{"id":1799,"date":"2023-10-12T00:32:13","date_gmt":"2023-10-12T00:32:13","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-inverse-functions\/"},"modified":"2024-08-29T02:01:01","modified_gmt":"2024-08-29T02:01:01","slug":"introduction-inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-inverse-functions\/","title":{"raw":"Inversion","rendered":"Inversion"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul class=\"ul1\">\r\n \t<li>Determine whether a function is one-to-one.<\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Verify inverse functions.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s1\">Find or evaluate the inverse of a function.<\/span><\/li>\r\n \t<li class=\"li3\"><span class=\"s8\">Use the graph of a one-to-one function to graph its inverse function on the same axes.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165135358875\">A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.<\/p>\r\nIf some physical machines can run in two directions, we might ask whether some of the function \"machines\" we have been studying can also run backwards. Figure 1\u00a0provides a visual representation of this question. In this section, we will consider the reverse nature of functions.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205509\/CNX_Precalc_Figure_01_07_0012.jpg\" alt=\"Diagram of a function and what would be its inverse.\" width=\"731\" height=\"305\" \/> <b>Figure 1.<\/b> Can a function \"machine\" operate in reverse?[\/caption]\r\n<h2>Determine whether a function is one-to-one<\/h2>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191009\/CNX_Precalc_Figure_01_00_001n2.jpg\" alt=\"Figure of a bull and a graph of market prices.\" width=\"975\" height=\"307\" \/>\r\n\r\nSome functions have a given output value that corresponds to two or more input values. For example, in the following stock chart the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000.\r\n\r\nHowever, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in.\r\n<table summary=\"Two columns and five rows. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th>Letter grade<\/th>\r\n<th>Grade point average<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>A<\/td>\r\n<td>4.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>B<\/td>\r\n<td>3.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>C<\/td>\r\n<td>2.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>D<\/td>\r\n<td>1.0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThis grading system represents a one-to-one function, because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter.\r\n\r\nTo visualize this concept, let\u2019s look again at the two simple functions sketched in (a) and (b)\u00a0below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18190946\/CNX_Precalc_Figure_01_01_0012.jpg\" alt=\"Three relations that demonstrate what constitute a function.\" width=\"975\" height=\"243\" \/>\r\n\r\nThe function in part (a) shows a relationship that is not a one-to-one function because inputs [latex]q[\/latex] and [latex]r[\/latex] both give output [latex]n[\/latex]. The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.\r\n<div class=\"textbox\">\r\n<h3>A General Note: One-to-One Function<\/h3>\r\nA one-to-one function is a function in which each output value corresponds to exactly one input value.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determining Whether a Relationship Is a One-to-One Function<\/h3>\r\nIs the area of a circle a function of its radius? If yes, is the function one-to-one?\r\n\r\n[reveal-answer q=\"380432\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380432\"]\r\n\r\nA circle of radius [latex]r[\/latex] has a unique area measure given by [latex]A=\\pi {r}^{2}[\/latex], so for any input, [latex]r[\/latex], there is only one output, [latex]A[\/latex]. The area is a function of radius [latex]r[\/latex].\r\n\r\nIf the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure [latex]A[\/latex] is given by the formula [latex]A=\\pi {r}^{2}[\/latex]. Because areas and radii are positive numbers, there is exactly one solution: [latex]r=\\sqrt{\\frac{A}{\\pi }}[\/latex]. So the area of a circle is a one-to-one function of the circle\u2019s radius.\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\n<ol>\r\n \t<li>Is a balance a function of the bank account number?<\/li>\r\n \t<li>Is a bank account number a function of the balance?<\/li>\r\n \t<li>Is a balance a one-to-one function of the bank account number?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"997233\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"997233\"]\r\n<ol>\r\n \t<li><span class=\"s1\">yes, because each bank account (input) has a single balance (output) at any given time.<\/span><\/li>\r\n \t<li><span class=\"s1\">no, because several bank accounts (inputs) may have the same balance (output).<\/span><\/li>\r\n \t<li><span class=\"s1\">no, because the more than one bank account (input) can have the same balance (output).<\/span><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom7\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15800&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Evaluating and Solving Functions<\/h2>\r\nWhen we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function [latex]f\\left(x\\right)=5 - 3{x}^{2}[\/latex] can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.\r\n<div class=\"textbox\">\r\n<h3><strong>How To: EVALUATE A FUNCTION Given ITS FORMula.\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Replace the input variable in the formula with the value provided.<\/li>\r\n \t<li>Calculate the result.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating Functions<\/h3>\r\nGiven the function [latex]h\\left(p\\right)={p}^{2}+2p[\/latex], evaluate [latex]h\\left(4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"768180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"768180\"]\r\n<p style=\"text-align: left;\">To evaluate [latex]h\\left(4\\right)[\/latex], we substitute the value 4 for the input variable [latex]p[\/latex] in the given function.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}h\\left(p\\right)&amp;={p}^{2}+2p \\\\ h\\left(4\\right)&amp;={\\left(4\\right)}^{2}+2\\left(4\\right) \\\\ &amp;=16+8 \\\\ &amp;=24 \\end{align}[\/latex]<\/p>\r\nTherefore, for an input of 4, we have an output of 24 or [latex]h(4)=24[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/Ehkzu5Uv7O0\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating Functions at Specific Values<\/h3>\r\nFor the function, [latex]f\\left(x\\right)={x}^{2}+3x - 4[\/latex], evaluate each of the following.\r\n<ol>\r\n \t<li>[latex]f\\left(2\\right)[\/latex]<\/li>\r\n \t<li>[latex]f(a)[\/latex]<\/li>\r\n \t<li>[latex]f(a+h)[\/latex]<\/li>\r\n \t<li>[latex]\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"645951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"645951\"]\r\n\r\nReplace the [latex]x[\/latex]\u00a0in the function with each specified value.\r\n<ol>\r\n \t<li>Because the input value is a number, 2, we can use algebra to simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}f\\left(2\\right)&amp;={2}^{2}+3\\left(2\\right)-4 \\\\ &amp;=4+6 - 4 \\\\ &amp;=6\\hfill \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>In this case, the input value is a letter so we cannot simplify the answer any further.\r\n<div style=\"text-align: center;\">[latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/div><\/li>\r\n \t<li>With an input value of [latex]a+h[\/latex], we must use the distributive property.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}f\\left(a+h\\right)&amp;={\\left(a+h\\right)}^{2}+3\\left(a+h\\right)-4 \\\\[2mm] &amp;={a}^{2}+2ah+{h}^{2}+3a+3h - 4 \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that\r\n<div style=\"text-align: center;\">[latex]f\\left(a+h\\right)={a}^{2}+2ah+{h}^{2}+3a+3h - 4[\/latex]<\/div>\r\nand we know that\r\n<div style=\"text-align: center;\">[latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/div>\r\nNow we combine the results and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}&amp;=\\dfrac{\\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\\right)-\\left({a}^{2}+3a - 4\\right)}{h} \\\\[2mm] &amp;=\\dfrac{2ah+{h}^{2}+3h}{h}\\\\[2mm] &amp;=\\frac{h\\left(2a+h+3\\right)}{h}&amp;&amp;\\text{Factor out }h. \\\\[2mm] &amp;=2a+h+3&amp;&amp;\\text{Simplify}.\\end{align}[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1647&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven the function [latex]g\\left(m\\right)=\\sqrt{m - 4}[\/latex], evaluate [latex]g\\left(5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"273881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"273881\"]\r\n\r\n[latex]g\\left(5\\right)=\\sqrt{5- 4}=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=97486&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/GLOmTED1UwA\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Functions<\/h3>\r\nGiven the function [latex]h\\left(p\\right)={p}^{2}+2p[\/latex], solve for [latex]h\\left(p\\right)=3[\/latex].\r\n\r\n[reveal-answer q=\"119909\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"119909\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;h\\left(p\\right)=3\\\\ &amp;{p}^{2}+2p=3 &amp;&amp;\\text{Substitute the original function }h\\left(p\\right)={p}^{2}+2p. \\\\ &amp;{p}^{2}+2p - 3=0 &amp;&amp;\\text{Subtract 3 from each side}. \\\\ &amp;\\left(p+3\\text{)(}p - 1\\right)=0 &amp;&amp;\\text{Factor}. \\end{align}[\/latex]<\/p>\r\nIf [latex]\\left(p+3\\right)\\left(p - 1\\right)=0[\/latex], either [latex]\\left(p+3\\right)=0[\/latex] or [latex]\\left(p - 1\\right)=0[\/latex] (or both of them equal 0). We will set each factor equal to 0 and solve for [latex]p[\/latex] in each case.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;p+3=0, &amp;&amp;p=-3 \\\\ &amp;p - 1=0, &amp;&amp;p=1\\hfill \\end{align}[\/latex]<\/p>\r\nThis gives us two solutions. The output [latex]h\\left(p\\right)=3[\/latex] when the input is either [latex]p=1[\/latex] or [latex]p=-3[\/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\/18190959\/CNX_Precalc_Figure_01_01_0062.jpg\" alt=\"Graph of a parabola with labeled points (-3, 3), (1, 3), and (4, 24).\" width=\"487\" height=\"459\" \/>\r\n\r\nWe can also verify by graphing as in Figure 5. The graph verifies that [latex]h\\left(1\\right)=h\\left(-3\\right)=3[\/latex] and [latex]h\\left(4\\right)=24[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/watch?v=NTmgEF_nZSc\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven the function [latex]g\\left(m\\right)=\\sqrt{m - 4}[\/latex], solve [latex]g\\left(m\\right)=2[\/latex].\r\n\r\n[reveal-answer q=\"480629\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"480629\"]\r\n\r\n[latex]m=8[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15766&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Evaluating Functions Expressed in Formulas<\/h2>\r\nSome functions are defined by mathematical rules or procedures expressed in <strong>equation<\/strong> form. If it is possible to express the function output with a <strong>formula<\/strong> involving the input quantity, then we can define a function in algebraic form. For example, the equation [latex]2n+6p=12[\/latex] expresses a functional relationship between [latex]n[\/latex]\u00a0and [latex]p[\/latex]. We can rewrite it to decide if [latex]p[\/latex] is a function of [latex]n[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a function in equation form, write its algebraic formula.<\/h3>\r\n<ol>\r\n \t<li>Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves <em>only<\/em> the input variable.<\/li>\r\n \t<li>Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding an Equation of a Function<\/h3>\r\nExpress the relationship [latex]2n+6p=12[\/latex] as a function [latex]p=f\\left(n\\right)[\/latex], if possible.\r\n\r\n[reveal-answer q=\"938453\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"938453\"]\r\n\r\nTo express the relationship in this form, we need to be able to write the relationship where [latex]p[\/latex] is a function of [latex]n[\/latex], which means writing it as [latex]p=[\/latex] expression involving [latex]n[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;2n+6p=12\\\\[1mm] &amp;6p=12 - 2n &amp;&amp;\\text{Subtract }2n\\text{ from both sides}. \\\\[1mm] &amp;p=\\frac{12 - 2n}{6} &amp;&amp;\\text{Divide both sides by 6 and simplify}. \\\\[1mm] &amp;p=\\frac{12}{6}-\\frac{2n}{6} \\\\[1mm] &amp;p=2-\\frac{1}{3}n \\end{align}[\/latex]<\/p>\r\nTherefore, [latex]p[\/latex] as a function of [latex]n[\/latex] is written as\r\n<p style=\"text-align: center;\">[latex]p=f\\left(n\\right)=2-\\frac{1}{3}n[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nIt is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/watch?v=lHTLjfPpFyQ\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Expressing the Equation of a Circle as a Function<\/h3>\r\nDoes the equation [latex]{x}^{2}+{y}^{2}=1[\/latex] represent a function with [latex]x[\/latex] as input and [latex]y[\/latex] as output? If so, express the relationship as a function [latex]y=f\\left(x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"557070\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"557070\"]\r\n\r\nFirst we subtract [latex]{x}^{2}[\/latex] from both sides.\r\n<p style=\"text-align: center;\">[latex]{y}^{2}=1-{x}^{2}[\/latex]<\/p>\r\nWe now try to solve for [latex]y[\/latex] in this equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y&amp;=\\pm \\sqrt{1-{x}^{2}} \\\\[1mm] &amp;=\\sqrt{1-{x}^{2}}\\hspace{3mm}\\text{and}\\hspace{3mm}-\\sqrt{1-{x}^{2}} \\end{align}[\/latex]<\/p>\r\nWe get two outputs corresponding to the same input, so this relationship cannot be represented as a single function [latex]y=f\\left(x\\right)[\/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\nIf [latex]x - 8{y}^{3}=0[\/latex], express [latex]y[\/latex] as a function of [latex]x[\/latex].\r\n\r\n[reveal-answer q=\"933974\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"933974\"][latex]y=f\\left(x\\right)=\\cfrac{\\sqrt[3]{x}}{2}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=111699&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Are there relationships expressed by an equation that do represent a function but which still cannot be represented by an algebraic formula?<\/strong>\r\n\r\n<em>Yes, this can happen. For example, given the equation [latex]x=y+{2}^{y}[\/latex], if we want to express [latex]y[\/latex] as a function of [latex]x[\/latex], there is no simple algebraic formula involving only [latex]x[\/latex] that equals [latex]y[\/latex]. However, each [latex]x[\/latex] does determine a unique value for [latex]y[\/latex], and there are mathematical procedures by which [latex]y[\/latex] can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for [latex]y[\/latex] as a function of [latex]x[\/latex], even though the formula cannot be written explicitly.<\/em>\r\n\r\n<\/div>\r\n<h2>Evaluating a Function Given in Tabular Form<\/h2>\r\nAs we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy\u2019s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours.\r\n\r\nThe function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See the table below.\r\n<table summary=\"Six rows and two columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th>Pet<\/th>\r\n<th>Memory span in hours<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Puppy<\/td>\r\n<td>0.008<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Adult dog<\/td>\r\n<td>0.083<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Cat<\/td>\r\n<td>16<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Goldfish<\/td>\r\n<td>2160<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Beta fish<\/td>\r\n<td>3600<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAt times, evaluating a function in table form may be more useful than using equations. Here let us call the function [latex]P[\/latex].\r\n\r\nThe <strong>domain<\/strong> of the function is the type of pet and the range is a real number representing the number of hours the pet\u2019s memory span lasts. We can evaluate the function [latex]P[\/latex] at the input value of \"goldfish.\" We would write [latex]P\\left(\\text{goldfish}\\right)=2160[\/latex]. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function [latex]P[\/latex] seems ideally suited to this function, more so than writing it in paragraph or function form.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a function represented by a table, identify specific output and input values.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Find the given input in the row (or column) of input values.<\/li>\r\n \t<li>Identify the corresponding output value paired with that input value.<\/li>\r\n \t<li>Find the given output values in the row (or column) of output values, noting every time that output value appears.<\/li>\r\n \t<li>Identify the input value(s) corresponding to the given output value.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating and Solving a Tabular Function<\/h3>\r\nUsing the table below,\r\n<ol>\r\n \t<li>Evaluate [latex]g\\left(3\\right)[\/latex].<\/li>\r\n \t<li>Solve [latex]g\\left(n\\right)=6[\/latex].<\/li>\r\n<\/ol>\r\n<table summary=\"Two rows and six columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]n[\/latex]<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]g(n)[\/latex]<\/strong><\/td>\r\n<td>8<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>6<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"15206\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"15206\"]\r\n<ul>\r\n \t<li>Evaluating [latex]g\\left(3\\right)[\/latex] means determining the output value of the function [latex]g[\/latex] for the input value of [latex]n=3[\/latex]. The table output value corresponding to [latex]n=3[\/latex] is 7, so [latex]g\\left(3\\right)=7[\/latex].<\/li>\r\n \t<li>Solving [latex]g\\left(n\\right)=6[\/latex] means identifying the input values, [latex]n[\/latex], that produce an output value of 6. The table below shows two solutions: [latex]n=2[\/latex] and [latex]n=4[\/latex].<\/li>\r\n<\/ul>\r\n<table summary=\"Two rows and six columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]n[\/latex]<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]g(n)[\/latex]<\/strong><\/td>\r\n<td>8<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>6<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen we input 2 into the function [latex]g[\/latex], our output is 6. When we input 4 into the function [latex]g[\/latex], our output is also 6.\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 table from the previous example, evaluate [latex]g\\left(1\\right)[\/latex] .\r\n\r\n[reveal-answer q=\"724802\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"724802\"][latex]g\\left(1\\right)=8[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3751&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Finding Function Values from a Graph<\/h2>\r\nEvaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s).\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Reading Function Values from a Graph<\/h3>\r\nGiven the graph below,\r\n<ol>\r\n \t<li>Evaluate [latex]f\\left(2\\right)[\/latex].<\/li>\r\n \t<li>Solve [latex]f\\left(x\\right)=4[\/latex].<\/li>\r\n<\/ol>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191001\/CNX_Precalc_Figure_01_01_0072.jpg\" alt=\"Graph of a positive parabola centered at (1, 0).\" width=\"487\" height=\"445\" \/>\r\n[reveal-answer q=\"915833\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"915833\"]\r\n<ol>\r\n \t<li>To evaluate [latex]f\\left(2\\right)[\/latex], locate the point on the curve where [latex]x=2[\/latex], then read the [latex]y[\/latex]-coordinate of that point. The point has coordinates [latex]\\left(2,1\\right)[\/latex], so [latex]f\\left(2\\right)=1[\/latex].<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191004\/CNX_Precalc_Figure_01_01_0082.jpg\" alt=\"Graph of a positive parabola centered at (1, 0) with the labeled point (2, 1) where f(2) =1.\" width=\"487\" height=\"445\" \/><\/li>\r\n \t<li>To solve [latex]f\\left(x\\right)=4[\/latex], we find the output value [latex]4[\/latex] on the vertical axis. Moving horizontally along the line [latex]y=4[\/latex], we locate two points of the curve with output value [latex]4:[\/latex] [latex]\\left(-1,4\\right)[\/latex] and [latex]\\left(3,4\\right)[\/latex]. These points represent the two solutions to [latex]f\\left(x\\right)=4:[\/latex] [latex]x=-1[\/latex] or [latex]x=3[\/latex]. This means [latex]f\\left(-1\\right)=4[\/latex] and [latex]f\\left(3\\right)=4[\/latex], or when the input is [latex]-1[\/latex] or [latex]\\text{3,}[\/latex] the output is [latex]\\text{4}\\text{.}[\/latex] See the graph below.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191006\/CNX_Precalc_Figure_01_01_0092.jpg\" alt=\"Graph of an upward-facing\u00a0parabola with a vertex at (0,1) and\u00a0labeled points at (-1, 4) and (3,4). A\u00a0line at y = 4 intersects the parabola at the labeled points.\" width=\"487\" height=\"445\" \/><\/li>\r\n<\/ol>\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 graph, solve [latex]f\\left(x\\right)=1[\/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\/18191004\/CNX_Precalc_Figure_01_01_0082.jpg\" alt=\"Graph of a positive parabola centered at (1, 0) with the labeled point (2, 1) where f(2) =1.\" width=\"487\" height=\"445\" \/>\r\n[reveal-answer q=\"529772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"529772\"]\r\n\r\n[latex]x=0[\/latex] or [latex]x=2[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom9\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2471&amp;theme=oea&amp;iframe_resize_id=mom9\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe>\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2886&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"600\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Identify Functions Using Graphs<\/h2>\r\nAs we have seen in examples above, we can represent a function using a graph. Graphs display many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. We typically construct graphs with the input values along the horizontal axis and the output values along the vertical axis.\r\n\r\nThe most common graphs name the input value [latex]x[\/latex] and the output value [latex]y[\/latex], and we say [latex]y[\/latex] is a function of [latex]x[\/latex], or [latex]y=f\\left(x\\right)[\/latex] when the function is named [latex]f[\/latex]. The graph of the function is the set of all points [latex]\\left(x,y\\right)[\/latex] in the plane that satisfies the equation [latex]y=f\\left(x\\right)[\/latex]. If the function is defined for only a few input values, then the graph of the function is only a few points, where the <em>x<\/em>-coordinate of each point is an input value and the <em>y<\/em>-coordinate of each point is the corresponding output value. For example, the black dots on the graph in the graph below\u00a0tell us that [latex]f\\left(0\\right)=2[\/latex] and [latex]f\\left(6\\right)=1[\/latex]. However, the set of all points [latex]\\left(x,y\\right)[\/latex] satisfying [latex]y=f\\left(x\\right)[\/latex] is a curve. The curve shown includes [latex]\\left(0,2\\right)[\/latex] and [latex]\\left(6,1\\right)[\/latex] because the curve passes through those points.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191012\/CNX_Precalc_Figure_01_01_0112.jpg\" alt=\"Graph of a polynomial.\" width=\"731\" height=\"442\" \/>\r\n\r\nThe <strong>vertical line test<\/strong> can be used to determine whether a graph represents a function. A vertical line includes all points with a particular [latex]x[\/latex] value. The [latex]y[\/latex] value of a point where a vertical line intersects a graph represents an output for that input [latex]x[\/latex] value. If we can draw <em>any<\/em> vertical line that intersects a graph more than once, then the graph does <em>not<\/em> define a function because that [latex]x[\/latex] value has more than one output. A function has only one output value for each input 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\/18191014\/CNX_Precalc_Figure_01_01_0122.jpg\" alt=\"Three graphs visually showing what is and is not a function.\" width=\"975\" height=\"271\" \/>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a graph, use the vertical line test to determine if the graph represents a function.<\/h3>\r\n<ol>\r\n \t<li>Inspect the graph to see if any vertical line drawn would intersect the curve more than once.<\/li>\r\n \t<li>If there is any such line, the graph does not represent a function.<\/li>\r\n \t<li>If no vertical line can intersect the curve more than once, the graph does represent a function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying the Vertical Line Test<\/h3>\r\nWhich of the graphs represent(s) a function [latex]y=f\\left(x\\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\/18191017\/CNX_Precalc_Figure_01_01_013abc.jpg\" alt=\"Graph of a polynomial.\" width=\"975\" height=\"418\" \/>\r\n[reveal-answer q=\"689864\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689864\"]\r\n\r\nIf any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of the graph above. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most <em>x<\/em>-values, a vertical line would intersect the graph at more than one point.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191020\/CNX_Precalc_Figure_01_01_016.jpg\" alt=\"Graph of a circle.\" width=\"487\" height=\"445\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/5Z8DaZPJLKY\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nDoes the graph below represent a function?\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191023\/CNX_Precalc_Figure_01_01_017.jpg\" alt=\"Graph of absolute value function.\" width=\"487\" height=\"366\" \/>\r\n[reveal-answer q=\"783855\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"783855\"]\r\n\r\nYes.\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=40676&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"650\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>The Horizontal Line Test<\/h2>\r\nOnce we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the <strong>horizontal line test<\/strong>. Draw horizontal lines through the graph. A horizontal line includes all points with a particular [latex]y[\/latex] value. The [latex]x[\/latex] value of a point where a vertical line intersects a function represents the input for that output [latex]y[\/latex] value. If we can draw <em>any<\/em> horizontal line that intersects a graph more than once, then the graph does <em>not<\/em> represent a one-to-one function because that [latex]y[\/latex] value has more than one input.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.<\/h3>\r\n<ol>\r\n \t<li>Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.<\/li>\r\n \t<li>If there is any such line, the function is not one-to-one.<\/li>\r\n \t<li>If no horizontal line can intersect the curve more than once, the function is one-to-one.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying the Horizontal Line Test<\/h3>\r\nConsider the functions (a), and (b)shown in\u00a0the graphs\u00a0below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191017\/CNX_Precalc_Figure_01_01_013abc.jpg\" alt=\"Graph of a polynomial.\" width=\"975\" height=\"418\" \/>\r\n\r\nAre either of the functions one-to-one?\r\n\r\n[reveal-answer q=\"173050\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"173050\"]\r\n\r\nThe function in (a) is not one-to-one. The horizontal line shown below intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191025\/CNX_Precalc_Figure_01_01_010.jpg\" width=\"487\" height=\"445\" \/>\r\n\r\nThe function in (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=111715&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"450\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/tbSGdcSN8RE\r\n<h2>Characteristics of Inverse Functions<\/h2>\r\nSuppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the <strong>Celsius<\/strong> scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees <strong>Fahrenheit<\/strong> to degrees Celsius. She finds the formula [latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex] and substitutes 75 for [latex]F[\/latex] to calculate [latex]\\frac{5}{9}\\left(75 - 32\\right)\\approx {24}^{ \\circ} {C}[\/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\/18205512\/CNX_Precalc_Figure_01_07_0022.jpg\" alt=\"A forecast of Monday\u2019s through Thursday\u2019s weather.\" width=\"731\" height=\"226\" \/>\r\n\r\nKnowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week\u2019s weather forecast\u00a0for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.\r\n\r\nAt first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for [latex]F[\/latex] after substituting a value for [latex]C[\/latex]. For example, to convert 26 degrees Celsius, she could write\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;26=\\frac{5}{9}\\left(F - 32\\right) \\\\[1.5mm] &amp;26\\cdot \\frac{9}{5}=F - 32 \\\\[1.5mm] &amp;F=26\\cdot \\frac{9}{5}+32\\approx 79 \\end{align}[\/latex]<\/p>\r\nAfter considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.\r\n\r\nThe formula for which Betty is searching corresponds to the idea of an <strong>inverse function<\/strong>, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.\r\n\r\nGiven a function [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as \"[latex]f[\/latex] inverse of [latex]x[\/latex].\" The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em>not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.\r\n\r\nThe \"exponent-like\" notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[\/latex], so [latex]{f}^{-1}\\circ f[\/latex] equals the identity function, that is,\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(y\\right)=x[\/latex]<\/p>\r\nThis holds for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. Informally, this means that inverse functions \"undo\" each other. However, just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.\r\n\r\nGiven a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether either [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] or [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)\r\n\r\nFor example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/p>\r\nand\r\n<p style=\"text-align: center;\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/p>\r\nA few coordinate pairs from the graph of the function [latex]y=4x[\/latex] are (\u22122, \u22128), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\\frac{1}{4}x[\/latex] are (\u22128, \u22122), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Inverse Function<\/h3>\r\nFor any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].\r\n\r\nThe notation [latex]{f}^{-1}[\/latex] is read \"[latex]f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]\".\r\n\r\nKeep in mind that [latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex] and not all functions have inverses.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\r\nIf for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?\r\n\r\n[reveal-answer q=\"348517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348517\"]\r\n\r\nThe inverse function reverses the input and output quantities, so if\r\n<p style=\"text-align: center;\">[latex]f\\left(2\\right)=4[\/latex], then [latex]{f}^{-1}\\left(4\\right)=2[\/latex]<\/p>\r\n<p style=\"text-align: left;\">and if<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(5\\right)=12[\/latex], then [latex]{f}^{-1}\\left(12\\right)=5[\/latex]<\/p>\r\nAlternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.\r\n<table summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\r\n<thead>\r\n<tr>\r\n<th>[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\r\n<th>[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\left(5,12\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(12,5\\right)[\/latex]<\/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\nGiven that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]\r\n\r\n[reveal-answer q=\"664604\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664604\"]\r\n\r\n[latex]h(2)=6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/TSztRfzmk0M\r\n<div class=\"textbox\">\r\n<h3>How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\r\n<ol>\r\n \t<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>If both statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Testing Inverse Relationships Algebraically<\/h3>\r\nIf [latex]f\\left(x\\right)=\\dfrac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\dfrac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"421291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421291\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&amp;=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }\\\\[1.5mm]&amp;={ x }+{ 2 } -{ 2 }\\\\[1.5mm]&amp;={ x } \\end{align}[\/latex]<\/p>\r\nso\r\n<p style=\"text-align: center;\">[latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex]<\/p>\r\nThis is enough to answer yes to the question, but we can also verify the other formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)&amp;=\\frac{1}{\\frac{1}{x}-2+2}\\\\[1.5mm] &amp;=\\frac{1}{\\frac{1}{x}} \\\\[1.5mm] &amp;=x \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice the inverse operations are in reverse order of the operations from the original 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\nIf [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"226656\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"226656\"]\r\n\r\nYes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determining Inverse Relationships for Power Functions<\/h3>\r\nIf [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"637419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"637419\"]\r\n\r\n[latex]f\\left(g\\left(x\\right)\\right)=\\left(\\frac{1}{3}x\\right)^3=\\dfrac{{x}^{3}}{27}\\ne x[\/latex]\r\n\r\nNo, the functions are not inverses.\r\n<h4>Analysis of the Solution<\/h4>\r\nThe correct inverse to [latex]x^3[\/latex] is the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.\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\nIf [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]\r\n\r\n[reveal-answer q=\"384680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384680\"]\r\n\r\nYes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Determine the Domain and Range of an Inverse Function<\/h2>\r\nThe outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.\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\/18205514\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/> Domain and range of a function and its inverse[\/caption]\r\n\r\nWhen a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square \"undoes\" a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex].\r\n\r\nWe can look at this problem from the other side, starting with the square (parent quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the \"inverse\" is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.\r\n\r\nIn many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).\r\n\r\nIf [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+1[\/latex].\r\n<ul>\r\n \t<li>The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>Is it possible for a function to have more than one inverse?<\/strong>\r\n\r\n<em>No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Domain and Range of Inverse Functions<\/h3>\r\nThe range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\nThe domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3><strong>How To: Given a function, find the domain and range of its inverse.\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\r\n \t<li>If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Inverses of PARENT Functions<\/h3>\r\nIdentify which of the\u00a0parent functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The\u00a0parent functions are reviewed below. We restrict the domain in such a fashion that the function assumes all <em>y<\/em>-values exactly once.\r\n<table summary=\"A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1\/x; the reciprocal squared function is f(x)=1\/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1\/3).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>Constant<\/td>\r\n<td>Identity<\/td>\r\n<td>Quadratic<\/td>\r\n<td>Cubic<\/td>\r\n<td>Reciprocal<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)=c[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=x[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Reciprocal squared<\/td>\r\n<td>Cube root<\/td>\r\n<td>Square root<\/td>\r\n<td>Absolute value<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]<\/td>\r\n<td>[latex]f\\left(x\\right)=|x|[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"229708\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"229708\"]\r\n\r\nThe constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.\r\n\r\nThe absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.\r\n\r\nThe reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18205517\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/> (a) Absolute value (b) Reciprocal squared[\/caption]\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\nThe domain of the function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of the function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.\r\n\r\n[reveal-answer q=\"146498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"146498\"]\r\n\r\nThe domain of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(-\\infty \\text{,}-2\\right)[\/latex] and the range of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Define and Graph an Inverse<\/h2>\r\nOnce we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.\r\n<h2>Inverting Tabular Functions<\/h2>\r\nSuppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.\r\n\r\nEach row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Interpreting the Inverse of a Tabular Function<\/h3>\r\nA function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].\r\n<table summary=\"Two rows and five columns. The first row is labeled\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"61261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"61261\"]\r\n\r\nThe inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.\r\n\r\nAlternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/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\nUsing the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].\r\n<table summary=\"Two rows and five columns. The first row is labeled\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\r\n<td>30<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\r\n<td>20<\/td>\r\n<td>40<\/td>\r\n<td>50<\/td>\r\n<td>60<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"401025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"401025\"]\r\n\r\na.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.\r\n\r\nb. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\r\nThe domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.\r\n<div class=\"textbox\">\r\n<h3>How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\r\n<ol>\r\n \t<li>Find the desired input of the inverse function on the [latex]y[\/latex]-axis of the given graph.<\/li>\r\n \t<li>Read the inverse function\u2019s output from the [latex]x[\/latex]-axis of the given graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\r\nA function [latex]g\\left(x\\right)[\/latex] is given\u00a0below. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\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\/18205520\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/>\r\n\r\n[reveal-answer q=\"334632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"334632\"]\r\n\r\nTo evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the [latex]y[\/latex]-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].\r\n\r\nTo evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/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\/18205521\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/>\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 graph in the previous example, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"350455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350455\"]\r\n\r\na. 3; b. 5.6\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Inverses of Functions Represented by Formulas<\/h2>\r\nSometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014for example, [latex]y[\/latex] as a function of [latex]x-[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].\r\n<div class=\"textbox\">\r\n<h3>How To: Given a function represented by a formula, find the inverse.<\/h3>\r\n<ol>\r\n \t<li>Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\r\n \t<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\r\n \t<li>Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\r\n \t<li>Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Inverting the Fahrenheit-to-Celsius Function<\/h3>\r\nFind a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.\r\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"625400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625400\"]\r\n<p style=\"text-align: center;\">[latex]{ C }=\\frac{5}{9}\\left(F - 32\\right)[\/latex]\r\n[latex]C\\cdot \\frac{9}{5}=F - 32[\/latex]\r\n[latex]F=\\frac{9}{5}C+32[\/latex]<\/p>\r\nBy solving in general, we have uncovered the inverse function. If\r\n<p style=\"text-align: center;\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\r\nthen\r\n<p style=\"text-align: center;\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\r\nIn this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.\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\nSolve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]\r\n\r\n[reveal-answer q=\"875458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"875458\"]\r\n\r\n[latex]x=3y+5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving to Find an Inverse Function<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=\\dfrac{2}{x - 3}+4[\/latex].\r\n\r\n[reveal-answer q=\"903464\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"903464\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=\\frac{2}{x - 3}+4 &amp;&amp; \\text{Change }f(x)\\text{ to }y. \\\\[1.5mm]&amp;x=\\frac{2}{y - 3}+4 &amp;&amp; \\text{Switch }x\\text{ and }y. \\\\[1.5mm] &amp;y - 4=\\frac{2}{x - 3} &amp;&amp; \\text{Subtract 4 from both sides}. \\\\[1.5mm] &amp;y - 3=\\frac{2}{x - 4} &amp;&amp; \\text{Multiply both sides by }y - 3\\text{ and divide by }x - 4. \\\\[1.5mm] &amp;y=\\frac{2}{x - 4}+3 &amp;&amp; \\text{Add 3 to both sides}.\\\\[-3mm]&amp;\\end{align}[\/latex]<\/p>\r\nSo [latex]{f}^{-1}\\left(x\\right)=\\dfrac{2}{x - 4}+3[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.\r\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\r\n<td>3<\/td>\r\n<td>2<\/td>\r\n<td>5<\/td>\r\n<td>[latex]y[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving to Find an Inverse with Radicals<\/h3>\r\nFind the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].\r\n\r\n[reveal-answer q=\"444147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"444147\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=2+\\sqrt{x - 4}\\\\[1.5mm]&amp;x=2+\\sqrt{y - 4}\\\\[1.5mm] &amp;{\\left(x - 2\\right)}^{2}=y - 4 \\\\[1.5mm] &amp;y={\\left(x- 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\r\nSo [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].\r\n\r\nThe domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/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\nWhat is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt{x}[\/latex]? State the domains of both the function and the inverse function.\r\nUse an online graphing tool to graph the function, its inverse, and [latex]f(x) = x[\/latex] to check whether you are correct.\r\n\r\n[reveal-answer q=\"327817\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"327817\"]\r\n\r\n[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of\u00a0 [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{ f}^{-1}:\\left(-\\infty ,2\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graph a Function's Inverse<\/h2>\r\nNow that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as below.\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\/18205523\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/> Quadratic function with domain restricted to [0, \u221e).[\/caption]<strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.\r\n\r\nWe already know that the inverse of the\u00a0parent quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]\r\n\r\nWe notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown below.\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\/18205525\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/> Square and square-root functions on the non-negative domain[\/caption]\r\n\r\nThis relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\r\nGiven the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\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\/18205527\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/>\r\n[reveal-answer q=\"694066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"694066\"]\r\n\r\nThis is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].\r\n\r\nIf we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in the graph below.\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\/18205529\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/> The function and its inverse, showing reflection about the identity line[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>Is there any function that is equal to its own inverse?<\/strong>\r\n\r\n<em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em>\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/p>\r\n<em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em>\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>A graph represents a one-to-one function if any horizontal line drawn on the graph intersects the graph at no more than one point.<\/li>\r\n \t<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then [latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex].<\/li>\r\n \t<li>Each of the\u00a0parent functions, except [latex]y=c[\/latex] has an inverse. Some need a restricted domain.<\/li>\r\n \t<li>For a function to have an inverse, it must be one-to-one (pass the horizontal line test).<\/li>\r\n \t<li>A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.<\/li>\r\n \t<li>For a tabular function, exchange the input and output rows to obtain the inverse.<\/li>\r\n \t<li>The inverse of a function can be determined at specific points on its graph.<\/li>\r\n \t<li>To find the inverse of a function [latex]y=f\\left(x\\right)[\/latex], switch the variables [latex]x[\/latex] and [latex]y[\/latex]. Then solve for [latex]y[\/latex] as a function of [latex]x[\/latex].<\/li>\r\n \t<li>The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137932588\" class=\"definition\">\r\n \t<dt><strong>horizontal line test<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134149777\">a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134149782\" class=\"definition\">\r\n \t<dt><strong>independent variable<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134149787\">an input variable<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135511353\" class=\"definition\"><\/dl>\r\n<dl id=\"fs-id1165135511364\" class=\"definition\"><\/dl>\r\n<dl id=\"fs-id1165137441703\" class=\"definition\">\r\n \t<dt>inverse function<\/dt>\r\n \t<dd id=\"fs-id1165137441708\">for any one-to-one function [latex]f\\left(x\\right)[\/latex], the inverse is a function [latex]{f}^{-1}\\left(x\\right)[\/latex] such that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]; this also implies that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135511364\" class=\"definition\">\r\n \t<dt><strong>one-to-one function<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135511369\">a function for which each value of the output is associated with a unique input value<\/dd>\r\n<\/dl>\r\n<section id=\"fs-id1165137660004\"><\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n<li>Determine whether a function is one-to-one.<\/li>\n<li class=\"li2\"><span class=\"s1\">Verify inverse functions.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one.<\/span><\/li>\n<li class=\"li2\"><span class=\"s1\">Find or evaluate the inverse of a function.<\/span><\/li>\n<li class=\"li3\"><span class=\"s8\">Use the graph of a one-to-one function to graph its inverse function on the same axes.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165135358875\">A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating.<\/p>\n<p>If some physical machines can run in two directions, we might ask whether some of the function &#8220;machines&#8221; we have been studying can also run backwards. Figure 1\u00a0provides a visual representation of this question. In this section, we will consider the reverse nature of functions.<\/p>\n<div style=\"width: 741px\" 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\/18205509\/CNX_Precalc_Figure_01_07_0012.jpg\" alt=\"Diagram of a function and what would be its inverse.\" width=\"731\" height=\"305\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Can a function &#8220;machine&#8221; operate in reverse?<\/p>\n<\/div>\n<h2>Determine whether a function is one-to-one<\/h2>\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\/18191009\/CNX_Precalc_Figure_01_00_001n2.jpg\" alt=\"Figure of a bull and a graph of market prices.\" width=\"975\" height=\"307\" \/><\/p>\n<p>Some functions have a given output value that corresponds to two or more input values. For example, in the following stock chart the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000.<\/p>\n<p>However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in.<\/p>\n<table summary=\"Two columns and five rows. The first column is labeled,\">\n<thead>\n<tr>\n<th>Letter grade<\/th>\n<th>Grade point average<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>A<\/td>\n<td>4.0<\/td>\n<\/tr>\n<tr>\n<td>B<\/td>\n<td>3.0<\/td>\n<\/tr>\n<tr>\n<td>C<\/td>\n<td>2.0<\/td>\n<\/tr>\n<tr>\n<td>D<\/td>\n<td>1.0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>This grading system represents a one-to-one function, because each letter input yields one particular grade point average output and each grade point average corresponds to one input letter.<\/p>\n<p>To visualize this concept, let\u2019s look again at the two simple functions sketched in (a) and (b)\u00a0below.<\/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\/18190946\/CNX_Precalc_Figure_01_01_0012.jpg\" alt=\"Three relations that demonstrate what constitute a function.\" width=\"975\" height=\"243\" \/><\/p>\n<p>The function in part (a) shows a relationship that is not a one-to-one function because inputs [latex]q[\/latex] and [latex]r[\/latex] both give output [latex]n[\/latex]. The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: One-to-One Function<\/h3>\n<p>A one-to-one function is a function in which each output value corresponds to exactly one input value.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Whether a Relationship Is a One-to-One Function<\/h3>\n<p>Is the area of a circle a function of its radius? If yes, is the function one-to-one?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380432\">Show Solution<\/span><\/p>\n<div id=\"q380432\" class=\"hidden-answer\" style=\"display: none\">\n<p>A circle of radius [latex]r[\/latex] has a unique area measure given by [latex]A=\\pi {r}^{2}[\/latex], so for any input, [latex]r[\/latex], there is only one output, [latex]A[\/latex]. The area is a function of radius [latex]r[\/latex].<\/p>\n<p>If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure [latex]A[\/latex] is given by the formula [latex]A=\\pi {r}^{2}[\/latex]. Because areas and radii are positive numbers, there is exactly one solution: [latex]r=\\sqrt{\\frac{A}{\\pi }}[\/latex]. So the area of a circle is a one-to-one function of the circle\u2019s radius.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<ol>\n<li>Is a balance a function of the bank account number?<\/li>\n<li>Is a bank account number a function of the balance?<\/li>\n<li>Is a balance a one-to-one function of the bank account number?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q997233\">Show Solution<\/span><\/p>\n<div id=\"q997233\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><span class=\"s1\">yes, because each bank account (input) has a single balance (output) at any given time.<\/span><\/li>\n<li><span class=\"s1\">no, because several bank accounts (inputs) may have the same balance (output).<\/span><\/li>\n<li><span class=\"s1\">no, because the more than one bank account (input) can have the same balance (output).<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom7\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15800&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Evaluating and Solving Functions<\/h2>\n<p>When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function [latex]f\\left(x\\right)=5 - 3{x}^{2}[\/latex] can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.<\/p>\n<div class=\"textbox\">\n<h3><strong>How To: EVALUATE A FUNCTION Given ITS FORMula.<br \/>\n<\/strong><\/h3>\n<ol>\n<li>Replace the input variable in the formula with the value provided.<\/li>\n<li>Calculate the result.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating Functions<\/h3>\n<p>Given the function [latex]h\\left(p\\right)={p}^{2}+2p[\/latex], evaluate [latex]h\\left(4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q768180\">Show Solution<\/span><\/p>\n<div id=\"q768180\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">To evaluate [latex]h\\left(4\\right)[\/latex], we substitute the value 4 for the input variable [latex]p[\/latex] in the given function.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}h\\left(p\\right)&={p}^{2}+2p \\\\ h\\left(4\\right)&={\\left(4\\right)}^{2}+2\\left(4\\right) \\\\ &=16+8 \\\\ &=24 \\end{align}[\/latex]<\/p>\n<p>Therefore, for an input of 4, we have an output of 24 or [latex]h(4)=24[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Evaluating Functions Using Function Notation (L9.3)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/Ehkzu5Uv7O0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating Functions at Specific Values<\/h3>\n<p>For the function, [latex]f\\left(x\\right)={x}^{2}+3x - 4[\/latex], evaluate each of the following.<\/p>\n<ol>\n<li>[latex]f\\left(2\\right)[\/latex]<\/li>\n<li>[latex]f(a)[\/latex]<\/li>\n<li>[latex]f(a+h)[\/latex]<\/li>\n<li>[latex]\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q645951\">Show Solution<\/span><\/p>\n<div id=\"q645951\" class=\"hidden-answer\" style=\"display: none\">\n<p>Replace the [latex]x[\/latex]\u00a0in the function with each specified value.<\/p>\n<ol>\n<li>Because the input value is a number, 2, we can use algebra to simplify.\n<div style=\"text-align: center;\">[latex]\\begin{align}f\\left(2\\right)&={2}^{2}+3\\left(2\\right)-4 \\\\ &=4+6 - 4 \\\\ &=6\\hfill \\end{align}[\/latex]<\/div>\n<\/li>\n<li>In this case, the input value is a letter so we cannot simplify the answer any further.\n<div style=\"text-align: center;\">[latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/div>\n<\/li>\n<li>With an input value of [latex]a+h[\/latex], we must use the distributive property.\n<div style=\"text-align: center;\">[latex]\\begin{align}f\\left(a+h\\right)&={\\left(a+h\\right)}^{2}+3\\left(a+h\\right)-4 \\\\[2mm] &={a}^{2}+2ah+{h}^{2}+3a+3h - 4 \\end{align}[\/latex]<\/div>\n<\/li>\n<li>In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that\n<div style=\"text-align: center;\">[latex]f\\left(a+h\\right)={a}^{2}+2ah+{h}^{2}+3a+3h - 4[\/latex]<\/div>\n<p>and we know that<\/p>\n<div style=\"text-align: center;\">[latex]f\\left(a\\right)={a}^{2}+3a - 4[\/latex]<\/div>\n<p>Now we combine the results and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}&=\\dfrac{\\left({a}^{2}+2ah+{h}^{2}+3a+3h - 4\\right)-\\left({a}^{2}+3a - 4\\right)}{h} \\\\[2mm] &=\\dfrac{2ah+{h}^{2}+3h}{h}\\\\[2mm] &=\\frac{h\\left(2a+h+3\\right)}{h}&&\\text{Factor out }h. \\\\[2mm] &=2a+h+3&&\\text{Simplify}.\\end{align}[\/latex]<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1647&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the function [latex]g\\left(m\\right)=\\sqrt{m - 4}[\/latex], evaluate [latex]g\\left(5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q273881\">Show Solution<\/span><\/p>\n<div id=\"q273881\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]g\\left(5\\right)=\\sqrt{5- 4}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=97486&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex: Find Function Inputs for a Given Quadratic Function Output\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GLOmTED1UwA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Functions<\/h3>\n<p>Given the function [latex]h\\left(p\\right)={p}^{2}+2p[\/latex], solve for [latex]h\\left(p\\right)=3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q119909\">Show Solution<\/span><\/p>\n<div id=\"q119909\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&h\\left(p\\right)=3\\\\ &{p}^{2}+2p=3 &&\\text{Substitute the original function }h\\left(p\\right)={p}^{2}+2p. \\\\ &{p}^{2}+2p - 3=0 &&\\text{Subtract 3 from each side}. \\\\ &\\left(p+3\\text{)(}p - 1\\right)=0 &&\\text{Factor}. \\end{align}[\/latex]<\/p>\n<p>If [latex]\\left(p+3\\right)\\left(p - 1\\right)=0[\/latex], either [latex]\\left(p+3\\right)=0[\/latex] or [latex]\\left(p - 1\\right)=0[\/latex] (or both of them equal 0). We will set each factor equal to 0 and solve for [latex]p[\/latex] in each case.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&p+3=0, &&p=-3 \\\\ &p - 1=0, &&p=1\\hfill \\end{align}[\/latex]<\/p>\n<p>This gives us two solutions. The output [latex]h\\left(p\\right)=3[\/latex] when the input is either [latex]p=1[\/latex] or [latex]p=-3[\/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\/18190959\/CNX_Precalc_Figure_01_01_0062.jpg\" alt=\"Graph of a parabola with labeled points (-3, 3), (1, 3), and (4, 24).\" width=\"487\" height=\"459\" \/><\/p>\n<p>We can also verify by graphing as in Figure 5. The graph verifies that [latex]h\\left(1\\right)=h\\left(-3\\right)=3[\/latex] and [latex]h\\left(4\\right)=24[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Finding Function Values\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/NTmgEF_nZSc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the function [latex]g\\left(m\\right)=\\sqrt{m - 4}[\/latex], solve [latex]g\\left(m\\right)=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q480629\">Show Solution<\/span><\/p>\n<div id=\"q480629\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]m=8[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15766&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Evaluating Functions Expressed in Formulas<\/h2>\n<p>Some functions are defined by mathematical rules or procedures expressed in <strong>equation<\/strong> form. If it is possible to express the function output with a <strong>formula<\/strong> involving the input quantity, then we can define a function in algebraic form. For example, the equation [latex]2n+6p=12[\/latex] expresses a functional relationship between [latex]n[\/latex]\u00a0and [latex]p[\/latex]. We can rewrite it to decide if [latex]p[\/latex] is a function of [latex]n[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a function in equation form, write its algebraic formula.<\/h3>\n<ol>\n<li>Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves <em>only<\/em> the input variable.<\/li>\n<li>Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding an Equation of a Function<\/h3>\n<p>Express the relationship [latex]2n+6p=12[\/latex] as a function [latex]p=f\\left(n\\right)[\/latex], if possible.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q938453\">Show Solution<\/span><\/p>\n<div id=\"q938453\" class=\"hidden-answer\" style=\"display: none\">\n<p>To express the relationship in this form, we need to be able to write the relationship where [latex]p[\/latex] is a function of [latex]n[\/latex], which means writing it as [latex]p=[\/latex] expression involving [latex]n[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&2n+6p=12\\\\[1mm] &6p=12 - 2n &&\\text{Subtract }2n\\text{ from both sides}. \\\\[1mm] &p=\\frac{12 - 2n}{6} &&\\text{Divide both sides by 6 and simplify}. \\\\[1mm] &p=\\frac{12}{6}-\\frac{2n}{6} \\\\[1mm] &p=2-\\frac{1}{3}n \\end{align}[\/latex]<\/p>\n<p>Therefore, [latex]p[\/latex] as a function of [latex]n[\/latex] is written as<\/p>\n<p style=\"text-align: center;\">[latex]p=f\\left(n\\right)=2-\\frac{1}{3}n[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>It is important to note that not every relationship expressed by an equation can also be expressed as a function with a formula.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex: Write a Linear Relation as a Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/lHTLjfPpFyQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Expressing the Equation of a Circle as a Function<\/h3>\n<p>Does the equation [latex]{x}^{2}+{y}^{2}=1[\/latex] represent a function with [latex]x[\/latex] as input and [latex]y[\/latex] as output? If so, express the relationship as a function [latex]y=f\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q557070\">Show Solution<\/span><\/p>\n<div id=\"q557070\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we subtract [latex]{x}^{2}[\/latex] from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]{y}^{2}=1-{x}^{2}[\/latex]<\/p>\n<p>We now try to solve for [latex]y[\/latex] in this equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y&=\\pm \\sqrt{1-{x}^{2}} \\\\[1mm] &=\\sqrt{1-{x}^{2}}\\hspace{3mm}\\text{and}\\hspace{3mm}-\\sqrt{1-{x}^{2}} \\end{align}[\/latex]<\/p>\n<p>We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function [latex]y=f\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]x - 8{y}^{3}=0[\/latex], express [latex]y[\/latex] as a function of [latex]x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q933974\">Show Solution<\/span><\/p>\n<div id=\"q933974\" class=\"hidden-answer\" style=\"display: none\">[latex]y=f\\left(x\\right)=\\cfrac{\\sqrt[3]{x}}{2}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=111699&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Are there relationships expressed by an equation that do represent a function but which still cannot be represented by an algebraic formula?<\/strong><\/p>\n<p><em>Yes, this can happen. For example, given the equation [latex]x=y+{2}^{y}[\/latex], if we want to express [latex]y[\/latex] as a function of [latex]x[\/latex], there is no simple algebraic formula involving only [latex]x[\/latex] that equals [latex]y[\/latex]. However, each [latex]x[\/latex] does determine a unique value for [latex]y[\/latex], and there are mathematical procedures by which [latex]y[\/latex] can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for [latex]y[\/latex] as a function of [latex]x[\/latex], even though the formula cannot be written explicitly.<\/em><\/p>\n<\/div>\n<h2>Evaluating a Function Given in Tabular Form<\/h2>\n<p>As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy\u2019s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours.<\/p>\n<p>The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See the table below.<\/p>\n<table summary=\"Six rows and two columns. The first column is labeled,\">\n<thead>\n<tr>\n<th>Pet<\/th>\n<th>Memory span in hours<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Puppy<\/td>\n<td>0.008<\/td>\n<\/tr>\n<tr>\n<td>Adult dog<\/td>\n<td>0.083<\/td>\n<\/tr>\n<tr>\n<td>Cat<\/td>\n<td>16<\/td>\n<\/tr>\n<tr>\n<td>Goldfish<\/td>\n<td>2160<\/td>\n<\/tr>\n<tr>\n<td>Beta fish<\/td>\n<td>3600<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>At times, evaluating a function in table form may be more useful than using equations. Here let us call the function [latex]P[\/latex].<\/p>\n<p>The <strong>domain<\/strong> of the function is the type of pet and the range is a real number representing the number of hours the pet\u2019s memory span lasts. We can evaluate the function [latex]P[\/latex] at the input value of &#8220;goldfish.&#8221; We would write [latex]P\\left(\\text{goldfish}\\right)=2160[\/latex]. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function [latex]P[\/latex] seems ideally suited to this function, more so than writing it in paragraph or function form.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a function represented by a table, identify specific output and input values.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Find the given input in the row (or column) of input values.<\/li>\n<li>Identify the corresponding output value paired with that input value.<\/li>\n<li>Find the given output values in the row (or column) of output values, noting every time that output value appears.<\/li>\n<li>Identify the input value(s) corresponding to the given output value.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating and Solving a Tabular Function<\/h3>\n<p>Using the table below,<\/p>\n<ol>\n<li>Evaluate [latex]g\\left(3\\right)[\/latex].<\/li>\n<li>Solve [latex]g\\left(n\\right)=6[\/latex].<\/li>\n<\/ol>\n<table summary=\"Two rows and six columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>[latex]n[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g(n)[\/latex]<\/strong><\/td>\n<td>8<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>6<\/td>\n<td>8<\/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=\"q15206\">Show Solution<\/span><\/p>\n<div id=\"q15206\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>Evaluating [latex]g\\left(3\\right)[\/latex] means determining the output value of the function [latex]g[\/latex] for the input value of [latex]n=3[\/latex]. The table output value corresponding to [latex]n=3[\/latex] is 7, so [latex]g\\left(3\\right)=7[\/latex].<\/li>\n<li>Solving [latex]g\\left(n\\right)=6[\/latex] means identifying the input values, [latex]n[\/latex], that produce an output value of 6. The table below shows two solutions: [latex]n=2[\/latex] and [latex]n=4[\/latex].<\/li>\n<\/ul>\n<table summary=\"Two rows and six columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>[latex]n[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g(n)[\/latex]<\/strong><\/td>\n<td>8<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>6<\/td>\n<td>8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When we input 2 into the function [latex]g[\/latex], our output is 6. When we input 4 into the function [latex]g[\/latex], our output is also 6.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the table from the previous example, evaluate [latex]g\\left(1\\right)[\/latex] .<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q724802\">Show Solution<\/span><\/p>\n<div id=\"q724802\" class=\"hidden-answer\" style=\"display: none\">[latex]g\\left(1\\right)=8[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3751&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Finding Function Values from a Graph<\/h2>\n<p>Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s).<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Reading Function Values from a Graph<\/h3>\n<p>Given the graph below,<\/p>\n<ol>\n<li>Evaluate [latex]f\\left(2\\right)[\/latex].<\/li>\n<li>Solve [latex]f\\left(x\\right)=4[\/latex].<\/li>\n<\/ol>\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\/18191001\/CNX_Precalc_Figure_01_01_0072.jpg\" alt=\"Graph of a positive parabola centered at (1, 0).\" width=\"487\" height=\"445\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q915833\">Show Solution<\/span><\/p>\n<div id=\"q915833\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>To evaluate [latex]f\\left(2\\right)[\/latex], locate the point on the curve where [latex]x=2[\/latex], then read the [latex]y[\/latex]-coordinate of that point. The point has coordinates [latex]\\left(2,1\\right)[\/latex], so [latex]f\\left(2\\right)=1[\/latex].<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191004\/CNX_Precalc_Figure_01_01_0082.jpg\" alt=\"Graph of a positive parabola centered at (1, 0) with the labeled point (2, 1) where f(2) =1.\" width=\"487\" height=\"445\" \/><\/li>\n<li>To solve [latex]f\\left(x\\right)=4[\/latex], we find the output value [latex]4[\/latex] on the vertical axis. Moving horizontally along the line [latex]y=4[\/latex], we locate two points of the curve with output value [latex]4:[\/latex] [latex]\\left(-1,4\\right)[\/latex] and [latex]\\left(3,4\\right)[\/latex]. These points represent the two solutions to [latex]f\\left(x\\right)=4:[\/latex] [latex]x=-1[\/latex] or [latex]x=3[\/latex]. This means [latex]f\\left(-1\\right)=4[\/latex] and [latex]f\\left(3\\right)=4[\/latex], or when the input is [latex]-1[\/latex] or [latex]\\text{3,}[\/latex] the output is [latex]\\text{4}\\text{.}[\/latex] See the graph below.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191006\/CNX_Precalc_Figure_01_01_0092.jpg\" alt=\"Graph of an upward-facing\u00a0parabola with a vertex at (0,1) and\u00a0labeled points at (-1, 4) and (3,4). A\u00a0line at y = 4 intersects the parabola at the labeled points.\" width=\"487\" height=\"445\" \/><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the graph, solve [latex]f\\left(x\\right)=1[\/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\/18191004\/CNX_Precalc_Figure_01_01_0082.jpg\" alt=\"Graph of a positive parabola centered at (1, 0) with the labeled point (2, 1) where f(2) =1.\" width=\"487\" height=\"445\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q529772\">Show Solution<\/span><\/p>\n<div id=\"q529772\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=0[\/latex] or [latex]x=2[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom9\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2471&amp;theme=oea&amp;iframe_resize_id=mom9\" width=\"100%\" height=\"550\" data-mce-fragment=\"1\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2886&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"600\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Identify Functions Using Graphs<\/h2>\n<p>As we have seen in examples above, we can represent a function using a graph. Graphs display many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. We typically construct graphs with the input values along the horizontal axis and the output values along the vertical axis.<\/p>\n<p>The most common graphs name the input value [latex]x[\/latex] and the output value [latex]y[\/latex], and we say [latex]y[\/latex] is a function of [latex]x[\/latex], or [latex]y=f\\left(x\\right)[\/latex] when the function is named [latex]f[\/latex]. The graph of the function is the set of all points [latex]\\left(x,y\\right)[\/latex] in the plane that satisfies the equation [latex]y=f\\left(x\\right)[\/latex]. If the function is defined for only a few input values, then the graph of the function is only a few points, where the <em>x<\/em>-coordinate of each point is an input value and the <em>y<\/em>-coordinate of each point is the corresponding output value. For example, the black dots on the graph in the graph below\u00a0tell us that [latex]f\\left(0\\right)=2[\/latex] and [latex]f\\left(6\\right)=1[\/latex]. However, the set of all points [latex]\\left(x,y\\right)[\/latex] satisfying [latex]y=f\\left(x\\right)[\/latex] is a curve. The curve shown includes [latex]\\left(0,2\\right)[\/latex] and [latex]\\left(6,1\\right)[\/latex] because the curve passes through those points.<\/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\/18191012\/CNX_Precalc_Figure_01_01_0112.jpg\" alt=\"Graph of a polynomial.\" width=\"731\" height=\"442\" \/><\/p>\n<p>The <strong>vertical line test<\/strong> can be used to determine whether a graph represents a function. A vertical line includes all points with a particular [latex]x[\/latex] value. The [latex]y[\/latex] value of a point where a vertical line intersects a graph represents an output for that input [latex]x[\/latex] value. If we can draw <em>any<\/em> vertical line that intersects a graph more than once, then the graph does <em>not<\/em> define a function because that [latex]x[\/latex] value has more than one output. A function has only one output value for each input 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\/18191014\/CNX_Precalc_Figure_01_01_0122.jpg\" alt=\"Three graphs visually showing what is and is not a function.\" width=\"975\" height=\"271\" \/><\/p>\n<div class=\"textbox\">\n<h3>How To: Given a graph, use the vertical line test to determine if the graph represents a function.<\/h3>\n<ol>\n<li>Inspect the graph to see if any vertical line drawn would intersect the curve more than once.<\/li>\n<li>If there is any such line, the graph does not represent a function.<\/li>\n<li>If no vertical line can intersect the curve more than once, the graph does represent a function.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Applying the Vertical Line Test<\/h3>\n<p>Which of the graphs represent(s) a function [latex]y=f\\left(x\\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\/18191017\/CNX_Precalc_Figure_01_01_013abc.jpg\" alt=\"Graph of a polynomial.\" width=\"975\" height=\"418\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q689864\">Show Solution<\/span><\/p>\n<div id=\"q689864\" class=\"hidden-answer\" style=\"display: none\">\n<p>If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of the graph above. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most <em>x<\/em>-values, a vertical line would intersect the graph at more than one point.<\/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\/18191020\/CNX_Precalc_Figure_01_01_016.jpg\" alt=\"Graph of a circle.\" width=\"487\" height=\"445\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex 1: Use the Vertical Line Test to Determine if a Graph Represents a Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/5Z8DaZPJLKY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Does the graph below represent a function?<\/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\/18191023\/CNX_Precalc_Figure_01_01_017.jpg\" alt=\"Graph of absolute value function.\" width=\"487\" height=\"366\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q783855\">Show Solution<\/span><\/p>\n<div id=\"q783855\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes.<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=40676&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"650\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>The Horizontal Line Test<\/h2>\n<p>Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the <strong>horizontal line test<\/strong>. Draw horizontal lines through the graph. A horizontal line includes all points with a particular [latex]y[\/latex] value. The [latex]x[\/latex] value of a point where a vertical line intersects a function represents the input for that output [latex]y[\/latex] value. If we can draw <em>any<\/em> horizontal line that intersects a graph more than once, then the graph does <em>not<\/em> represent a one-to-one function because that [latex]y[\/latex] value has more than one input.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.<\/h3>\n<ol>\n<li>Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.<\/li>\n<li>If there is any such line, the function is not one-to-one.<\/li>\n<li>If no horizontal line can intersect the curve more than once, the function is one-to-one.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Applying the Horizontal Line Test<\/h3>\n<p>Consider the functions (a), and (b)shown in\u00a0the graphs\u00a0below.<\/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\/18191017\/CNX_Precalc_Figure_01_01_013abc.jpg\" alt=\"Graph of a polynomial.\" width=\"975\" height=\"418\" \/><\/p>\n<p>Are either of the functions one-to-one?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q173050\">Show Solution<\/span><\/p>\n<div id=\"q173050\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function in (a) is not one-to-one. The horizontal line shown below intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)<\/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\/18191025\/CNX_Precalc_Figure_01_01_010.jpg\" width=\"487\" height=\"445\" alt=\"image\" \/><\/p>\n<p>The function in (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=111715&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"450\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex 1:  Determine if the Graph of a Relation is a One-to-One Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tbSGdcSN8RE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Characteristics of Inverse Functions<\/h2>\n<p>Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the <strong>Celsius<\/strong> scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees <strong>Fahrenheit<\/strong> to degrees Celsius. She finds the formula [latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex] and substitutes 75 for [latex]F[\/latex] to calculate [latex]\\frac{5}{9}\\left(75 - 32\\right)\\approx {24}^{ \\circ} {C}[\/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\/18205512\/CNX_Precalc_Figure_01_07_0022.jpg\" alt=\"A forecast of Monday\u2019s through Thursday\u2019s weather.\" width=\"731\" height=\"226\" \/><\/p>\n<p>Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week\u2019s weather forecast\u00a0for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.<\/p>\n<p>At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for [latex]F[\/latex] after substituting a value for [latex]C[\/latex]. For example, to convert 26 degrees Celsius, she could write<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&26=\\frac{5}{9}\\left(F - 32\\right) \\\\[1.5mm] &26\\cdot \\frac{9}{5}=F - 32 \\\\[1.5mm] &F=26\\cdot \\frac{9}{5}+32\\approx 79 \\end{align}[\/latex]<\/p>\n<p>After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.<\/p>\n<p>The formula for which Betty is searching corresponds to the idea of an <strong>inverse function<\/strong>, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.<\/p>\n<p>Given a function [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as &#8220;[latex]f[\/latex] inverse of [latex]x[\/latex].&#8221; The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em>not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\n<p>The &#8220;exponent-like&#8221; notation comes from an analogy between function composition and multiplication: just as [latex]{a}^{-1}a=1[\/latex] (1 is the identity element for multiplication) for any nonzero number [latex]a[\/latex], so [latex]{f}^{-1}\\circ f[\/latex] equals the identity function, that is,<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(y\\right)=x[\/latex]<\/p>\n<p>This holds for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. Informally, this means that inverse functions &#8220;undo&#8221; each other. However, just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.<\/p>\n<p>Given a function [latex]f\\left(x\\right)[\/latex], we can verify whether some other function [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex] by checking whether either [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex] or [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)<\/p>\n<p>For example, [latex]y=4x[\/latex] and [latex]y=\\frac{1}{4}x[\/latex] are inverse functions.<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{-1}\\circ f\\right)\\left(x\\right)={f}^{-1}\\left(4x\\right)=\\frac{1}{4}\\left(4x\\right)=x[\/latex]<\/p>\n<p>and<\/p>\n<p style=\"text-align: center;\">[latex]\\left({f}^{}\\circ {f}^{-1}\\right)\\left(x\\right)=f\\left(\\frac{1}{4}x\\right)=4\\left(\\frac{1}{4}x\\right)=x[\/latex]<\/p>\n<p>A few coordinate pairs from the graph of the function [latex]y=4x[\/latex] are (\u22122, \u22128), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function [latex]y=\\frac{1}{4}x[\/latex] are (\u22128, \u22122), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Inverse Function<\/h3>\n<p>For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex]. This can also be written as [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]. It also follows that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex] if [latex]{f}^{-1}[\/latex] is the inverse of [latex]f[\/latex].<\/p>\n<p>The notation [latex]{f}^{-1}[\/latex] is read &#8220;[latex]f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x[\/latex]&#8220;.<\/p>\n<p>Keep in mind that [latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex] and not all functions have inverses.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying an Inverse Function for a Given Input-Output Pair<\/h3>\n<p>If for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348517\">Show Solution<\/span><\/p>\n<div id=\"q348517\" class=\"hidden-answer\" style=\"display: none\">\n<p>The inverse function reverses the input and output quantities, so if<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(2\\right)=4[\/latex], then [latex]{f}^{-1}\\left(4\\right)=2[\/latex]<\/p>\n<p style=\"text-align: left;\">and if<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(5\\right)=12[\/latex], then [latex]{f}^{-1}\\left(12\\right)=5[\/latex]<\/p>\n<p>Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\n<table summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\n<thead>\n<tr>\n<th>[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\n<th>[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\left(5,12\\right)[\/latex]<\/td>\n<td>[latex]\\left(12,5\\right)[\/latex]<\/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>Given that [latex]{h}^{-1}\\left(6\\right)=2[\/latex], what are the corresponding input and output values of the original function [latex]h?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q664604\">Show Solution<\/span><\/p>\n<div id=\"q664604\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]h(2)=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex:  Find an Inverse Function From a Table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TSztRfzmk0M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox\">\n<h3>How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\n<ol>\n<li>Determine whether [latex]f\\left(g\\left(x\\right)\\right)=x[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>If both statements are true, then [latex]g={f}^{-1}[\/latex] and [latex]f={g}^{-1}[\/latex]. If either statement is false, then [latex]g\\ne {f}^{-1}[\/latex] and [latex]f\\ne {g}^{-1}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Testing Inverse Relationships Algebraically<\/h3>\n<p>If [latex]f\\left(x\\right)=\\dfrac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\dfrac{1}{x}-2[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421291\">Show Solution<\/span><\/p>\n<div id=\"q421291\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align} g\\left(f\\left(x\\right)\\right)&=\\frac{1}{\\left(\\frac{1}{x+2}\\right)}{-2 }\\\\[1.5mm]&={ x }+{ 2 } -{ 2 }\\\\[1.5mm]&={ x } \\end{align}[\/latex]<\/p>\n<p>so<\/p>\n<p style=\"text-align: center;\">[latex]g={f}^{-1}\\text{ and }f={g}^{-1}[\/latex]<\/p>\n<p>This is enough to answer yes to the question, but we can also verify the other formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} f\\left(g\\left(x\\right)\\right)&=\\frac{1}{\\frac{1}{x}-2+2}\\\\[1.5mm] &=\\frac{1}{\\frac{1}{x}} \\\\[1.5mm] &=x \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice the inverse operations are in reverse order of the operations from the original function.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]f\\left(x\\right)={x}^{3}-4[\/latex] and [latex]g\\left(x\\right)=\\sqrt[3]{x+4}[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q226656\">Show Solution<\/span><\/p>\n<div id=\"q226656\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determining Inverse Relationships for Power Functions<\/h3>\n<p>If [latex]f\\left(x\\right)={x}^{3}[\/latex] (the cube function) and [latex]g\\left(x\\right)=\\frac{1}{3}x[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q637419\">Show Solution<\/span><\/p>\n<div id=\"q637419\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(g\\left(x\\right)\\right)=\\left(\\frac{1}{3}x\\right)^3=\\dfrac{{x}^{3}}{27}\\ne x[\/latex]<\/p>\n<p>No, the functions are not inverses.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The correct inverse to [latex]x^3[\/latex] is the cube root [latex]\\sqrt[3]{x}={x}^{\\frac{1}{3}}[\/latex], that is, the one-third is an exponent, not a multiplier.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{3}\\text{and}g\\left(x\\right)=\\sqrt[3]{x}+1[\/latex], is [latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384680\">Show Solution<\/span><\/p>\n<div id=\"q384680\" class=\"hidden-answer\" style=\"display: none\">\n<p>Yes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Determine the Domain and Range of an Inverse Function<\/h2>\n<p>The outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.<\/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\/18205514\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" \/><\/p>\n<p class=\"wp-caption-text\">Domain and range of a function and its inverse<\/p>\n<\/div>\n<p>When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex] is [latex]{f}^{-1}\\left(x\\right)={x}^{2}[\/latex], because a square &#8220;undoes&#8221; a square root; but the square is only the inverse of the square root on the domain [latex]\\left[0,\\infty \\right)[\/latex], since that is the range of [latex]f\\left(x\\right)=\\sqrt{x}[\/latex].<\/p>\n<p>We can look at this problem from the other side, starting with the square (parent quadratic) function [latex]f\\left(x\\right)={x}^{2}[\/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and \u20133. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the &#8220;inverse&#8221; is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.<\/p>\n<p>In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\n<p>If [latex]f\\left(x\\right)={\\left(x - 1\\right)}^{2}[\/latex] on [latex]\\left[1,\\infty \\right)[\/latex], then the inverse function is [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}+1[\/latex].<\/p>\n<ul>\n<li>The domain of [latex]f[\/latex] = range of [latex]{f}^{-1}[\/latex] = [latex]\\left[1,\\infty \\right)[\/latex].<\/li>\n<li>The domain of [latex]{f}^{-1}[\/latex] = range of [latex]f[\/latex] = [latex]\\left[0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<div class=\"textbox\">\n<p><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Is it possible for a function to have more than one inverse?<\/strong><\/p>\n<p><em>No. If two supposedly different functions, say, [latex]g[\/latex] and [latex]h[\/latex], both meet the definition of being inverses of another function [latex]f[\/latex], then you can prove that [latex]g=h[\/latex]. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Domain and Range of Inverse Functions<\/h3>\n<p>The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p>The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3><strong>How To: Given a function, find the domain and range of its inverse.<br \/>\n<\/strong><\/h3>\n<ol>\n<li>If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.<\/li>\n<li>If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Inverses of PARENT Functions<\/h3>\n<p>Identify which of the\u00a0parent functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The\u00a0parent functions are reviewed below. We restrict the domain in such a fashion that the function assumes all <em>y<\/em>-values exactly once.<\/p>\n<table summary=\"A list of the toolkit function. The constant function is f(x) = c where c is the constant; the identity function is f(x) = x; the absolute function is f(x)=|x|; the quadratic function is f(x) = x^2; the cubic function is f(x)=x^3; the reciprocal function is f(x)=1\/x; the reciprocal squared function is f(x)=1\/x^2; the square root function is f(x)=sqrt(x); the cube root function is f(x) = x^(1\/3).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>Constant<\/td>\n<td>Identity<\/td>\n<td>Quadratic<\/td>\n<td>Cubic<\/td>\n<td>Reciprocal<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)=c[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=x[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)={x}^{2}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Reciprocal squared<\/td>\n<td>Cube root<\/td>\n<td>Square root<\/td>\n<td>Absolute value<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\sqrt[3]{x}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=\\sqrt{x}[\/latex]<\/td>\n<td>[latex]f\\left(x\\right)=|x|[\/latex]<\/td>\n<td><\/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=\"q229708\">Show Solution<\/span><\/p>\n<div id=\"q229708\" class=\"hidden-answer\" style=\"display: none\">\n<p>The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.<\/p>\n<p>The absolute value function can be restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], where it is equal to the identity function.<\/p>\n<p>The reciprocal-squared function can be restricted to the domain [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs.\u00a0They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.<\/p>\n<div style=\"width: 985px\" 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\/18205517\/CNX_Precalc_Figure_01_07_004ab2.jpg\" alt=\"Graph of an absolute function.\" width=\"975\" height=\"404\" \/><\/p>\n<p class=\"wp-caption-text\">(a) Absolute value (b) Reciprocal squared<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The domain of the function [latex]f[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex] and the range of the function [latex]f[\/latex] is [latex]\\left(\\mathrm{-\\infty },-2\\right)[\/latex]. Find the domain and range of the inverse function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q146498\">Show Solution<\/span><\/p>\n<div id=\"q146498\" class=\"hidden-answer\" style=\"display: none\">\n<p>The domain of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(-\\infty \\text{,}-2\\right)[\/latex] and the range of the function [latex]{f}^{-1}[\/latex] is [latex]\\left(1,\\infty \\right)[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Define and Graph an Inverse<\/h2>\n<p>Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases.<\/p>\n<h2>Inverting Tabular Functions<\/h2>\n<p>Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range.<\/p>\n<p>Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Interpreting the Inverse of a Tabular Function<\/h3>\n<p>A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes. Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/p>\n<table summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>60<\/td>\n<td>70<\/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=\"q61261\">Show Solution<\/span><\/p>\n<div id=\"q61261\" class=\"hidden-answer\" style=\"display: none\">\n<p>The inverse function takes an output of [latex]f[\/latex] and returns an input for [latex]f[\/latex]. So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p>\n<p>Alternatively, recall that the definition of the inverse was that if [latex]f\\left(a\\right)=b[\/latex], then [latex]{f}^{-1}\\left(b\\right)=a[\/latex]. By this definition, if we are given [latex]{f}^{-1}\\left(70\\right)=a[\/latex], then we are looking for a value [latex]a[\/latex] so that [latex]f\\left(a\\right)=70[\/latex]. In this case, we are looking for a [latex]t[\/latex] so that [latex]f\\left(t\\right)=70[\/latex], which is when [latex]t=90[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the table below,\u00a0find and interpret (a) [latex]\\text{ }f\\left(60\\right)[\/latex], and (b) [latex]\\text{ }{f}^{-1}\\left(60\\right)[\/latex].<\/p>\n<table summary=\"Two rows and five columns. The first row is labeled\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]t\\text{ (minutes)}[\/latex]<\/td>\n<td>30<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/td>\n<td>90<\/td>\n<\/tr>\n<tr>\n<td>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex]<\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>50<\/td>\n<td>60<\/td>\n<td>70<\/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=\"q401025\">Show Solution<\/span><\/p>\n<div id=\"q401025\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]f\\left(60\\right)=50[\/latex]. In 60 minutes, 50 miles are traveled.<\/p>\n<p>b. [latex]{f}^{-1}\\left(60\\right)=70[\/latex]. To travel 60 miles, it will take 70 minutes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Evaluating the Inverse of a Function, Given a Graph of the Original Function<\/h2>\n<p>The domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the <em>vertical<\/em> extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the <em>horizontal<\/em> extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function\u2019s graph.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the graph of a function, evaluate its inverse at specific points.<\/h3>\n<ol>\n<li>Find the desired input of the inverse function on the [latex]y[\/latex]-axis of the given graph.<\/li>\n<li>Read the inverse function\u2019s output from the [latex]x[\/latex]-axis of the given graph.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Function and Its Inverse from a Graph at Specific Points<\/h3>\n<p>A function [latex]g\\left(x\\right)[\/latex] is given\u00a0below. Find [latex]g\\left(3\\right)[\/latex] and [latex]{g}^{-1}\\left(3\\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\/18205520\/CNX_Precalc_Figure_01_07_0062.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q334632\">Show Solution<\/span><\/p>\n<div id=\"q334632\" class=\"hidden-answer\" style=\"display: none\">\n<p>To evaluate [latex]g\\left(3\\right)[\/latex], we find 3 on the <em>x<\/em>-axis and find the corresponding output value on the [latex]y[\/latex]-axis. The point [latex]\\left(3,1\\right)[\/latex] tells us that [latex]g\\left(3\\right)=1[\/latex].<\/p>\n<p>To evaluate [latex]{g}^{-1}\\left(3\\right)[\/latex], recall that by definition [latex]{g}^{-1}\\left(3\\right)[\/latex] means the value of <em>x<\/em> for which [latex]g\\left(x\\right)=3[\/latex]. By looking for the output value 3 on the vertical axis, we find the point [latex]\\left(5,3\\right)[\/latex] on the graph, which means [latex]g\\left(5\\right)=3[\/latex], so by definition, [latex]{g}^{-1}\\left(3\\right)=5[\/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\/18205521\/CNX_Precalc_Figure_01_07_0072.jpg\" alt=\"Graph of g(x).\" width=\"487\" height=\"254\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the graph in the previous example, (a) find [latex]{g}^{-1}\\left(1\\right)[\/latex], and (b) estimate [latex]{g}^{-1}\\left(4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q350455\">Show Solution<\/span><\/p>\n<div id=\"q350455\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. 3; b. 5.6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Inverses of Functions Represented by Formulas<\/h2>\n<p>Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula\u2014for example, [latex]y[\/latex] as a function of [latex]x-[\/latex] we can often find the inverse function by solving to obtain [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a function represented by a formula, find the inverse.<\/h3>\n<ol>\n<li>Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\n<li>Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\n<li>Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\n<li>Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Inverting the Fahrenheit-to-Celsius Function<\/h3>\n<p>Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.<\/p>\n<p style=\"text-align: center;\">[latex]C=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625400\">Show Solution<\/span><\/p>\n<div id=\"q625400\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]{ C }=\\frac{5}{9}\\left(F - 32\\right)[\/latex]<br \/>\n[latex]C\\cdot \\frac{9}{5}=F - 32[\/latex]<br \/>\n[latex]F=\\frac{9}{5}C+32[\/latex]<\/p>\n<p>By solving in general, we have uncovered the inverse function. If<\/p>\n<p style=\"text-align: center;\">[latex]C=h\\left(F\\right)=\\frac{5}{9}\\left(F - 32\\right)[\/latex],<\/p>\n<p>then<\/p>\n<p style=\"text-align: center;\">[latex]F={h}^{-1}\\left(C\\right)=\\frac{9}{5}C+32[\/latex].<\/p>\n<p>In this case, we introduced a function [latex]h[\/latex] to represent the conversion because the input and output variables are descriptive, and writing [latex]{C}^{-1}[\/latex] could get confusing.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve for [latex]x[\/latex] in terms of [latex]y[\/latex] given [latex]y=\\frac{1}{3}\\left(x - 5\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q875458\">Show Solution<\/span><\/p>\n<div id=\"q875458\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=3y+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving to Find an Inverse Function<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=\\dfrac{2}{x - 3}+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q903464\">Show Solution<\/span><\/p>\n<div id=\"q903464\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=\\frac{2}{x - 3}+4 && \\text{Change }f(x)\\text{ to }y. \\\\[1.5mm]&x=\\frac{2}{y - 3}+4 && \\text{Switch }x\\text{ and }y. \\\\[1.5mm] &y - 4=\\frac{2}{x - 3} && \\text{Subtract 4 from both sides}. \\\\[1.5mm] &y - 3=\\frac{2}{x - 4} && \\text{Multiply both sides by }y - 3\\text{ and divide by }x - 4. \\\\[1.5mm] &y=\\frac{2}{x - 4}+3 && \\text{Add 3 to both sides}.\\\\[-3mm]&\\end{align}[\/latex]<\/p>\n<p>So [latex]{f}^{-1}\\left(x\\right)=\\dfrac{2}{x - 4}+3[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The domain and range of [latex]f[\/latex] exclude the values 3 and 4, respectively. [latex]f[\/latex] and [latex]{f}^{-1}[\/latex] are equal at two points but are not the same function, as we can see by creating\u00a0the table below.<\/p>\n<table id=\"Table_01_07_05\" summary=\"The values of f(x) are: f(1)=3, f(2)=2, and f(5)=5. So f^(-1)(y)=y.\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]{f}^{-1}\\left(y\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex] <\/strong><\/td>\n<td>3<\/td>\n<td>2<\/td>\n<td>5<\/td>\n<td>[latex]y[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving to Find an Inverse with Radicals<\/h3>\n<p>Find the inverse of the function [latex]f\\left(x\\right)=2+\\sqrt{x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q444147\">Show Solution<\/span><\/p>\n<div id=\"q444147\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=2+\\sqrt{x - 4}\\\\[1.5mm]&x=2+\\sqrt{y - 4}\\\\[1.5mm] &{\\left(x - 2\\right)}^{2}=y - 4 \\\\[1.5mm] &y={\\left(x- 2\\right)}^{2}+4 \\end{align}[\/latex]<\/p>\n<p>So [latex]{f}^{-1}\\left(x\\right)={\\left(x - 2\\right)}^{2}+4[\/latex].<\/p>\n<p>The domain of [latex]f[\/latex] is [latex]\\left[4,\\infty \\right)[\/latex]. Notice that the range of [latex]f[\/latex] is [latex]\\left[2,\\infty \\right)[\/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[\/latex] is also [latex]\\left[2,\\infty \\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The formula we found for [latex]{f}^{-1}\\left(x\\right)[\/latex] looks like it would be valid for all real [latex]x[\/latex]. However, [latex]{f}^{-1}[\/latex] itself must have an inverse (namely, [latex]f[\/latex] ) so we have to restrict the domain of [latex]{f}^{-1}[\/latex] to [latex]\\left[2,\\infty \\right)[\/latex] in order to make [latex]{f}^{-1}[\/latex] a one-to-one function. This domain of [latex]{f}^{-1}[\/latex] is exactly the range of [latex]f[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>What is the inverse of the function [latex]f\\left(x\\right)=2-\\sqrt{x}[\/latex]? State the domains of both the function and the inverse function.<br \/>\nUse an online graphing tool to graph the function, its inverse, and [latex]f(x) = x[\/latex] to check whether you are correct.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q327817\">Show Solution<\/span><\/p>\n<div id=\"q327817\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{f}^{-1}\\left(x\\right)={\\left(2-x\\right)}^{2}[\/latex]; domain of\u00a0 [latex]f:\\left[0,\\infty \\right)[\/latex]; domain of [latex]{ f}^{-1}:\\left(-\\infty ,2\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graph a Function&#8217;s Inverse<\/h2>\n<p>Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function [latex]f\\left(x\\right)={x}^{2}[\/latex] restricted to the domain [latex]\\left[0,\\infty \\right)[\/latex], on which this function is one-to-one, and graph it as below.<\/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\/18205523\/CNX_Precalc_Figure_01_07_0082.jpg\" alt=\"Graph of f(x).\" width=\"487\" height=\"254\" \/><\/p>\n<p class=\"wp-caption-text\">Quadratic function with domain restricted to [0, \u221e).<\/p>\n<\/div>\n<p><strong>Restricting the domain<\/strong> to [latex]\\left[0,\\infty \\right)[\/latex] makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain.<\/p>\n<p>We already know that the inverse of the\u00a0parent quadratic function is the square root function, that is, [latex]{f}^{-1}\\left(x\\right)=\\sqrt{x}[\/latex]. What happens if we graph both [latex]f\\text{ }[\/latex] and [latex]{f}^{-1}[\/latex] on the same set of axes, using the [latex]x\\text{-}[\/latex] axis for the input to both [latex]f\\text{ and }{f}^{-1}?[\/latex]<\/p>\n<p>We notice a distinct relationship: The graph of [latex]{f}^{-1}\\left(x\\right)[\/latex] is the graph of [latex]f\\left(x\\right)[\/latex] reflected about the diagonal line [latex]y=x[\/latex], which we will call the identity line, shown below.<\/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\/18205525\/CNX_Precalc_Figure_01_07_0092.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"251\" \/><\/p>\n<p class=\"wp-caption-text\">Square and square-root functions on the non-negative domain<\/p>\n<\/div>\n<p>This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Inverse of a Function Using Reflection about the Identity Line<\/h3>\n<p>Given the graph of [latex]f\\left(x\\right)[\/latex], sketch a graph of [latex]{f}^{-1}\\left(x\\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\/18205527\/CNX_Precalc_Figure_01_07_0102.jpg\" alt=\"Graph of f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q694066\">Show Solution<\/span><\/p>\n<div id=\"q694066\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of [latex]\\left(0,\\infty \\right)[\/latex] and range of [latex]\\left(-\\infty ,\\infty \\right)[\/latex], so the inverse will have a domain of [latex]\\left(-\\infty ,\\infty \\right)[\/latex] and range of [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<p>If we reflect this graph over the line [latex]y=x[\/latex], the point [latex]\\left(1,0\\right)[\/latex] reflects to [latex]\\left(0,1\\right)[\/latex] and the point [latex]\\left(4,2\\right)[\/latex] reflects to [latex]\\left(2,4\\right)[\/latex]. Sketching the inverse on the same axes as the original graph gives us\u00a0the result in the graph below.<\/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\/18205529\/CNX_Precalc_Figure_01_07_0112.jpg\" alt=\"Graph of f(x) and f^(-1)(x).\" width=\"487\" height=\"363\" \/><\/p>\n<p class=\"wp-caption-text\">The function and its inverse, showing reflection about the identity line<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Is there any function that is equal to its own inverse?<\/strong><\/p>\n<p><em>Yes. If [latex]f={f}^{-1}[\/latex], then [latex]f\\left(f\\left(x\\right)\\right)=x[\/latex], and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{\\frac{1}{x}}=x[\/latex]<\/p>\n<p><em>Any function [latex]f\\left(x\\right)=c-x[\/latex], where [latex]c[\/latex] is a constant, is also equal to its own inverse.<\/em><\/p>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>A graph represents a one-to-one function if any horizontal line drawn on the graph intersects the graph at no more than one point.<\/li>\n<li>If [latex]g\\left(x\\right)[\/latex] is the inverse of [latex]f\\left(x\\right)[\/latex], then [latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex].<\/li>\n<li>Each of the\u00a0parent functions, except [latex]y=c[\/latex] has an inverse. Some need a restricted domain.<\/li>\n<li>For a function to have an inverse, it must be one-to-one (pass the horizontal line test).<\/li>\n<li>A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.<\/li>\n<li>For a tabular function, exchange the input and output rows to obtain the inverse.<\/li>\n<li>The inverse of a function can be determined at specific points on its graph.<\/li>\n<li>To find the inverse of a function [latex]y=f\\left(x\\right)[\/latex], switch the variables [latex]x[\/latex] and [latex]y[\/latex]. Then solve for [latex]y[\/latex] as a function of [latex]x[\/latex].<\/li>\n<li>The graph of an inverse function is the reflection of the graph of the original function across the line [latex]y=x[\/latex].<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137932588\" class=\"definition\">\n<dt><strong>horizontal line test<\/strong><\/dt>\n<dd id=\"fs-id1165134149777\">a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134149782\" class=\"definition\">\n<dt><strong>independent variable<\/strong><\/dt>\n<dd id=\"fs-id1165134149787\">an input variable<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135511353\" class=\"definition\"><\/dl>\n<dl id=\"fs-id1165135511364\" class=\"definition\"><\/dl>\n<dl id=\"fs-id1165137441703\" class=\"definition\">\n<dt>inverse function<\/dt>\n<dd id=\"fs-id1165137441708\">for any one-to-one function [latex]f\\left(x\\right)[\/latex], the inverse is a function [latex]{f}^{-1}\\left(x\\right)[\/latex] such that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex]; this also implies that [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]{f}^{-1}[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135511364\" class=\"definition\">\n<dt><strong>one-to-one function<\/strong><\/dt>\n<dd id=\"fs-id1165135511369\">a function for which each value of the output is associated with a unique input value<\/dd>\n<\/dl>\n<section id=\"fs-id1165137660004\"><\/section>\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-1799\">\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><\/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><\/ul><div class=\"license-attribution-dropdown-subheading\">All 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