{"id":93,"date":"2021-02-03T20:52:49","date_gmt":"2021-02-03T20:52:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=93"},"modified":"2022-03-11T21:34:36","modified_gmt":"2022-03-11T21:34:36","slug":"combining-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/combining-functions\/","title":{"raw":"Combining Functions","rendered":"Combining Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Make new functions from two or more given functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Composite Functions<\/h2>\r\n<p id=\"fs-id1170572234066\">Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when we combine functions in different ways, using basic mathematical operations to create new functions. For example, if the cost for a company to manufacture [latex]x[\/latex] items is described by the function [latex]C(x)[\/latex] and the revenue created by the sale of [latex]x[\/latex] items is described by the function [latex]R(x)[\/latex], then the profit on the manufacture and sale of [latex]x[\/latex] items is defined as [latex]P(x)=R(x)-C(x)[\/latex]. Using the difference between two functions, we created a new function.<\/p>\r\n<p id=\"fs-id1170572135596\">Alternatively, we can create a new function by composing two functions. For example, given the functions [latex]f(x)=x^2[\/latex] and [latex]g(x)=3x+1[\/latex], the <strong>composite function<\/strong> [latex]f\\circ g[\/latex] is defined such that<\/p>\r\n\r\n<div id=\"fs-id1170572247955\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f\\circ g)(x)=f(g(x))=(g(x))^2=(3x+1)^2[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572140952\">The composite function [latex]g\\circ f[\/latex] is defined such that<\/p>\r\n\r\n<div id=\"fs-id1170572481486\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))=3f(x)+1=3x^2+1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572450916\">Note that these two new functions are different from each other.<\/p>\r\n\r\n<div id=\"fs-id1170572450919\" class=\"bc-section section\" style=\"text-align: left;\">\r\n<h3>Combining Functions with Mathematical Operators<\/h3>\r\n<p id=\"fs-id1170572167715\">To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Given two functions [latex]f[\/latex] and [latex]g,[\/latex] we can define four new functions:<\/p>\r\n\r\n<div id=\"fs-id1170572167731\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}(f+g)(x)=f(x)+g(x)\\hfill &amp; &amp; &amp; \\text{Sum}\\hfill \\\\ (f-g)(x)=f(x)-g(x)\\hfill &amp; &amp; &amp; \\text{Difference}\\hfill \\\\ (f\u00b7g)(x)=f(x)g(x)\\hfill &amp; &amp; &amp; \\text{Product}\\hfill \\\\ \\Big(\\frac{f}{g}\\Big)(x)=\\frac{f(x)}{g(x)} \\, \\text{for} \\, g(x)\\ne 0\\hfill &amp; &amp; &amp; \\text{Quotient}\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Often these functions have more than one term, so when you perform operations on them you will need to remember how to work with polynomials.<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Given multiple polynomials, add or subtract them to simplify the expressions<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Combine like terms.<\/li>\r\n \t<li>Simplify and write in standard form. Standard form means you start with the leading term, and write the rest of the terms in descending order by degree.<\/li>\r\n<\/ol>\r\n<strong>BE CAREFUL WHEN SUBTRACTING POLYNOMIALS!<\/strong>\r\n\r\nWhen subtracting a polynomial from another, be careful to subtract\u00a0<em>each term<\/em> in the second from the first. That is, use the distributive property to distribute the minus sign through the second polynomial.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\left(3x^2-2x+9\\right)-\\left(x^2-4x+5\\right)\\text{}\\hfill &amp;\\text{Distribute the negative in front of the parenthesis} \\hfill \\\\\r\n3x^2-2x+9 -x^2 -\\left(-4x\\right) - 5\\hfill &amp; \\text{Be careful when subtracting a negative}.\\hfill \\\\ 3x^2 - x^2 -2x+4x+9-5\\hfill &amp; \\text{Rearrange terms in descending order of degree} \\hfill \\\\ 2x^2 +2x +4 \\hfill &amp; \\text{Combine like terms}. \\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Given two binomials, Multiplying Using FOIL<\/h3>\r\n<ol>\r\n \t<li>Multiply the first terms of each binomial.<\/li>\r\n \t<li>Multiply the outer terms of the binomials.<\/li>\r\n \t<li>Multiply the inner terms of the binomials.<\/li>\r\n \t<li>Multiply the last terms of each binomial.<\/li>\r\n \t<li>Add the products.<\/li>\r\n \t<li>Combine like terms and simplify.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21205149\/CNX_CAT_Figure_01_04_003.jpg\" alt=\"Two quantities in parentheses are being multiplied, the first being: a times x plus b and the second being: c times x plus d. This expression equals ac times x squared plus ad times x plus bc times x plus bd. The terms ax and cx are labeled: First Terms. The terms ax and d are labeled: Outer Terms. The terms b and cx are labeled: Inner Terms. The terms b and d are labeled: Last Terms.\" \/><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170572549843\" class=\"textbox exercises\">\r\n<h3>Example: Combining Functions Using Mathematical Operations<\/h3>\r\n<p id=\"fs-id1170572549852\">Given the functions [latex]f(x)=2x-3[\/latex] and [latex]g(x)=x^2-1[\/latex], find each of the following functions and state its domain.<\/p>\r\n\r\n<ol id=\"fs-id1170572169320\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex](f+g)(x)[\/latex]<\/li>\r\n \t<li>[latex](f-g)(x)[\/latex]<\/li>\r\n \t<li>[latex](f\u00b7g)(x)[\/latex]<\/li>\r\n \t<li>[latex]\\Big(\\frac{f}{g}\\Big)(x)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572228247\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572228247\"]\r\n<ol id=\"fs-id1170572228247\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex](f+g)(x)=(2x-3)+(x^2-1)=x^2+2x-4[\/latex]. The domain of this function is the interval [latex](\u2212\\infty ,\\infty )[\/latex].<\/li>\r\n \t<li>[latex](f-g)(x)=(2x-3)-(x^2-1)=\u2212x^2+2x-2[\/latex]. The domain of this function is the interval [latex](\u2212\\infty ,\\infty)[\/latex].<\/li>\r\n \t<li>[latex](f\u00b7g)(x)=(2x-3)(x^2-1)=2x^3-3x^2-2x+3[\/latex]. The domain of this function is the interval [latex](\u2212\\infty ,\\infty )[\/latex].<\/li>\r\n \t<li>[latex]\\Big(\\frac{f}{g}\\Big)(x)=\\dfrac{2x-3}{x^2-1}[\/latex]. The domain of this function is [latex]\\{x|x\\ne \\text{\u00b1}1\\}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Combining Functions Using Mathematical Operations[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qL2tyJhmrkg?controls=0&amp;start=1025&amp;end=1238&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>&nbsp;\r\n\r\n[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.1ReviewofFunctions1025to1238_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.1 Review of Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n\r\nYou can think of the domain of a function as all possible inputs that will produce a valid output. When you think about it this way, you will see that fractional functions that have a variable in the denominator (when a denominator is zero, you have an undefined value) or a variable under an even root radical (you cannot evaluate a negative value under an even square root), you will have restrictions in your domain.\r\n<div class=\"textbox examples\">\r\n<h3>Recall:\u00a0Given a function written in an equation form that includes a fraction, find the domain.<\/h3>\r\n<ol>\r\n \t<li>Identify the input values.<\/li>\r\n \t<li>Identify any restrictions on the input. If there is a denominator in the function\u2019s formula, set the denominator equal to zero and solve for [latex]x[\/latex] . These are the values that cannot be inputs in the function.<\/li>\r\n \t<li>Write the domain in interval form, making sure to exclude any restricted values from the domain.<\/li>\r\n<\/ol>\r\nFind the domain of the function [latex]f\\left(x\\right)=\\dfrac{x+1}{2-x}[\/latex].\r\n\r\n[reveal-answer q=\"759017\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"759017\"]\r\n\r\nWhen there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2-x&amp;=0 \\\\ -x&amp;=-2 \\\\ x&amp;=2 \\end{align}[\/latex]<\/p>\r\nNow, we will exclude 2 from the domain. The answers are all real numbers where [latex]x&lt;2[\/latex] or [latex]x&gt;2[\/latex]. We can use a symbol known as the union, [latex]\\cup [\/latex], to combine the two sets. In interval notation, we write the solution: [latex]\\left(\\mathrm{-\\infty },2\\right)\\cup \\left(2,\\infty \\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\/18193532\/CNX_Precalc_Figure_01_02_028n2.jpg\" alt=\"Line graph of x=!2.\" width=\"487\" height=\"164\" \/>\r\n\r\nIn interval form, the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,2\\right)\\cup \\left(2,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall:\u00a0Given a function written in equation form including an even root, find the domain.<\/h3>\r\n<ol>\r\n \t<li>Identify the input values.<\/li>\r\n \t<li>Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/li>\r\n \t<li>The solution(s) are the domain of the function. If possible, write the answer in interval form.<\/li>\r\n<\/ol>\r\nFind the domain of the function [latex]f\\left(x\\right)=\\sqrt{7-x}[\/latex].\r\n\r\n[reveal-answer q=\"722013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"722013\"]\r\n\r\nWhen there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.\r\n\r\nSet the radicand greater than or equal to zero and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}7-x&amp;\\ge 0 \\\\ -x&amp;\\ge -7 \\\\ x&amp;\\le 7 \\end{align}[\/latex]<\/p>\r\nNow, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to [latex]7[\/latex], or [latex]\\left(-\\infty ,7\\right][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572548529\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572548538\">For [latex]f(x)=x^2+3[\/latex] and [latex]g(x)=2x-5[\/latex], find [latex]\\left(\\frac{f}{g}\\right)(x)[\/latex] and state its domain.<\/p>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"265378\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"265378\"]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1165043323895\">The new function [latex]\\left(\\frac{f}{g}\\right)(x)[\/latex] is a quotient of two functions. For what values of [latex]x[\/latex] is the denominator zero?<\/p>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"fs-id1170572482389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572482389\"]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1170572482389\">[latex]\\Big(\\frac{f}{g}\\Big)(x)=\\dfrac{x^2+3}{2x-5}[\/latex]. The domain is [latex]\\{x|x\\ne \\frac{5}{2}\\}[\/latex].<\/p>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572212495\" class=\"bc-section section\">\r\n<h3>Function Composition<\/h3>\r\n<p id=\"fs-id1170572212500\">When we compose functions, we take a function of a function. For example, suppose the temperature [latex]T[\/latex] on a given day is described as a function of time [latex]t[\/latex] (measured in hours after midnight) as in Figure 6. Suppose the cost [latex]C[\/latex], to heat or cool a building for 1 hour, can be described as a function of the temperature [latex]T[\/latex]. Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating [latex]C(T(t))[\/latex]. We have defined a new function, denoted [latex]C\\circ T[\/latex], which is defined such that [latex](C\\circ T)(t)=C(T(t))[\/latex] for all [latex]t[\/latex] in the domain of [latex]T[\/latex]. This new function is called a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makes sense to define this new function [latex](C\\circ T)(t)[\/latex]. It does not make sense to consider [latex](T\\circ C)(t)[\/latex], because temperature is not a function of cost.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572479276\">Consider the function [latex]f[\/latex] with domain [latex]A[\/latex] and range [latex]B[\/latex], and the function [latex]g[\/latex] with domain [latex]D[\/latex] and range [latex]E[\/latex]. If [latex]B[\/latex] is a subset of [latex]D[\/latex], then the <strong>composite function<\/strong> [latex](g\\circ f)(x)[\/latex] is the function with domain [latex]A[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1170572176830\" class=\"equation\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170572217742\">A composite function [latex]g\\circ f[\/latex] can be viewed in two steps. First, the function [latex]f[\/latex] maps each input [latex]x[\/latex] in the domain of [latex]f[\/latex] to its output [latex]f(x)[\/latex] in the range of [latex]f[\/latex]. Second, since the range of [latex]f[\/latex] is a subset of the domain of [latex]g[\/latex], the output [latex]f(x)[\/latex] is an element in the domain of [latex]g[\/latex], and therefore it is mapped to an output [latex]g(f(x))[\/latex] in the range of [latex]g[\/latex]. In Figure 12, we see a visual image of a composite function.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"437\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202134\/CNX_Calc_Figure_01_01_011.jpg\" alt=\"An image with three items. The first item is a blue bubble that has two labels: \u201cdomain of f\u201d and \u201cdomain of g of f\u201d. This item contains the numbers 1, 2, and 3. The second item is two bubbles: an orange bubble labeled \u201cdomain of g\u201d and a blue bubble that is completely contained within the orange bubble and is labeled \u201crange of f\u201d. The blue bubble contains the numbers 0 and 1, which are thus also contained within the larger orange bubble. The orange bubble contains two numbers not contained within the smaller blue bubble, which are 2 and 3. The third item is two bubbles: an orange bubble labeled \u201crange of g\u201d and a blue bubble that is completely contained within the orange bubble and is labeled \u201crange of g of f\u201d. The blue bubble contains the numbers 4 and 5, which are thus also contained within the larger orange bubble. The orange bubble contains one number not contained within the smaller blue bubble, which is the number 3. The first item points has a blue arrow with the label \u201cf\u201d that points to the blue bubble in the second item. The orange bubble in the second item has an orange arrow labeled \u201cg\u201d that points the orange bubble in the third item. The first item has a blue arrow labeled \u201cg of f\u201d which points to the blue bubble in the third item. There are three blue arrows pointing from numbers in the first item to the numbers contained in the blue bubble of the second item. The first blue arrow points from the 1 to the 0, the second blue arrow points from the 2 to the 1, and the third blue arrow points from the 3 to the 0. There are 4 orange arrows pointing from the numbers contained in the orange bubble in the second item, including those also contained in the blue bubble of the second item, to the numbers contained in the orange bubble of the third item, including the numbers in the blue bubble of the third item. The first orange arrow points from 2 to 3, the second orange arrow points from 3 to 5, the third orange arrow points from 0 to 4, and the fourth orange arrow points from 1 to 5.\" width=\"437\" height=\"252\" \/> Figure 12. For the composite function [latex]g\\circ f[\/latex], we have [latex](g\\circ f)(1)=4, \\, (g\\circ f)(2)=5,[\/latex] and [latex](g\\circ f)(3)=4[\/latex].[\/caption]<\/div>\r\nWhen composing functions, don't forget to work from the inside out. Also, keep in mind that in general,\u00a0 [latex]f\\left(g\\left(x\\right)\\right)\\ne g\\left(f\\left(x\\right)\\right)[\/latex] for all [latex]x[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Recall: How to perform a composition of functions<\/h3>\r\nIt is important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18195616\/CNX_Precalc_Figure_01_04_0012.jpg\" alt=\"Explanation of the composite function. g(x), the output of g is the input of f. X is the input of g.\" width=\"487\" height=\"171\" \/>\r\n\r\nIn general [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] are different functions. In other words in many cases [latex]f\\left(g\\left(x\\right)\\right)\\ne g\\left(f\\left(x\\right)\\right)[\/latex] for all [latex]x[\/latex].\r\n\r\nFor example if [latex]f\\left(x\\right)={x}^{2}[\/latex] and [latex]g\\left(x\\right)=x+2[\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&amp;=f\\left(x+2\\right) \\\\[2mm] &amp;={\\left(x+2\\right)}^{2} \\\\[2mm] &amp;={x}^{2}+4x+4\\hfill \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">but<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&amp;=g\\left({x}^{2}\\right) \\\\[2mm] \\text{ }&amp;={x}^{2}+2\\hfill \\end{align}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572212495\" class=\"bc-section section\">\r\n<div id=\"fs-id1170572481349\" class=\"textbox exercises\">\r\n<h3>Example: Compositions of Functions Defined by Formulas<\/h3>\r\n<p id=\"fs-id1170572481359\">Consider the functions [latex]f(x)=x^2+1[\/latex] and [latex]g(x)=\\frac{1}{x}[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170572482111\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find [latex](g\\circ f)(x)[\/latex] and state its domain and range.<\/li>\r\n \t<li>Evaluate [latex](g\\circ f)(4)[\/latex] and [latex](g\\circ f)\\left(-\\frac{1}{2}\\right)[\/latex].<\/li>\r\n \t<li>Find [latex](f\\circ g)(x)[\/latex] and state its domain and range.<\/li>\r\n \t<li>Evaluate [latex](f\\circ g)(4)[\/latex] and [latex](f\\circ g)\\left(-\\frac{1}{2}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572222946\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572222946\"]\r\n<ol id=\"fs-id1170572222946\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>We can find the formula for [latex](g\\circ f)(x)[\/latex] in two different ways. We could write\r\n<div id=\"fs-id1170571109518\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))=g(x^2+1)=\\dfrac{1}{x^2+1}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nAlternatively, we could write\r\n<div id=\"fs-id1170573521539\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))=\\dfrac{1}{f(x)}=\\dfrac{1}{x^2+1}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nSince [latex]x^2+1\\ne 0[\/latex] for all real numbers [latex]x[\/latex], the domain of [latex](g\\circ f)(x)[\/latex] is the set of all real numbers. Since [latex]0&lt;\\frac{1}{(x^2+1)}\\le 1[\/latex], the range is, at most, the interval [latex](0,1][\/latex]. To show that the range is this entire interval, we let [latex]y=\\frac{1}{(x^2+1)}[\/latex] and solve this equation for [latex]x[\/latex] to show that for all [latex]y[\/latex] in the interval [latex](0,1][\/latex], there exists a real number [latex]x[\/latex] such that [latex]y=\\frac{1}{(x^2+1)}[\/latex]. Solving this equation for [latex]x[\/latex], we see that [latex]x^2+1=\\frac{1}{y}[\/latex], which implies that\r\n<div id=\"fs-id1170573368782\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\u00b1\\sqrt{\\frac{1}{y}-1}[\/latex].<\/div>\r\n&nbsp;\r\n\r\nIf [latex]y[\/latex] is in the interval [latex](0,1][\/latex], the expression under the radical is nonnegative, and therefore there exists a real number [latex]x[\/latex] such that [latex]\\frac{1}{(x^2+1)}=y[\/latex]. We conclude that the range of [latex]g\\circ f[\/latex] is the interval [latex](0,1][\/latex].<\/li>\r\n \t<li>[latex](g\\circ f)(4)=g(f(4))=g(4^2+1)=g(17)=\\frac{1}{17}[\/latex]\r\n[latex](g\\circ f)(-\\frac{1}{2})=g(f(-\\frac{1}{2}))=g((-\\frac{1}{2})^2+1)=g(\\frac{5}{4})=\\frac{4}{5}[\/latex]<\/li>\r\n \t<li>We can find a formula for [latex](f\\circ g)(x)[\/latex] in two ways. First, we could write\r\n<div id=\"fs-id1170573750183\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f\\circ g)(x)=f(g(x))=f\\left(\\frac{1}{x}\\right)=\\left(\\frac{1}{x}\\right)^2+1[\/latex]<\/div>\r\n&nbsp;\r\n\r\nAlternatively, we could write\r\n<div id=\"fs-id1170571123790\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f\\circ g)(x)=f(g(x))=(g(x))^2+1=\\left(\\frac{1}{x}\\right)^2+1[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe domain of [latex]f\\circ g[\/latex] is the set of all real numbers [latex]x[\/latex] such that [latex]x\\ne 0[\/latex]. To find the range of [latex]f[\/latex], we need to find all values [latex]y[\/latex] for which there exists a real number [latex]x\\ne 0[\/latex] such that\r\n<div id=\"fs-id1170571118828\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left(\\frac{1}{x}\\right)^2+1=y[\/latex]<\/div>\r\n&nbsp;\r\n\r\nSolving this equation for [latex]x[\/latex], we see that we need [latex]x[\/latex] to satisfy\r\n<div id=\"fs-id1170571118874\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left(\\frac{1}{x}\\right)^2=y-1[\/latex],<\/div>\r\n&nbsp;\r\n\r\nwhich simplifies to\r\n<div id=\"fs-id1170571330548\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{x}=\u00b1\\sqrt{y-1}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nFinally, we obtain\r\n<div id=\"fs-id1170571282415\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\u00b1\\frac{1}{\\sqrt{y-1}}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nSince [latex]\\frac{1}{\\sqrt{y-1}}[\/latex] is a real number if and only if [latex]y&gt;1[\/latex], the range of [latex]f[\/latex] is the set [latex]\\{y|y\\ge 1\\}[\/latex].<\/li>\r\n \t<li>[latex](f\\circ g)(4)=f(g(4))=f\\left(\\frac{1}{4}\\right)=\\left(\\frac{1}{4}\\right)^2+1=\\frac{17}{16}[\/latex]\r\n[latex](f\\circ g)\\left(-\\frac{1}{2}\\right)=f\\left(g\\left(-\\frac{1}{2}\\right)\\right)=f(-2)=(-2)^2+1=5[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572173194\">In the box above, we can see that [latex](f\\circ g)(x)\\ne (g\\circ f)(x)[\/latex]. This tells us, in general terms, that the order in which we compose functions matters.<\/p>\r\n\r\n<div id=\"fs-id1170572483657\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572483665\">Let [latex]f(x)=2-5x[\/latex]<\/p>\r\nLet [latex]g(x)=\\sqrt{x}[\/latex]\r\n\r\nFind [latex](f\\circ g)(x)[\/latex]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"fs-id1170572481428\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572481428\"]\r\n<p id=\"fs-id1170572481428\">[latex](f\\circ g)(x)=2-5\\sqrt{x}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572173789\" class=\"textbox exercises\">\r\n<h3>Example: Composition of Functions Defined by Tables<\/h3>\r\n<p id=\"fs-id1170572173799\">Consider the functions [latex]f[\/latex] and [latex]g[\/latex] described below.<\/p>\r\n\r\n<table id=\"fs-id1170572173822\" class=\"column-header\" summary=\"A table with 2 rows and 8 columns. The first row is labeled \u201cx\u201d and has the values \u201c-3; -2; -1; 0; 1; 2; 3; 4\u201d. The second row is labeled \u201cf(x)\u201d and has the values \u201c0; 4; 2; 4; -2; 0; -2; 4\u201d.\">\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]\\mathbf{x}[\/latex]<\/td>\r\n<td>-3<\/td>\r\n<td>-2<\/td>\r\n<td>-1<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\mathbf{f(x)}[\/latex]<\/td>\r\n<td>0<\/td>\r\n<td>4<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>-2<\/td>\r\n<td>0<\/td>\r\n<td>-2<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table id=\"fs-id1170572550754\" class=\"column-header\" summary=\"A table with 2 rows and 5 columns. The first row is labeled \u201cx\u201d and has the values \u201c-4; -2; 0; 2; 4\u201d. The second row is labeled \u201cg(x)\u201d and has the values \u201c1; 0; 3; 0; 5\u201d.\">\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]\\mathbf{x}[\/latex]<\/td>\r\n<td>-4<\/td>\r\n<td>-2<\/td>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\mathbf{g(x)}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>0<\/td>\r\n<td>3<\/td>\r\n<td>0<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol id=\"fs-id1170572242030\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Evaluate [latex](g\\circ f)(3)[\/latex] and [latex](g\\circ f)(0)[\/latex].<\/li>\r\n \t<li>State the domain and range of [latex](g\\circ f)(x)[\/latex].<\/li>\r\n \t<li>Evaluate [latex](f\\circ f)(3)[\/latex] and [latex](f\\circ f)(1)[\/latex].<\/li>\r\n \t<li>State the domain and range of [latex](f\\circ f)(x)[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572552174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572552174\"]\r\n<ol id=\"fs-id1170572552174\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex](g\\circ f)(3)=g(f(3))=g(-2)=0[\/latex]\r\n[latex](g\\circ f)(0)=g(4)=5[\/latex]<\/li>\r\n \t<li>The domain of [latex]g\\circ f[\/latex] is the set [latex]\\{-3,-2,-1,0,1,2,3,4\\}[\/latex]. Since the range of [latex]f[\/latex] is the set [latex]\\{-2,0,2,4\\}[\/latex], the range of [latex]g\\circ f[\/latex] is the set [latex]\\{0,3,5\\}[\/latex].<\/li>\r\n \t<li>[latex](f\\circ f)(3)=f(f(3))=f(-2)=4[\/latex]\r\n[latex](f\\circ f)(1)=f(f(1))=f(-2)=4[\/latex]<\/li>\r\n \t<li>The domain of [latex]f\\circ f[\/latex] is the set [latex]\\{-3,-2,-1,0,1,2,3,4\\}[\/latex]. Since the range of [latex]f[\/latex] is the set [latex]\\{-2,0,2,4\\}[\/latex], the range of [latex]f\\circ f[\/latex] is the set [latex]\\{0,4\\}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Composition of Functions Defined by Tables[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qL2tyJhmrkg?controls=0&amp;start=1503&amp;end=1757&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.1ReviewofFunctions1503to1757_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.1 Review of Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170572173737\" class=\"textbox exercises\">\r\n<h3>Example: Application Involving a Composite Function<\/h3>\r\n<p id=\"fs-id1170572173747\">A store is advertising a sale of [latex]20\\%[\/latex] off all merchandise. Caroline has a coupon that entitles her to an additional [latex]15\\%[\/latex] off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of [latex]x[\/latex] dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by using a composite function.<\/p>\r\n[reveal-answer q=\"fs-id1170572454320\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572454320\"]\r\n<p id=\"fs-id1170572454320\">Since the sale price is [latex]20\\%[\/latex] off the original price, if an item is [latex]x[\/latex] dollars, its sale price is given by [latex]f(x)=0.80x[\/latex]. Since the coupon entitles an individual to [latex]15\\%[\/latex] off the price of any item, if an item is [latex]y[\/latex] dollars, the price, after applying the coupon, is given by [latex]g(y)=0.85y[\/latex]. Therefore, if the price is originally [latex]x[\/latex] dollars, its sale price will be [latex]f(x)=0.80x[\/latex] and then its final price after the coupon will be [latex]g(f(x))=0.85(0.80x)=0.68x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170572480210\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170572480218\">If items are on sale for [latex]10\\%[\/latex] off their original price, and a customer has a coupon for an additional [latex]30\\%[\/latex] off, what will be the final price for an item that is originally [latex]x[\/latex] dollars, after applying the coupon to the sale price?<\/p>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"324566\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"324566\"]\r\n\r\nThe sale price of an item with an original price of [latex]x[\/latex] dollars is [latex]f(x)=0.90x[\/latex]. The coupon price for an item that is [latex]y[\/latex] dollars is [latex]g(y)=0.70y[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"fs-id1170572213178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572213178\"]\r\n<p id=\"fs-id1170572213178\">[latex](g\\circ f)(x)=0.63x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]33467[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Make new functions from two or more given functions<\/li>\n<\/ul>\n<\/div>\n<h2>Composite Functions<\/h2>\n<p id=\"fs-id1170572234066\">Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when we combine functions in different ways, using basic mathematical operations to create new functions. For example, if the cost for a company to manufacture [latex]x[\/latex] items is described by the function [latex]C(x)[\/latex] and the revenue created by the sale of [latex]x[\/latex] items is described by the function [latex]R(x)[\/latex], then the profit on the manufacture and sale of [latex]x[\/latex] items is defined as [latex]P(x)=R(x)-C(x)[\/latex]. Using the difference between two functions, we created a new function.<\/p>\n<p id=\"fs-id1170572135596\">Alternatively, we can create a new function by composing two functions. For example, given the functions [latex]f(x)=x^2[\/latex] and [latex]g(x)=3x+1[\/latex], the <strong>composite function<\/strong> [latex]f\\circ g[\/latex] is defined such that<\/p>\n<div id=\"fs-id1170572247955\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f\\circ g)(x)=f(g(x))=(g(x))^2=(3x+1)^2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572140952\">The composite function [latex]g\\circ f[\/latex] is defined such that<\/p>\n<div id=\"fs-id1170572481486\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))=3f(x)+1=3x^2+1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572450916\">Note that these two new functions are different from each other.<\/p>\n<div id=\"fs-id1170572450919\" class=\"bc-section section\" style=\"text-align: left;\">\n<h3>Combining Functions with Mathematical Operators<\/h3>\n<p id=\"fs-id1170572167715\">To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Given two functions [latex]f[\/latex] and [latex]g,[\/latex] we can define four new functions:<\/p>\n<div id=\"fs-id1170572167731\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cccc}(f+g)(x)=f(x)+g(x)\\hfill & & & \\text{Sum}\\hfill \\\\ (f-g)(x)=f(x)-g(x)\\hfill & & & \\text{Difference}\\hfill \\\\ (f\u00b7g)(x)=f(x)g(x)\\hfill & & & \\text{Product}\\hfill \\\\ \\Big(\\frac{f}{g}\\Big)(x)=\\frac{f(x)}{g(x)} \\, \\text{for} \\, g(x)\\ne 0\\hfill & & & \\text{Quotient}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<div>Often these functions have more than one term, so when you perform operations on them you will need to remember how to work with polynomials.<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Given multiple polynomials, add or subtract them to simplify the expressions<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Combine like terms.<\/li>\n<li>Simplify and write in standard form. Standard form means you start with the leading term, and write the rest of the terms in descending order by degree.<\/li>\n<\/ol>\n<p><strong>BE CAREFUL WHEN SUBTRACTING POLYNOMIALS!<\/strong><\/p>\n<p>When subtracting a polynomial from another, be careful to subtract\u00a0<em>each term<\/em> in the second from the first. That is, use the distributive property to distribute the minus sign through the second polynomial.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cc}\\left(3x^2-2x+9\\right)-\\left(x^2-4x+5\\right)\\text{}\\hfill &\\text{Distribute the negative in front of the parenthesis} \\hfill \\\\  3x^2-2x+9 -x^2 -\\left(-4x\\right) - 5\\hfill & \\text{Be careful when subtracting a negative}.\\hfill \\\\ 3x^2 - x^2 -2x+4x+9-5\\hfill & \\text{Rearrange terms in descending order of degree} \\hfill \\\\ 2x^2 +2x +4 \\hfill & \\text{Combine like terms}. \\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Given two binomials, Multiplying Using FOIL<\/h3>\n<ol>\n<li>Multiply the first terms of each binomial.<\/li>\n<li>Multiply the outer terms of the binomials.<\/li>\n<li>Multiply the inner terms of the binomials.<\/li>\n<li>Multiply the last terms of each binomial.<\/li>\n<li>Add the products.<\/li>\n<li>Combine like terms and simplify.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21205149\/CNX_CAT_Figure_01_04_003.jpg\" alt=\"Two quantities in parentheses are being multiplied, the first being: a times x plus b and the second being: c times x plus d. This expression equals ac times x squared plus ad times x plus bc times x plus bd. The terms ax and cx are labeled: First Terms. The terms ax and d are labeled: Outer Terms. The terms b and cx are labeled: Inner Terms. The terms b and d are labeled: Last Terms.\" \/><\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170572549843\" class=\"textbox exercises\">\n<h3>Example: Combining Functions Using Mathematical Operations<\/h3>\n<p id=\"fs-id1170572549852\">Given the functions [latex]f(x)=2x-3[\/latex] and [latex]g(x)=x^2-1[\/latex], find each of the following functions and state its domain.<\/p>\n<ol id=\"fs-id1170572169320\" style=\"list-style-type: lower-alpha;\">\n<li>[latex](f+g)(x)[\/latex]<\/li>\n<li>[latex](f-g)(x)[\/latex]<\/li>\n<li>[latex](f\u00b7g)(x)[\/latex]<\/li>\n<li>[latex]\\Big(\\frac{f}{g}\\Big)(x)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572228247\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572228247\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572228247\" style=\"list-style-type: lower-alpha;\">\n<li>[latex](f+g)(x)=(2x-3)+(x^2-1)=x^2+2x-4[\/latex]. The domain of this function is the interval [latex](\u2212\\infty ,\\infty )[\/latex].<\/li>\n<li>[latex](f-g)(x)=(2x-3)-(x^2-1)=\u2212x^2+2x-2[\/latex]. The domain of this function is the interval [latex](\u2212\\infty ,\\infty)[\/latex].<\/li>\n<li>[latex](f\u00b7g)(x)=(2x-3)(x^2-1)=2x^3-3x^2-2x+3[\/latex]. The domain of this function is the interval [latex](\u2212\\infty ,\\infty )[\/latex].<\/li>\n<li>[latex]\\Big(\\frac{f}{g}\\Big)(x)=\\dfrac{2x-3}{x^2-1}[\/latex]. The domain of this function is [latex]\\{x|x\\ne \\text{\u00b1}1\\}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Combining Functions Using Mathematical Operations<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qL2tyJhmrkg?controls=0&amp;start=1025&amp;end=1238&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.1ReviewofFunctions1025to1238_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.1 Review of Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<p>You can think of the domain of a function as all possible inputs that will produce a valid output. When you think about it this way, you will see that fractional functions that have a variable in the denominator (when a denominator is zero, you have an undefined value) or a variable under an even root radical (you cannot evaluate a negative value under an even square root), you will have restrictions in your domain.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall:\u00a0Given a function written in an equation form that includes a fraction, find the domain.<\/h3>\n<ol>\n<li>Identify the input values.<\/li>\n<li>Identify any restrictions on the input. If there is a denominator in the function\u2019s formula, set the denominator equal to zero and solve for [latex]x[\/latex] . These are the values that cannot be inputs in the function.<\/li>\n<li>Write the domain in interval form, making sure to exclude any restricted values from the domain.<\/li>\n<\/ol>\n<p>Find the domain of the function [latex]f\\left(x\\right)=\\dfrac{x+1}{2-x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q759017\">Show Solution<\/span><\/p>\n<div id=\"q759017\" class=\"hidden-answer\" style=\"display: none\">\n<p>When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2-x&=0 \\\\ -x&=-2 \\\\ x&=2 \\end{align}[\/latex]<\/p>\n<p>Now, we will exclude 2 from the domain. The answers are all real numbers where [latex]x<2[\/latex] or [latex]x>2[\/latex]. We can use a symbol known as the union, [latex]\\cup[\/latex], to combine the two sets. In interval notation, we write the solution: [latex]\\left(\\mathrm{-\\infty },2\\right)\\cup \\left(2,\\infty \\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\/18193532\/CNX_Precalc_Figure_01_02_028n2.jpg\" alt=\"Line graph of x=!2.\" width=\"487\" height=\"164\" \/><\/p>\n<p>In interval form, the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,2\\right)\\cup \\left(2,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall:\u00a0Given a function written in equation form including an even root, find the domain.<\/h3>\n<ol>\n<li>Identify the input values.<\/li>\n<li>Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/li>\n<li>The solution(s) are the domain of the function. If possible, write the answer in interval form.<\/li>\n<\/ol>\n<p>Find the domain of the function [latex]f\\left(x\\right)=\\sqrt{7-x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q722013\">Show Solution<\/span><\/p>\n<div id=\"q722013\" class=\"hidden-answer\" style=\"display: none\">\n<p>When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.<\/p>\n<p>Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}7-x&\\ge 0 \\\\ -x&\\ge -7 \\\\ x&\\le 7 \\end{align}[\/latex]<\/p>\n<p>Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to [latex]7[\/latex], or [latex]\\left(-\\infty ,7\\right][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572548529\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572548538\">For [latex]f(x)=x^2+3[\/latex] and [latex]g(x)=2x-5[\/latex], find [latex]\\left(\\frac{f}{g}\\right)(x)[\/latex] and state its domain.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265378\">Hint<\/span><\/p>\n<div id=\"q265378\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043323895\">The new function [latex]\\left(\\frac{f}{g}\\right)(x)[\/latex] is a quotient of two functions. For what values of [latex]x[\/latex] is the denominator zero?<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572482389\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572482389\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572482389\">[latex]\\Big(\\frac{f}{g}\\Big)(x)=\\dfrac{x^2+3}{2x-5}[\/latex]. The domain is [latex]\\{x|x\\ne \\frac{5}{2}\\}[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572212495\" class=\"bc-section section\">\n<h3>Function Composition<\/h3>\n<p id=\"fs-id1170572212500\">When we compose functions, we take a function of a function. For example, suppose the temperature [latex]T[\/latex] on a given day is described as a function of time [latex]t[\/latex] (measured in hours after midnight) as in Figure 6. Suppose the cost [latex]C[\/latex], to heat or cool a building for 1 hour, can be described as a function of the temperature [latex]T[\/latex]. Combining these two functions, we can describe the cost of heating or cooling a building as a function of time by evaluating [latex]C(T(t))[\/latex]. We have defined a new function, denoted [latex]C\\circ T[\/latex], which is defined such that [latex](C\\circ T)(t)=C(T(t))[\/latex] for all [latex]t[\/latex] in the domain of [latex]T[\/latex]. This new function is called a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makes sense to define this new function [latex](C\\circ T)(t)[\/latex]. It does not make sense to consider [latex](T\\circ C)(t)[\/latex], because temperature is not a function of cost.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572479276\">Consider the function [latex]f[\/latex] with domain [latex]A[\/latex] and range [latex]B[\/latex], and the function [latex]g[\/latex] with domain [latex]D[\/latex] and range [latex]E[\/latex]. If [latex]B[\/latex] is a subset of [latex]D[\/latex], then the <strong>composite function<\/strong> [latex](g\\circ f)(x)[\/latex] is the function with domain [latex]A[\/latex] such that<\/p>\n<div id=\"fs-id1170572176830\" class=\"equation\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1170572217742\">A composite function [latex]g\\circ f[\/latex] can be viewed in two steps. First, the function [latex]f[\/latex] maps each input [latex]x[\/latex] in the domain of [latex]f[\/latex] to its output [latex]f(x)[\/latex] in the range of [latex]f[\/latex]. Second, since the range of [latex]f[\/latex] is a subset of the domain of [latex]g[\/latex], the output [latex]f(x)[\/latex] is an element in the domain of [latex]g[\/latex], and therefore it is mapped to an output [latex]g(f(x))[\/latex] in the range of [latex]g[\/latex]. In Figure 12, we see a visual image of a composite function.<\/p>\n<div style=\"width: 447px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202134\/CNX_Calc_Figure_01_01_011.jpg\" alt=\"An image with three items. The first item is a blue bubble that has two labels: \u201cdomain of f\u201d and \u201cdomain of g of f\u201d. This item contains the numbers 1, 2, and 3. The second item is two bubbles: an orange bubble labeled \u201cdomain of g\u201d and a blue bubble that is completely contained within the orange bubble and is labeled \u201crange of f\u201d. The blue bubble contains the numbers 0 and 1, which are thus also contained within the larger orange bubble. The orange bubble contains two numbers not contained within the smaller blue bubble, which are 2 and 3. The third item is two bubbles: an orange bubble labeled \u201crange of g\u201d and a blue bubble that is completely contained within the orange bubble and is labeled \u201crange of g of f\u201d. The blue bubble contains the numbers 4 and 5, which are thus also contained within the larger orange bubble. The orange bubble contains one number not contained within the smaller blue bubble, which is the number 3. The first item points has a blue arrow with the label \u201cf\u201d that points to the blue bubble in the second item. The orange bubble in the second item has an orange arrow labeled \u201cg\u201d that points the orange bubble in the third item. The first item has a blue arrow labeled \u201cg of f\u201d which points to the blue bubble in the third item. There are three blue arrows pointing from numbers in the first item to the numbers contained in the blue bubble of the second item. The first blue arrow points from the 1 to the 0, the second blue arrow points from the 2 to the 1, and the third blue arrow points from the 3 to the 0. There are 4 orange arrows pointing from the numbers contained in the orange bubble in the second item, including those also contained in the blue bubble of the second item, to the numbers contained in the orange bubble of the third item, including the numbers in the blue bubble of the third item. The first orange arrow points from 2 to 3, the second orange arrow points from 3 to 5, the third orange arrow points from 0 to 4, and the fourth orange arrow points from 1 to 5.\" width=\"437\" height=\"252\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 12. For the composite function [latex]g\\circ f[\/latex], we have [latex](g\\circ f)(1)=4, \\, (g\\circ f)(2)=5,[\/latex] and [latex](g\\circ f)(3)=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>When composing functions, don&#8217;t forget to work from the inside out. Also, keep in mind that in general,\u00a0 [latex]f\\left(g\\left(x\\right)\\right)\\ne g\\left(f\\left(x\\right)\\right)[\/latex] for all [latex]x[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: How to perform a composition of functions<\/h3>\n<p>It is important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside.<\/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\/18195616\/CNX_Precalc_Figure_01_04_0012.jpg\" alt=\"Explanation of the composite function. g(x), the output of g is the input of f. X is the input of g.\" width=\"487\" height=\"171\" \/><\/p>\n<p>In general [latex]f\\circ g[\/latex] and [latex]g\\circ f[\/latex] are different functions. In other words in many cases [latex]f\\left(g\\left(x\\right)\\right)\\ne g\\left(f\\left(x\\right)\\right)[\/latex] for all [latex]x[\/latex].<\/p>\n<p>For example if [latex]f\\left(x\\right)={x}^{2}[\/latex] and [latex]g\\left(x\\right)=x+2[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&=f\\left(x+2\\right) \\\\[2mm] &={\\left(x+2\\right)}^{2} \\\\[2mm] &={x}^{2}+4x+4\\hfill \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">but<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&=g\\left({x}^{2}\\right) \\\\[2mm] \\text{ }&={x}^{2}+2\\hfill \\end{align}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1170572212495\" class=\"bc-section section\">\n<div id=\"fs-id1170572481349\" class=\"textbox exercises\">\n<h3>Example: Compositions of Functions Defined by Formulas<\/h3>\n<p id=\"fs-id1170572481359\">Consider the functions [latex]f(x)=x^2+1[\/latex] and [latex]g(x)=\\frac{1}{x}[\/latex].<\/p>\n<ol id=\"fs-id1170572482111\" style=\"list-style-type: lower-alpha;\">\n<li>Find [latex](g\\circ f)(x)[\/latex] and state its domain and range.<\/li>\n<li>Evaluate [latex](g\\circ f)(4)[\/latex] and [latex](g\\circ f)\\left(-\\frac{1}{2}\\right)[\/latex].<\/li>\n<li>Find [latex](f\\circ g)(x)[\/latex] and state its domain and range.<\/li>\n<li>Evaluate [latex](f\\circ g)(4)[\/latex] and [latex](f\\circ g)\\left(-\\frac{1}{2}\\right)[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572222946\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572222946\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572222946\" style=\"list-style-type: lower-alpha;\">\n<li>We can find the formula for [latex](g\\circ f)(x)[\/latex] in two different ways. We could write\n<div id=\"fs-id1170571109518\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))=g(x^2+1)=\\dfrac{1}{x^2+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Alternatively, we could write<\/p>\n<div id=\"fs-id1170573521539\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](g\\circ f)(x)=g(f(x))=\\dfrac{1}{f(x)}=\\dfrac{1}{x^2+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>Since [latex]x^2+1\\ne 0[\/latex] for all real numbers [latex]x[\/latex], the domain of [latex](g\\circ f)(x)[\/latex] is the set of all real numbers. Since [latex]0<\\frac{1}{(x^2+1)}\\le 1[\/latex], the range is, at most, the interval [latex](0,1][\/latex]. To show that the range is this entire interval, we let [latex]y=\\frac{1}{(x^2+1)}[\/latex] and solve this equation for [latex]x[\/latex] to show that for all [latex]y[\/latex] in the interval [latex](0,1][\/latex], there exists a real number [latex]x[\/latex] such that [latex]y=\\frac{1}{(x^2+1)}[\/latex]. Solving this equation for [latex]x[\/latex], we see that [latex]x^2+1=\\frac{1}{y}[\/latex], which implies that\n\n\n<div id=\"fs-id1170573368782\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\u00b1\\sqrt{\\frac{1}{y}-1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>If [latex]y[\/latex] is in the interval [latex](0,1][\/latex], the expression under the radical is nonnegative, and therefore there exists a real number [latex]x[\/latex] such that [latex]\\frac{1}{(x^2+1)}=y[\/latex]. We conclude that the range of [latex]g\\circ f[\/latex] is the interval [latex](0,1][\/latex].<\/li>\n<li>[latex](g\\circ f)(4)=g(f(4))=g(4^2+1)=g(17)=\\frac{1}{17}[\/latex]<br \/>\n[latex](g\\circ f)(-\\frac{1}{2})=g(f(-\\frac{1}{2}))=g((-\\frac{1}{2})^2+1)=g(\\frac{5}{4})=\\frac{4}{5}[\/latex]<\/li>\n<li>We can find a formula for [latex](f\\circ g)(x)[\/latex] in two ways. First, we could write\n<div id=\"fs-id1170573750183\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f\\circ g)(x)=f(g(x))=f\\left(\\frac{1}{x}\\right)=\\left(\\frac{1}{x}\\right)^2+1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Alternatively, we could write<\/p>\n<div id=\"fs-id1170571123790\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f\\circ g)(x)=f(g(x))=(g(x))^2+1=\\left(\\frac{1}{x}\\right)^2+1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The domain of [latex]f\\circ g[\/latex] is the set of all real numbers [latex]x[\/latex] such that [latex]x\\ne 0[\/latex]. To find the range of [latex]f[\/latex], we need to find all values [latex]y[\/latex] for which there exists a real number [latex]x\\ne 0[\/latex] such that<\/p>\n<div id=\"fs-id1170571118828\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left(\\frac{1}{x}\\right)^2+1=y[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Solving this equation for [latex]x[\/latex], we see that we need [latex]x[\/latex] to satisfy<\/p>\n<div id=\"fs-id1170571118874\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\left(\\frac{1}{x}\\right)^2=y-1[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p>which simplifies to<\/p>\n<div id=\"fs-id1170571330548\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{x}=\u00b1\\sqrt{y-1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Finally, we obtain<\/p>\n<div id=\"fs-id1170571282415\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\u00b1\\frac{1}{\\sqrt{y-1}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Since [latex]\\frac{1}{\\sqrt{y-1}}[\/latex] is a real number if and only if [latex]y>1[\/latex], the range of [latex]f[\/latex] is the set [latex]\\{y|y\\ge 1\\}[\/latex].<\/li>\n<li>[latex](f\\circ g)(4)=f(g(4))=f\\left(\\frac{1}{4}\\right)=\\left(\\frac{1}{4}\\right)^2+1=\\frac{17}{16}[\/latex]<br \/>\n[latex](f\\circ g)\\left(-\\frac{1}{2}\\right)=f\\left(g\\left(-\\frac{1}{2}\\right)\\right)=f(-2)=(-2)^2+1=5[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170572173194\">In the box above, we can see that [latex](f\\circ g)(x)\\ne (g\\circ f)(x)[\/latex]. This tells us, in general terms, that the order in which we compose functions matters.<\/p>\n<div id=\"fs-id1170572483657\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572483665\">Let [latex]f(x)=2-5x[\/latex]<\/p>\n<p>Let [latex]g(x)=\\sqrt{x}[\/latex]<\/p>\n<p>Find [latex](f\\circ g)(x)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572481428\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572481428\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572481428\">[latex](f\\circ g)(x)=2-5\\sqrt{x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572173789\" class=\"textbox exercises\">\n<h3>Example: Composition of Functions Defined by Tables<\/h3>\n<p id=\"fs-id1170572173799\">Consider the functions [latex]f[\/latex] and [latex]g[\/latex] described below.<\/p>\n<table id=\"fs-id1170572173822\" class=\"column-header\" summary=\"A table with 2 rows and 8 columns. The first row is labeled \u201cx\u201d and has the values \u201c-3; -2; -1; 0; 1; 2; 3; 4\u201d. The second row is labeled \u201cf(x)\u201d and has the values \u201c0; 4; 2; 4; -2; 0; -2; 4\u201d.\">\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\\mathbf{x}[\/latex]<\/td>\n<td>-3<\/td>\n<td>-2<\/td>\n<td>-1<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\mathbf{f(x)}[\/latex]<\/td>\n<td>0<\/td>\n<td>4<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>-2<\/td>\n<td>0<\/td>\n<td>-2<\/td>\n<td>4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table id=\"fs-id1170572550754\" class=\"column-header\" summary=\"A table with 2 rows and 5 columns. The first row is labeled \u201cx\u201d and has the values \u201c-4; -2; 0; 2; 4\u201d. The second row is labeled \u201cg(x)\u201d and has the values \u201c1; 0; 3; 0; 5\u201d.\">\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\\mathbf{x}[\/latex]<\/td>\n<td>-4<\/td>\n<td>-2<\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\mathbf{g(x)}[\/latex]<\/td>\n<td>1<\/td>\n<td>0<\/td>\n<td>3<\/td>\n<td>0<\/td>\n<td>5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol id=\"fs-id1170572242030\" style=\"list-style-type: lower-alpha;\">\n<li>Evaluate [latex](g\\circ f)(3)[\/latex] and [latex](g\\circ f)(0)[\/latex].<\/li>\n<li>State the domain and range of [latex](g\\circ f)(x)[\/latex].<\/li>\n<li>Evaluate [latex](f\\circ f)(3)[\/latex] and [latex](f\\circ f)(1)[\/latex].<\/li>\n<li>State the domain and range of [latex](f\\circ f)(x)[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572552174\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572552174\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572552174\" style=\"list-style-type: lower-alpha;\">\n<li>[latex](g\\circ f)(3)=g(f(3))=g(-2)=0[\/latex]<br \/>\n[latex](g\\circ f)(0)=g(4)=5[\/latex]<\/li>\n<li>The domain of [latex]g\\circ f[\/latex] is the set [latex]\\{-3,-2,-1,0,1,2,3,4\\}[\/latex]. Since the range of [latex]f[\/latex] is the set [latex]\\{-2,0,2,4\\}[\/latex], the range of [latex]g\\circ f[\/latex] is the set [latex]\\{0,3,5\\}[\/latex].<\/li>\n<li>[latex](f\\circ f)(3)=f(f(3))=f(-2)=4[\/latex]<br \/>\n[latex](f\\circ f)(1)=f(f(1))=f(-2)=4[\/latex]<\/li>\n<li>The domain of [latex]f\\circ f[\/latex] is the set [latex]\\{-3,-2,-1,0,1,2,3,4\\}[\/latex]. Since the range of [latex]f[\/latex] is the set [latex]\\{-2,0,2,4\\}[\/latex], the range of [latex]f\\circ f[\/latex] is the set [latex]\\{0,4\\}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Composition of Functions Defined by Tables<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/qL2tyJhmrkg?controls=0&amp;start=1503&amp;end=1757&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.1ReviewofFunctions1503to1757_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.1 Review of Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170572173737\" class=\"textbox exercises\">\n<h3>Example: Application Involving a Composite Function<\/h3>\n<p id=\"fs-id1170572173747\">A store is advertising a sale of [latex]20\\%[\/latex] off all merchandise. Caroline has a coupon that entitles her to an additional [latex]15\\%[\/latex] off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of [latex]x[\/latex] dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by using a composite function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572454320\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572454320\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572454320\">Since the sale price is [latex]20\\%[\/latex] off the original price, if an item is [latex]x[\/latex] dollars, its sale price is given by [latex]f(x)=0.80x[\/latex]. Since the coupon entitles an individual to [latex]15\\%[\/latex] off the price of any item, if an item is [latex]y[\/latex] dollars, the price, after applying the coupon, is given by [latex]g(y)=0.85y[\/latex]. Therefore, if the price is originally [latex]x[\/latex] dollars, its sale price will be [latex]f(x)=0.80x[\/latex] and then its final price after the coupon will be [latex]g(f(x))=0.85(0.80x)=0.68x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572480210\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170572480218\">If items are on sale for [latex]10\\%[\/latex] off their original price, and a customer has a coupon for an additional [latex]30\\%[\/latex] off, what will be the final price for an item that is originally [latex]x[\/latex] dollars, after applying the coupon to the sale price?<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q324566\">Hint<\/span><\/p>\n<div id=\"q324566\" class=\"hidden-answer\" style=\"display: none\">\n<p>The sale price of an item with an original price of [latex]x[\/latex] dollars is [latex]f(x)=0.90x[\/latex]. The coupon price for an item that is [latex]y[\/latex] dollars is [latex]g(y)=0.70y[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170572213178\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170572213178\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572213178\">[latex](g\\circ f)(x)=0.63x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm33467\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33467&theme=oea&iframe_resize_id=ohm33467&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-93\">\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>1.1 Review of Functions. <strong>Authored by<\/strong>: Ryan Melton. <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>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"1.1 Review of Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-93","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/93","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":57,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/93\/revisions"}],"predecessor-version":[{"id":4738,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/93\/revisions\/4738"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/93\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=93"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=93"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=93"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=93"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}