 \\\\[1.5mm]&amp; f\\left(g\\left(3\\right)\\right)=f\\left(2\\right)=8\\end{align}[\/latex]<\/p>\r\nTo 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[\/latex], we can evaluate\r\n<p style=\"text-align: center;\">[latex]g\\left(f\\left(3\\right)\\right)=g\\left(3\\right)=2[\/latex]<\/p>\r\nThe table below shows the composite functions [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] as tables.\r\n<table summary=\"Two rows and five columns. When x=3, g(3)=2, f(g(3))=8, f(3)=3, and g(f(3))=2.\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>[latex]x[\/latex]<\/td>\r\n<td>[latex]g\\left(x\\right)[\/latex]<\/td>\r\n<td>[latex]f\\left(g\\left(x\\right)\\right)[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td>[latex]g\\left(f\\left(x\\right)\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<td>8<\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the table below, evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex] and [latex]g\\left(f\\left(4\\right)\\right)[\/latex].\r\n<table summary=\"Five rows and three columns. The first column is labeled,\"><colgroup> <col \/> <col \/> <col \/><\/colgroup>\r\n<thead>\r\n<tr>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]f\\left(x\\right)[\/latex]<\/th>\r\n<th>[latex]g\\left(x\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>1<\/td>\r\n<td>6<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>8<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4<\/td>\r\n<td>1<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"161706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"161706\"]\r\n\r\n[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]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3585&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Evaluating Composite Functions Using Graphs<\/h3>\r\nWhen 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.<strong>\r\n<\/strong><\/h3>\r\n<ol>\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 class=\"textbox exercises\">\r\n<h3>Example: Using a Graph to Evaluate a Composite Function<\/h3>\r\nUsing the graphs below, evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195618\/CNX_Precalc_Figure_01_04_002ab2.jpg\" alt=\"Explanation of the composite function.\" width=\"975\" height=\"543\" \/>\r\n\r\n[reveal-answer q=\"44734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44734\"]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195620\/CNX_Precalc_Figure_01_04_0042.jpg\" 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\" \/>\r\n\r\nTo evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex], we start with the inside evaluation.<span id=\"fs-id1165137644158\">\r\n<\/span>\r\n\r\nWe 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].\r\n<p style=\"text-align: center;\">[latex]f\\left(g\\left(1\\right)\\right)=f\\left(3\\right)[\/latex]<\/p>\r\nWe 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].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe figure\u00a0shows how we can mark the graphs with arrows to trace the path from the input value to the output value.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195623\/CNX_Precalc_Figure_01_04_0052.jpg\" alt=\"Two graphs of a positive and negative parabola.\" width=\"975\" height=\"520\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the graphs below, evaluate [latex]g\\left(f\\left(2\\right)\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195620\/CNX_Precalc_Figure_01_04_0042.jpg\" 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\" \/>\r\n\r\n[reveal-answer q=\"682475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"682475\"][latex]g\\left(f\\left(2\\right)\\right)=g\\left(5\\right)=3[\/latex][\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=69936&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Evaluating Composite Functions Using Formulas<\/h3>\r\nWhen 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.\r\n\r\nWhile 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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a formula for a composite function, evaluate the function.<strong>\r\n<\/strong><\/h3>\r\n<ol>\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 class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input<\/h3>\r\nGiven [latex]f\\left(t\\right)={t}^{2}-{t}[\/latex] and [latex]h\\left(x\\right)=3x+2[\/latex], evaluate [latex]f\\left(h\\left(1\\right)\\right)[\/latex].\r\n\r\n[reveal-answer q=\"345593\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"345593\"]\r\n\r\nBecause the inside expression is [latex]h\\left(1\\right)[\/latex], we start by evaluating [latex]h\\left(x\\right)[\/latex] at 1.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}h\\left(1\\right)&amp;=3\\left(1\\right)+2\\\\[2mm] h\\left(1\\right)&amp;=5\\end{align}[\/latex]<\/p>\r\nThen [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.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(h\\left(1\\right)\\right)&amp;=f\\left(5\\right)\\\\[2mm] f\\left(h\\left(1\\right)\\right)&amp;={5}^{2}-5\\\\[2mm] f\\left(h\\left(1\\right)\\right)&amp;=20\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nIt makes no difference what the input variables [latex]t[\/latex] and [latex]x[\/latex] were called in this problem because we evaluated for specific numerical values.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]f\\left(t\\right)={t}^{2}-t[\/latex] and [latex]h\\left(x\\right)=3x+2[\/latex], evaluate\r\n\r\nA) [latex]h\\left(f\\left(2\\right)\\right)[\/latex]\r\n\r\nB) [latex]h\\left(f\\left(-2\\right)\\right)[\/latex]\r\n\r\n[reveal-answer q=\"138476\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"138476\"]\r\n\r\nA. 8; B. 20\r\n\r\nYou can check your work with an online graphing tool. Enter the functions above into a graphing calculator as they are defined. In the next line enter [latex]h\\left(f\\left(2\\right)\\right)[\/latex]. You should see [latex]=8[\/latex] in the bottom right corner.\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16852&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Domain of a Composition<\/h2>\r\nAs we discussed previously, the <strong>domain of a composite function<\/strong> 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].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Domain of a Composite Function<\/h3>\r\nThe domain of a composite function [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 class=\"textbox\">\r\n<h3>How To: Given a function composition [latex]f\\left(g\\left(x\\right)\\right)[\/latex], determine its domain.<\/h3>\r\n<ol>\r\n \t<li>Find the domain of g.<\/li>\r\n \t<li>Find the domain of f.<\/li>\r\n \t<li>Find those inputs,\u00a0[latex]x[\/latex],\u00a0in the domain of [latex]g[\/latex]\u00a0for which [latex]g(x)[\/latex]\u00a0is in the domain of [latex]f[\/latex]. That is, exclude those inputs, [latex]x[\/latex], from the domain of [latex]g[\/latex]\u00a0for which [latex]g(x)[\/latex]\u00a0is 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 class=\"textbox exercises\">\r\n<h3>Example: Finding the Domain of a Composite Function<\/h3>\r\nFind the domain of\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\dfrac{5}{x - 1}\\text{ and }g\\left(x\\right)=\\dfrac{4}{3x - 2}[\/latex]<\/p>\r\n[reveal-answer q=\"924127\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924127\"]\r\n\r\nThe 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\dfrac{4}{3x - 2}=1\\hspace{5mm}&amp;&amp;\\text{Set}\\hspace{2mm}g(x)\\hspace{2mm}\\text{equal to 1} \\\\[2mm]&amp; 4=3x - 2 &amp;&amp;\\text{Multiply by}\\hspace{2mm} 3x-2\\\\[2mm]&amp; 6=3x&amp;&amp;\\text{Add 2 to both sides}\\\\[2mm]&amp; x=2&amp;&amp;\\text{Divide by 3} \\end{align}[\/latex]<\/p>\r\nSo 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\r\n<p style=\"text-align: center;\">[latex]x\\ne \\frac{2}{3}\\hspace{2mm}\\text{or}\\hspace{2mm}x\\ne 2[\/latex]<\/p>\r\nWe can write this in interval notation as\r\n<p style=\"text-align: center;\">[latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Domain of a Composite Function Involving Radicals<\/h3>\r\nFind the domain of\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\sqrt{x+2}\\text{ and }g\\left(x\\right)=\\sqrt{3-x}[\/latex]<\/p>\r\n[reveal-answer q=\"880067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"880067\"]\r\n\r\nBecause 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\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{\\sqrt{3-x}+2}[\/latex]<\/p>\r\nFor [latex]\\left(f\\circ g\\right)\\left(x\\right)[\/latex], we need [latex]\\sqrt{3-x}+2\\ge{0}[\/latex], since the radicand of a square root must be positive. Since square roots are positive, [latex]\\sqrt{3-x}\\ge{0}[\/latex], or [latex]3-x\\ge{0}[\/latex], which gives a domain of [latex]\\left(f\\circ g\\right)\\left(x\\right) = (-\\infty,3][\/latex].\r\n<h4>Analysis of the Solution<\/h4>\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].\r\n\r\nYou cannot rely on an algorithm to find the domain of a composite function. Rather, you will need to first ask yourself \"what is the domain of the inner function\", and determine whether this set will comply with the domain restrictions of the outer function. In this case, the set [latex](-\\infty,3][\/latex] ensures a non-negative output for the inner function, which will in turn ensure a positive input for the composite function.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the domain of\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\dfrac{1}{x - 2}\\text{ and }g\\left(x\\right)=\\sqrt{x+4}[\/latex]<\/p>\r\n[reveal-answer q=\"936114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"936114\"]\r\n\r\n[latex]\\left[-4,0\\right)\\cup \\left(0,\\infty \\right)[\/latex][\/hidden-answer]\r\n\r\n[ohm_question hide_question_numbers=1]1603[\/ohm_question]\r\n\r\n[ohm_question hide_question_numbers=1]1604[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nWe can use graphs to visualize the domain that results from a composition of two functions.\r\nGraph the two functions below with an online graphing tool.\r\n<ol>\r\n \t<li>[latex]f(x)=\\sqrt{3-x}[\/latex]<\/li>\r\n \t<li>[latex]g(t) = \\sqrt{x+4}[\/latex]<\/li>\r\n<\/ol>\r\nNext, create a new function, [latex]h(x) = g(f(x))[\/latex]. \u00a0Based on the graph, what is the domain of this function? Explain why [latex]g(f(x))[\/latex] and [latex]f(x)[\/latex] have the same domain.\r\n\r\nNow define another composition, [latex]p(x) = f(g(x)[\/latex]. \u00a0What is the domain of this function? Explain why you can evaluate\u00a0[latex]g(10)[\/latex], but not [latex]p(10)[\/latex].\r\n\r\n<\/div>\r\n&nbsp;\r\n<h3>Decompose a Composite Function<\/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 is almost always more than one way to <strong>decompose a composite function<\/strong>, so we may choose the decomposition that appears to be most obvious.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Decomposing a Function<\/h3>\r\nWrite [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.\r\n\r\n[reveal-answer q=\"702975\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"702975\"]\r\n\r\nWe 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\r\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=5-{x}^{2}\\hspace{2mm}\\text{and}\\hspace{2mm}g\\left(x\\right)=\\sqrt{x}[\/latex]<\/p>\r\nWe can check our answer by recomposing the functions.\r\n<p style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g\\left(5-{x}^{2}\\right)=\\sqrt{5-{x}^{2}}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nFor every composition there are infinitely many possible function pairs that will work. In this case, another function pair where\u00a0[latex]g\\left(h\\left(x\\right)\\right)=\\sqrt{5-{x}^{2}}[\/latex]\u00a0 is\u00a0 [latex]h(x)=x^2[\/latex] and [latex]g(x)=\\sqrt{5-x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite [latex]f\\left(x\\right)=\\dfrac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.\r\n\r\n[reveal-answer q=\"489928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"489928\"]\r\n\r\nPossible answer:\r\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\r\n[latex]h\\left(x\\right)=\\dfrac{4}{3-x}[\/latex]\r\n\r\n[latex]f=h\\circ g[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32917&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><span style=\"width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Equation<\/h2>\r\n<section id=\"fs-id1165135388428\" class=\"key-equations\">\r\n<table id=\"eip-id1165134118229\" summary=\"..\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>Composite function<\/td>\r\n<td>[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<\/section><section id=\"fs-id1165137768015\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137754842\">\r\n \t<li>We can perform algebraic operations on functions.<\/li>\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>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 \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<\/ul>\r\n<\/section><section id=\"fs-id1165137611819\" class=\"section-exercises\"><section class=\"section-exercises\"><section class=\"section-exercises\">\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137832421\" class=\"definition\">\r\n \t<dt><strong>composite function<\/strong><\/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<\/dl>\r\n<\/div>\r\n<\/section><\/section><\/section><img class=\"size-medium wp-image-2016 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5269\/2020\/06\/22204550\/stop-sign-with-hand-300x300.png\" alt=\"Stop Here\" width=\"300\" height=\"300\" \/>\r\n<h3 style=\"text-align: center;\"><span data-sheets-root=\"1\">STOP HERE and complete Homework 1.5 - Composition<\/span><\/h3>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li class=\"li2\"><span class=\"s1\">Combine functions using algebraic operations.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Create a new function by composition of functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Evaluate composite functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Find the domain of a composite function.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Decompose a composite function into its component functions.<\/span><\/li>\n<\/ul>\n<\/div>\n<p>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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195613\/CNX_Precalc_Figure_01_04_0062.jpg\" 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 class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>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 <strong>cost function<\/strong> at that temperature. We would write [latex]C\\left(T\\left(5\\right)\\right)[\/latex].<\/p>\n<p>By combining these two relationships into one function, we have performed function composition, which is the focus of this section.<\/p>\n<h2>Compositions of Functions<\/h2>\n<p id=\"fs-id1165137446477\">Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.<\/p>\n<p id=\"fs-id1165135533159\">Suppose we need to add two columns of numbers that represent a husband and wife\u2019s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year\u2019s incomes and then collecting all the data in a new column. If [latex]w\\left(y\\right)[\/latex] is the wife\u2019s income and [latex]h\\left(y\\right)[\/latex] is the husband\u2019s income in year [latex]y[\/latex], and we want [latex]T[\/latex] to represent the total income, then we can define a new function.<\/p>\n<div id=\"fs-id1165137641862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T\\left(y\\right)=h\\left(y\\right)+w\\left(y\\right)[\/latex]<\/div>\n<p id=\"fs-id1165135347766\">If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write<\/p>\n<div id=\"fs-id1165137665765\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=h+w[\/latex]<\/div>\n<p id=\"fs-id1165132957142\">Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.<\/p>\n<p id=\"fs-id1165137424116\">For two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] with real number outputs, we define new functions [latex]f+g,f-g,f\\cdot{g}[\/latex], and [latex]\\dfrac{f}{g}[\/latex] by the relations<\/p>\n<div id=\"fs-id1165137543139\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{}\\left(f+g\\right)\\left(x\\right)&=f\\left(x\\right)+g\\left(x\\right) \\\\[2mm] \\left(f-g\\right)\\left(x\\right)&=f\\left(x\\right)-g\\left(x\\right) \\\\[2mm] \\text{ }\\left(f\\cdot{g}\\right)\\left(x\\right)&=f\\left(x\\right)\\cdot{g}\\left(x\\right) \\\\[2mm] \\text{ }\\left(\\dfrac{f}{g}\\right)\\left(x\\right)&=\\dfrac{f\\left(x\\right)}{g\\left(x\\right)}\\\\& \\end{align}[\/latex]<\/div>\n<div><\/div>\n<div id=\"Example_01_04_01\" class=\"example\">\n<div id=\"fs-id1165137585443\" class=\"exercise\">\n<div class=\"textbox exercises\">\n<h3>Example: Performing Algebraic Operations on Functions<\/h3>\n<p>Find and simplify the functions [latex]\\left(g-f\\right)\\left(x\\right)[\/latex] and [latex]\\left(\\dfrac{g}{f}\\right)\\left(x\\right)[\/latex], given [latex]f\\left(x\\right)=x - 1[\/latex] and [latex]g\\left(x\\right)={x}^{2}-1[\/latex]. Give the domain of your result. Are they the same function?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q117163\">Show Solution<\/span><\/p>\n<div id=\"q117163\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137466263\">Begin by writing the general form, and then substitute the given functions.<\/p>\n<div id=\"fs-id1165135701567\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\left(g-f\\right)\\left(x\\right)&=g\\left(x\\right)-f\\left(x\\right) \\\\[2mm] \\left(g-f\\right)\\left(x\\right)&={x}^{2}-1-\\left(x - 1\\right)\\\\[2mm] \\text{ }&={x}^{2}-x \\\\[2mm] \\text{ }&=x\\left(x - 1\\right) \\\\[2mm]\\end{align}\\hspace{20mm}[\/latex] [latex]\\begin{align}\\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&=\\frac{g\\left(x\\right)}{f\\left(x\\right)} \\\\[2mm] \\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&=\\frac{{x}^{2}-1}{x - 1}\\\\[2mm] \\text{ }&=\\frac{\\left(x+1\\right)\\left(x - 1\\right)}{x - 1}\\text{ where }x\\ne 1 \\\\[2mm] \\text{ }&=x+1 \\end{align}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The domain [latex](g-f)(x)[\/latex] is all real numbers and the domain of [latex]\\left(\\dfrac{g}{f}\\right)(x)[\/latex] is [latex]x\\ne1[\/latex]. The functions are not the same.<\/p>\n<p id=\"fs-id1165135351612\"><em>Note<\/em>: For [latex]\\left(\\dfrac{g}{f}\\right)\\left(x\\right)[\/latex], the condition [latex]x\\ne 1[\/latex] is necessary because when [latex]x=1[\/latex], [latex]f(x)=1-1=0[\/latex], which makes the quotient function undefined (dividing by zero).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find and simplify the functions [latex]\\left(f\\cdot{g}\\right)\\left(x\\right)[\/latex] and [latex]\\left(f-g\\right)\\left(x\\right)[\/latex].<\/p>\n<div id=\"fs-id1165137434994\" class=\"equation unnumbered\">[latex]f\\left(x\\right)=x - 1\\text{ and }g\\left(x\\right)={x}^{2}-1[\/latex]<\/div>\n<p id=\"fs-id1165137434911\">Are they the same function?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q721147\">Show Solution<\/span><\/p>\n<div id=\"q721147\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(f\\cdot{g}\\right)\\left(x\\right)=f\\left(x\\right)g\\left(x\\right)=\\left(x - 1\\right)\\left({x}^{2}-1\\right)={x}^{3}-{x}^{2}-x+1[\/latex]<\/p>\n<p>[latex]\\left(f-g\\right)\\left(x\\right)=f\\left(x\\right)-g\\left(x\\right)=\\left(x - 1\\right)-\\left({x}^{2}-1\\right)=x-{x}^{2}[\/latex]<\/p>\n<p>No, the functions are not the same.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=111999&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Let&#8217;s explore what happens to the graphs of functions that are combined using algebraic operations. Use an online graphing tool to graph the following functions:<\/p>\n<ol>\n<li>[latex]f(x) = x^2+3x-4[\/latex]<\/li>\n<li>[latex]g(x) = \\dfrac{1}{x-1}[\/latex]<\/li>\n<\/ol>\n<p>Now, enter [latex]h(x) = f(x)+g(x)[\/latex] into the next line.<\/p>\n<p>Evaluate [latex]h(1)[\/latex], why do you think you get this result?<\/p>\n<\/div>\n<h3>Create a New Function Using a Composition<\/h3>\n<\/div>\n<\/div>\n<p>Performing algebraic operations on functions combines them into a new function, but we can also 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 combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/p>\n<p>We read the left-hand side as [latex]\"f[\/latex] composed with [latex]g[\/latex] at [latex]x,\"[\/latex] and the right-hand side as [latex]\"f[\/latex] of [latex]g[\/latex] of [latex]x.\"[\/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. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases [latex]f\\left(g\\left(x\\right)\\right)\\ne f\\left(x\\right)g\\left(x\\right)[\/latex].<\/p>\n<p>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].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195616\/CNX_Precalc_Figure_01_04_0012.jpg\" alt=\"Explanation of the composite function. g(x), the output of g is the input of f. X is the input of g.\" width=\"487\" height=\"171\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p>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].<\/p>\n<p>For example if [latex]f\\left(x\\right)={x}^{2}[\/latex] and [latex]g\\left(x\\right)=x+2[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&=f\\left(x+2\\right) \\\\[2mm] &={\\left(x+2\\right)}^{2} \\\\[2mm] &={x}^{2}+4x+4\\hfill \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">but<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&=g\\left({x}^{2}\\right) \\\\[2mm] \\text{ }&={x}^{2}+2\\hfill \\end{align}[\/latex]<\/p>\n<p>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>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. Additionally, in applied settings,\u00a0function composition usually only makes sense in one specific order.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Composition of Functions<\/h3>\n<p>When the output of one function is used as the input of another, we call the entire operation a composition of functions. 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<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/p>\n<p>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<p>It 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<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Define a Composition of Functions<\/h3>\n<p>Using the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex].<br \/>\n[latex]f\\left(x\\right)=2x+1\\\\g\\left(x\\right)=3-x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q822785\">Show Solution<\/span><\/p>\n<div id=\"q822785\" class=\"hidden-answer\" style=\"display: none\">\n<p>To determine [latex]f\\left(g\\left(x\\right)\\right)[\/latex], we substitute [latex]g\\left(x\\right)[\/latex] into [latex]f\\left(x\\right)[\/latex].<\/p>\n<p>[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&=2\\left(3-x\\right)+1 \\\\[2mm] &=6 - 2x+1 \\\\[2mm] &=7 - 2x \\end{align}[\/latex]<\/p>\n<p>Now we can determine\u00a0[latex]g\\left(f\\left(x\\right)\\right)[\/latex] by substituting [latex]f\\left(x\\right)[\/latex] into [latex]g\\left(x\\right)[\/latex].<\/p>\n<p>[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&=3-\\left(2x+1\\right) \\\\[2mm] &=3 - 2x - 1 \\\\[2mm] &=-2x+2 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32907&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Now we will explore the graph of a composition of functions. Use an online graphing tool to graph the following functions:<\/p>\n<ol>\n<li>[latex]f(x) = x^2+3x-4[\/latex]<\/li>\n<li>[latex]g(x) = \\dfrac{1}{x-1}[\/latex]<\/li>\n<\/ol>\n<p>Now define a new function:<\/p>\n<p style=\"padding-left: 30px;\">[latex]h(x) = f(f(x))[\/latex]<\/p>\n<p>Now zoom in (use the plus sign in the upper right-hand corner of the graph) on the point [latex](0.23607,-3.236)[\/latex], notice that both graphs pass through this point. Continue this pattern for a few more iterations: define a new function that contains another composition of [latex]f(x)[\/latex] with itself. For example: [latex]p(x) = f(f(f(x)))[\/latex], and so on. Continue to use the zoom as you iterate.<\/p>\n<p>The point\u00a0[latex](0.23607,-3.236)[\/latex] is called a fixed point of the function [latex]f(x)=x^2+3x-4[\/latex]. Fixed points are used in mathematical applications such as the page rank algorithm that Google uses to generate internet search results. Read more about <a href=\"https:\/\/en.wikipedia.org\/wiki\/Fixed_point_(mathematics)\">fixed points here<\/a>, and <a href=\"https:\/\/en.wikipedia.org\/wiki\/PageRank\">page rank here<\/a>.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Interpreting Composite Functions<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q542344\">Show Solution<\/span><\/p>\n<div id=\"q542344\" class=\"hidden-answer\" style=\"display: none\">\n<p>The inside expression in the composition is [latex]s\\left(3\\right)[\/latex]. Because the input to the s-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>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<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Investigating the Order of Function Composition<\/h3>\n<p>Suppose [latex]f\\left(x\\right)[\/latex] gives miles that can be driven in [latex]x[\/latex] hours and [latex]g\\left(y\\right)[\/latex] gives the 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736073\">Show Solution<\/span><\/p>\n<div id=\"q736073\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p>[latex]\\text{number of miles }=f\\left(\\text{number of hours}\\right)[\/latex]<\/p>\n<p>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<p>[latex]\\text{number of gallons }=g\\left(\\text{number of miles}\\right)[\/latex]<\/p>\n<p>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>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<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=57278&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>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?<\/strong><\/p>\n<p><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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The gravitational force on a planet a distance r 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q349230\">Show Solution<\/span><\/p>\n<div id=\"q349230\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 r from the Sun (due to gravity), but [latex]G\\left(a\\left(F\\right)\\right)[\/latex] does not make sense.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3674&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Evaluate a Composition of Functions<\/h2>\n<p>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<h3>Evaluating Composite Functions Using Tables<\/h3>\n<p>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 class=\"textbox exercises\">\n<h3>Example: Using a Table to Evaluate a Composite Function<\/h3>\n<p>Using the table below,\u00a0evaluate [latex]f\\left(g\\left(3\\right)\\right)[\/latex] and [latex]g\\left(f\\left(3\\right)\\right)[\/latex].<\/p>\n<table summary=\"Five rows and three columns. The first column is labeled,\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]f\\left(x\\right)[\/latex]<\/th>\n<th>[latex]g\\left(x\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1<\/td>\n<td>6<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>8<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>3<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>1<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q195147\">Show Solution<\/span><\/p>\n<div id=\"q195147\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]\\begin{align}&g\\left(3\\right)=2 \\\\[1.5mm]& f\\left(g\\left(3\\right)\\right)=f\\left(2\\right)=8\\end{align}[\/latex]<\/p>\n<p>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[\/latex], we can evaluate<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(f\\left(3\\right)\\right)=g\\left(3\\right)=2[\/latex]<\/p>\n<p>The table below shows the composite functions [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] as tables.<\/p>\n<table 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<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]x[\/latex]<\/td>\n<td>[latex]g\\left(x\\right)[\/latex]<\/td>\n<td>[latex]f\\left(g\\left(x\\right)\\right)[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)[\/latex]<\/td>\n<td>[latex]g\\left(f\\left(x\\right)\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>2<\/td>\n<td>8<\/td>\n<td>3<\/td>\n<td>2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the table below, evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex] and [latex]g\\left(f\\left(4\\right)\\right)[\/latex].<\/p>\n<table summary=\"Five rows and three columns. The first column is labeled,\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<thead>\n<tr>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]f\\left(x\\right)[\/latex]<\/th>\n<th>[latex]g\\left(x\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1<\/td>\n<td>6<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>8<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>3<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>1<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q161706\">Show Solution<\/span><\/p>\n<div id=\"q161706\" class=\"hidden-answer\" style=\"display: none\">\n<p>[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<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3585&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Evaluating Composite Functions Using Graphs<\/h3>\n<p>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 class=\"textbox\">\n<h3>How To: Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.<strong><br \/>\n<\/strong><\/h3>\n<ol>\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 class=\"textbox exercises\">\n<h3>Example: Using a Graph to Evaluate a Composite Function<\/h3>\n<p>Using the graphs below, evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195618\/CNX_Precalc_Figure_01_04_002ab2.jpg\" alt=\"Explanation of the composite function.\" width=\"975\" height=\"543\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44734\">Show Solution<\/span><\/p>\n<div id=\"q44734\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195620\/CNX_Precalc_Figure_01_04_0042.jpg\" 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>To evaluate [latex]f\\left(g\\left(1\\right)\\right)[\/latex], we start with the inside evaluation.<span id=\"fs-id1165137644158\"><br \/>\n<\/span><\/p>\n<p>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<p style=\"text-align: center;\">[latex]f\\left(g\\left(1\\right)\\right)=f\\left(3\\right)[\/latex]<\/p>\n<p>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<h4>Analysis of the Solution<\/h4>\n<p>The figure\u00a0shows how we can mark the graphs with arrows to trace the path from the input value to the output value.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195623\/CNX_Precalc_Figure_01_04_0052.jpg\" alt=\"Two graphs of a positive and negative parabola.\" width=\"975\" height=\"520\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the graphs below, evaluate [latex]g\\left(f\\left(2\\right)\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195620\/CNX_Precalc_Figure_01_04_0042.jpg\" 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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q682475\">Show Solution<\/span><\/p>\n<div id=\"q682475\" class=\"hidden-answer\" style=\"display: none\">[latex]g\\left(f\\left(2\\right)\\right)=g\\left(5\\right)=3[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=69936&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Evaluating Composite Functions Using Formulas<\/h3>\n<p>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>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 class=\"textbox\">\n<h3>How To: Given a formula for a composite function, evaluate the function.<strong><br \/>\n<\/strong><\/h3>\n<ol>\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 class=\"textbox exercises\">\n<h3>Example: Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input<\/h3>\n<p>Given [latex]f\\left(t\\right)={t}^{2}-{t}[\/latex] and [latex]h\\left(x\\right)=3x+2[\/latex], evaluate [latex]f\\left(h\\left(1\\right)\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q345593\">Show Solution<\/span><\/p>\n<div id=\"q345593\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the inside expression is [latex]h\\left(1\\right)[\/latex], we start by evaluating [latex]h\\left(x\\right)[\/latex] at 1.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}h\\left(1\\right)&=3\\left(1\\right)+2\\\\[2mm] h\\left(1\\right)&=5\\end{align}[\/latex]<\/p>\n<p>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=\"text-align: center;\">[latex]\\begin{align}f\\left(h\\left(1\\right)\\right)&=f\\left(5\\right)\\\\[2mm] f\\left(h\\left(1\\right)\\right)&={5}^{2}-5\\\\[2mm] f\\left(h\\left(1\\right)\\right)&=20\\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>It makes no difference what the input variables [latex]t[\/latex] and [latex]x[\/latex] were called in this problem because we evaluated for specific numerical values.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given [latex]f\\left(t\\right)={t}^{2}-t[\/latex] and [latex]h\\left(x\\right)=3x+2[\/latex], evaluate<\/p>\n<p>A) [latex]h\\left(f\\left(2\\right)\\right)[\/latex]<\/p>\n<p>B) [latex]h\\left(f\\left(-2\\right)\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q138476\">Show Solution<\/span><\/p>\n<div id=\"q138476\" class=\"hidden-answer\" style=\"display: none\">\n<p>A. 8; B. 20<\/p>\n<p>You can check your work with an online graphing tool. Enter the functions above into a graphing calculator as they are defined. In the next line enter [latex]h\\left(f\\left(2\\right)\\right)[\/latex]. You should see [latex]=8[\/latex] in the bottom right corner.<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16852&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Domain of a Composition<\/h2>\n<p>As we discussed previously, the <strong>domain of a composite function<\/strong> 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\">\n<h3>A General Note: Domain of a Composite Function<\/h3>\n<p>The domain of a composite function [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 class=\"textbox\">\n<h3>How To: Given a function composition [latex]f\\left(g\\left(x\\right)\\right)[\/latex], determine its domain.<\/h3>\n<ol>\n<li>Find the domain of g.<\/li>\n<li>Find the domain of f.<\/li>\n<li>Find those inputs,\u00a0[latex]x[\/latex],\u00a0in the domain of [latex]g[\/latex]\u00a0for which [latex]g(x)[\/latex]\u00a0is in the domain of [latex]f[\/latex]. That is, exclude those inputs, [latex]x[\/latex], from the domain of [latex]g[\/latex]\u00a0for which [latex]g(x)[\/latex]\u00a0is 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 class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Composite Function<\/h3>\n<p>Find the domain of<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\dfrac{5}{x - 1}\\text{ and }g\\left(x\\right)=\\dfrac{4}{3x - 2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924127\">Show Solution<\/span><\/p>\n<div id=\"q924127\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]\\begin{align}&\\dfrac{4}{3x - 2}=1\\hspace{5mm}&&\\text{Set}\\hspace{2mm}g(x)\\hspace{2mm}\\text{equal to 1} \\\\[2mm]& 4=3x - 2 &&\\text{Multiply by}\\hspace{2mm} 3x-2\\\\[2mm]& 6=3x&&\\text{Add 2 to both sides}\\\\[2mm]& x=2&&\\text{Divide by 3} \\end{align}[\/latex]<\/p>\n<p>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<\/p>\n<p style=\"text-align: center;\">[latex]x\\ne \\frac{2}{3}\\hspace{2mm}\\text{or}\\hspace{2mm}x\\ne 2[\/latex]<\/p>\n<p>We can write this in interval notation as<\/p>\n<p style=\"text-align: center;\">[latex]\\left(-\\infty ,\\frac{2}{3}\\right)\\cup \\left(\\frac{2}{3},2\\right)\\cup \\left(2,\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Composite Function Involving Radicals<\/h3>\n<p>Find the domain of<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\sqrt{x+2}\\text{ and }g\\left(x\\right)=\\sqrt{3-x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q880067\">Show Solution<\/span><\/p>\n<div id=\"q880067\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=\\sqrt{\\sqrt{3-x}+2}[\/latex]<\/p>\n<p>For [latex]\\left(f\\circ g\\right)\\left(x\\right)[\/latex], we need [latex]\\sqrt{3-x}+2\\ge{0}[\/latex], since the radicand of a square root must be positive. Since square roots are positive, [latex]\\sqrt{3-x}\\ge{0}[\/latex], or [latex]3-x\\ge{0}[\/latex], which gives a domain of [latex]\\left(f\\circ g\\right)\\left(x\\right) = (-\\infty,3][\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\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].<\/p>\n<p>You cannot rely on an algorithm to find the domain of a composite function. Rather, you will need to first ask yourself &#8220;what is the domain of the inner function&#8221;, and determine whether this set will comply with the domain restrictions of the outer function. In this case, the set [latex](-\\infty,3][\/latex] ensures a non-negative output for the inner function, which will in turn ensure a positive input for the composite function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain of<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)\\text{ where}f\\left(x\\right)=\\dfrac{1}{x - 2}\\text{ and }g\\left(x\\right)=\\sqrt{x+4}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q936114\">Show Solution<\/span><\/p>\n<div id=\"q936114\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left[-4,0\\right)\\cup \\left(0,\\infty \\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1603\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1603&theme=oea&iframe_resize_id=ohm1603\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm1604\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1604&theme=oea&iframe_resize_id=ohm1604\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>We can use graphs to visualize the domain that results from a composition of two functions.<br \/>\nGraph the two functions below with an online graphing tool.<\/p>\n<ol>\n<li>[latex]f(x)=\\sqrt{3-x}[\/latex]<\/li>\n<li>[latex]g(t) = \\sqrt{x+4}[\/latex]<\/li>\n<\/ol>\n<p>Next, create a new function, [latex]h(x) = g(f(x))[\/latex]. \u00a0Based on the graph, what is the domain of this function? Explain why [latex]g(f(x))[\/latex] and [latex]f(x)[\/latex] have the same domain.<\/p>\n<p>Now define another composition, [latex]p(x) = f(g(x)[\/latex]. \u00a0What is the domain of this function? Explain why you can evaluate\u00a0[latex]g(10)[\/latex], but not [latex]p(10)[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Decompose a Composite Function<\/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 is almost always more than one way to <strong>decompose a composite function<\/strong>, so we may choose the decomposition that appears to be most obvious.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Decomposing a Function<\/h3>\n<p>Write [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q702975\">Show Solution<\/span><\/p>\n<div id=\"q702975\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]h\\left(x\\right)=5-{x}^{2}\\hspace{2mm}\\text{and}\\hspace{2mm}g\\left(x\\right)=\\sqrt{x}[\/latex]<\/p>\n<p>We can check our answer by recomposing the functions.<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g\\left(5-{x}^{2}\\right)=\\sqrt{5-{x}^{2}}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>For every composition there are infinitely many possible function pairs that will work. In this case, another function pair where\u00a0[latex]g\\left(h\\left(x\\right)\\right)=\\sqrt{5-{x}^{2}}[\/latex]\u00a0 is\u00a0 [latex]h(x)=x^2[\/latex] and [latex]g(x)=\\sqrt{5-x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write [latex]f\\left(x\\right)=\\dfrac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q489928\">Show Solution<\/span><\/p>\n<div id=\"q489928\" class=\"hidden-answer\" style=\"display: none\">\n<p>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)=\\dfrac{4}{3-x}[\/latex]<\/p>\n<p>[latex]f=h\\circ g[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32917&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><span style=\"width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe><\/p>\n<\/div>\n<h2>Key Equation<\/h2>\n<section id=\"fs-id1165135388428\" class=\"key-equations\">\n<table id=\"eip-id1165134118229\" summary=\"..\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>Composite function<\/td>\n<td>[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<\/section>\n<section id=\"fs-id1165137768015\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137754842\">\n<li>We can perform algebraic operations on functions.<\/li>\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>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<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<\/ul>\n<\/section>\n<section id=\"fs-id1165137611819\" class=\"section-exercises\">\n<section class=\"section-exercises\">\n<section class=\"section-exercises\">\n<div>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137832421\" class=\"definition\">\n<dt><strong>composite function<\/strong><\/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<\/dl>\n<\/div>\n<\/section>\n<\/section>\n<\/section>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2016 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5269\/2020\/06\/22204550\/stop-sign-with-hand-300x300.png\" alt=\"Stop Here\" width=\"300\" height=\"300\" \/><\/p>\n<h3 style=\"text-align: center;\"><span data-sheets-root=\"1\">STOP HERE and complete Homework 1.5 &#8211; Composition<\/span><\/h3>\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-1795\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 111999. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 32907, 69936. <strong>Authored by<\/strong>: Smart, Jim. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 5727. <strong>Authored by<\/strong>: Pepe, Mike. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 3674. <strong>Authored by<\/strong>: Rasmussen, Melonie, mb Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 3585. <strong>Authored by<\/strong>: Reidel, Jessica. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 16852. <strong>Authored by<\/strong>: Pierpoint, William. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 15772. <strong>Authored by<\/strong>: Master, Course. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 1600. <strong>Authored by<\/strong>: WebWork-Rochester, mb Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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