{"id":3889,"date":"2021-05-20T18:29:55","date_gmt":"2021-05-20T18:29:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-the-chain-rule\/"},"modified":"2021-07-03T18:54:29","modified_gmt":"2021-07-03T18:54:29","slug":"review-for-the-chain-rule","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-the-chain-rule\/","title":{"raw":"Skills Review for The Chain Rule and Derivatives of Inverse Functions","rendered":"Skills Review for The Chain Rule and Derivatives of Inverse Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Evaluate composite functions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4611,&quot;3&quot;:{&quot;1&quot;:0},&quot;4&quot;:{&quot;1&quot;:2,&quot;2&quot;:13624051},&quot;12&quot;:0,&quot;15&quot;:&quot;Work Sans&quot;}\">Evaluate composite functions<\/span><\/li>\r\n \t<li>Factor out the greatest common factor of a polynomial<\/li>\r\n \t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Factor polynomials with negative or fractional exponents&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4611,&quot;3&quot;:{&quot;1&quot;:0},&quot;4&quot;:[null,2,16573901],&quot;12&quot;:0,&quot;15&quot;:&quot;Work Sans&quot;}\">Factor polynomials with negative or fractional exponents<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nIn <em>The Chain Rule<\/em> section, you will learn how to take the derivative of composed functions. Here we will review some basics about function composition and how it basically involves plugging one function into another. Knowledge about function composition, including how to plug one function into another, will also be useful in the\u00a0<em>Derivatives of Inverse Function<\/em> section as you learn about the Inverse Function Theorem. Lastly, to help with simplifying derivatives of functions that require the use of the chain rule, we will review factoring common binomial factors from algebraic expressions.\r\n<h2>Find Composite Functions<\/h2>\r\nWe can create functions by composing them. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:\r\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/p>\r\nWe read the left-hand side as [latex]\"f[\/latex] composed with [latex]g[\/latex] at [latex]x,\"[\/latex] and the right-hand side as [latex]\"f[\/latex] of [latex]g[\/latex] of [latex]x.\"[\/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\\circ [\/latex] is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value.\r\n\r\nIt is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function [latex]g[\/latex] takes the input [latex]x[\/latex] first and yields an output [latex]g\\left(x\\right)[\/latex]. Then the function [latex]f[\/latex] takes [latex]g\\left(x\\right)[\/latex] as an input and yields an output [latex]f\\left(g\\left(x\\right)\\right)[\/latex].\r\n\r\n<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\nThese expressions are not equal.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Find a CompositE Function<\/h3>\r\nUsing the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex].\r\n[latex]f\\left(x\\right)=2x+1\\\\g\\left(x\\right)=3-x[\/latex]\r\n[reveal-answer q=\"822785\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"822785\"]\r\n\r\nLet\u2019s begin by substituting [latex]g\\left(x\\right)[\/latex] into [latex]f\\left(x\\right)[\/latex].\r\n\r\n[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&amp;=2\\left(3-x\\right)+1 \\\\[2mm] &amp;=6 - 2x+1 \\\\[2mm] &amp;=7 - 2x \\end{align}[\/latex]\r\n\r\nNow we can substitute [latex]f\\left(x\\right)[\/latex] into [latex]g\\left(x\\right)[\/latex].\r\n\r\n[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&amp;=3-\\left(2x+1\\right) \\\\[2mm] &amp;=3 - 2x - 1 \\\\[2mm] &amp;=-2x+2 \\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]32907[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Factor Common Binomial Factors from Algebraic Expressions<\/h2>\r\nThe greatest common factor (GCF) of an algebraic expression can sometimes be a binomial expression. Here is an example of a GCF that is a binomial.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Example: Factoring Out a Common Binomial Factor<\/h3>\r\nFactor [latex]x\\left({b}^{2}-a\\right)+6\\left({b}^{2}-a\\right)[\/latex] by pulling out the GCF.\r\n\r\n[reveal-answer q=\"94532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"94532\"]\r\n\r\nThe common binomial factor is\u00a0[latex]\\left({b}^{2}-a\\right)[\/latex].\r\n\r\nWe will now factor out this common binomial factor to get [latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Example: Factoring Out a Common Binomial Factor<\/h3>\r\nFactor [latex]x\\left(y-9\\right)^{3}+6\\left(y-9\\right)^2[\/latex] by pulling out the GCF.\r\n\r\n[reveal-answer q=\"94533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"94533\"]\r\n\r\nThe greatest common factor is\u00a0[latex]\\left(y-9\\right)^{2}[\/latex]. Remember, as long as the factor is present in all terms, the greatest common factor is that factor to the lowest power present.\r\n\r\nWe will factor [latex]\\left(y-9\\right)^{2}[\/latex] from the expression.\r\n\r\nThis gives\u00a0[latex]\\left(y-9\\right)^{2}(x(y-9)+6)[\/latex].\r\n\r\nDistributing the\u00a0[latex]x[\/latex] gives\u00a0[latex]\\left(y-9\\right)^{2}(xy-9x+6)[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7888[\/ohm_question]\r\n\r\n<\/div>\r\nExpressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, [latex]2{x}^{\\frac{1}{4}}+5{x}^{\\frac{3}{4}}[\/latex] can be factored by pulling out [latex]{x}^{\\frac{1}{4}}[\/latex] and being rewritten as [latex]{x}^{\\frac{1}{4}}\\left(2+5{x}^{\\frac{1}{2}}\\right)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Factoring an Expression with Fractional or Negative Exponents<\/h3>\r\nFactor [latex]3x{\\left(x+2\\right)}^{\\frac{-1}{3}}+4{\\left(x+2\\right)}^{\\frac{2}{3}}[\/latex].\r\n\r\n[reveal-answer q=\"164967\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164967\"]\r\n\r\nFactor out the term with the lowest value of the exponent. In this case, that would be [latex]{\\left(x+2\\right)}^{-\\frac{1}{3}}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4\\left(x+2\\right)\\right)\\hfill &amp; \\text{Factor out the GCF}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4x+8\\right)\\hfill &amp; \\text{Simplify}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(7x+8\\right)\\hfill &amp; \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]2{\\left(5a - 1\\right)}^{\\frac{3}{4}}+7a{\\left(5a - 1\\right)}^{-\\frac{1}{4}}[\/latex].\r\n\r\n[reveal-answer q=\"989124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989124\"]\r\n\r\n[latex]{\\left(5a - 1\\right)}^{-\\frac{1}{4}}\\left(17a - 2\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Evaluate composite functions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4611,&quot;3&quot;:{&quot;1&quot;:0},&quot;4&quot;:{&quot;1&quot;:2,&quot;2&quot;:13624051},&quot;12&quot;:0,&quot;15&quot;:&quot;Work Sans&quot;}\">Evaluate composite functions<\/span><\/li>\n<li>Factor out the greatest common factor of a polynomial<\/li>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Factor polynomials with negative or fractional exponents&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4611,&quot;3&quot;:{&quot;1&quot;:0},&quot;4&quot;:[null,2,16573901],&quot;12&quot;:0,&quot;15&quot;:&quot;Work Sans&quot;}\">Factor polynomials with negative or fractional exponents<\/span><\/li>\n<\/ul>\n<\/div>\n<p>In <em>The Chain Rule<\/em> section, you will learn how to take the derivative of composed functions. Here we will review some basics about function composition and how it basically involves plugging one function into another. Knowledge about function composition, including how to plug one function into another, will also be useful in the\u00a0<em>Derivatives of Inverse Function<\/em> section as you learn about the Inverse Function Theorem. Lastly, to help with simplifying derivatives of functions that require the use of the chain rule, we will review factoring common binomial factors from algebraic expressions.<\/p>\n<h2>Find Composite Functions<\/h2>\n<p>We can create functions by composing them. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(f\\circ g\\right)\\left(x\\right)=f\\left(g\\left(x\\right)\\right)[\/latex]<\/p>\n<p>We read the left-hand side as [latex]\"f[\/latex] composed with [latex]g[\/latex] at [latex]x,\"[\/latex] and the right-hand side as [latex]\"f[\/latex] of [latex]g[\/latex] of [latex]x.\"[\/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\\circ[\/latex] is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value.<\/p>\n<p>It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function [latex]g[\/latex] takes the input [latex]x[\/latex] first and yields an output [latex]g\\left(x\\right)[\/latex]. Then the function [latex]f[\/latex] takes [latex]g\\left(x\\right)[\/latex] as an input and yields an output [latex]f\\left(g\\left(x\\right)\\right)[\/latex].<\/p>\n<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<p>These expressions are not equal.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Find a CompositE Function<\/h3>\n<p>Using the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex].<br \/>\n[latex]f\\left(x\\right)=2x+1\\\\g\\left(x\\right)=3-x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q822785\">Show Solution<\/span><\/p>\n<div id=\"q822785\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s begin by substituting [latex]g\\left(x\\right)[\/latex] into [latex]f\\left(x\\right)[\/latex].<\/p>\n<p>[latex]\\begin{align}f\\left(g\\left(x\\right)\\right)&=2\\left(3-x\\right)+1 \\\\[2mm] &=6 - 2x+1 \\\\[2mm] &=7 - 2x \\end{align}[\/latex]<\/p>\n<p>Now we can substitute [latex]f\\left(x\\right)[\/latex] into [latex]g\\left(x\\right)[\/latex].<\/p>\n<p>[latex]\\begin{align}g\\left(f\\left(x\\right)\\right)&=3-\\left(2x+1\\right) \\\\[2mm] &=3 - 2x - 1 \\\\[2mm] &=-2x+2 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm32907\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32907&theme=oea&iframe_resize_id=ohm32907&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Factor Common Binomial Factors from Algebraic Expressions<\/h2>\n<p>The greatest common factor (GCF) of an algebraic expression can sometimes be a binomial expression. Here is an example of a GCF that is a binomial.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Example: Factoring Out a Common Binomial Factor<\/h3>\n<p>Factor [latex]x\\left({b}^{2}-a\\right)+6\\left({b}^{2}-a\\right)[\/latex] by pulling out the GCF.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q94532\">Show Solution<\/span><\/p>\n<div id=\"q94532\" class=\"hidden-answer\" style=\"display: none\">\n<p>The common binomial factor is\u00a0[latex]\\left({b}^{2}-a\\right)[\/latex].<\/p>\n<p>We will now factor out this common binomial factor to get [latex]\\left({b}^{2}-a\\right)\\left(x+6\\right)[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Example: Factoring Out a Common Binomial Factor<\/h3>\n<p>Factor [latex]x\\left(y-9\\right)^{3}+6\\left(y-9\\right)^2[\/latex] by pulling out the GCF.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q94533\">Show Solution<\/span><\/p>\n<div id=\"q94533\" class=\"hidden-answer\" style=\"display: none\">\n<p>The greatest common factor is\u00a0[latex]\\left(y-9\\right)^{2}[\/latex]. Remember, as long as the factor is present in all terms, the greatest common factor is that factor to the lowest power present.<\/p>\n<p>We will factor [latex]\\left(y-9\\right)^{2}[\/latex] from the expression.<\/p>\n<p>This gives\u00a0[latex]\\left(y-9\\right)^{2}(x(y-9)+6)[\/latex].<\/p>\n<p>Distributing the\u00a0[latex]x[\/latex] gives\u00a0[latex]\\left(y-9\\right)^{2}(xy-9x+6)[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7888\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7888&theme=oea&iframe_resize_id=ohm7888&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, [latex]2{x}^{\\frac{1}{4}}+5{x}^{\\frac{3}{4}}[\/latex] can be factored by pulling out [latex]{x}^{\\frac{1}{4}}[\/latex] and being rewritten as [latex]{x}^{\\frac{1}{4}}\\left(2+5{x}^{\\frac{1}{2}}\\right)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Factoring an Expression with Fractional or Negative Exponents<\/h3>\n<p>Factor [latex]3x{\\left(x+2\\right)}^{\\frac{-1}{3}}+4{\\left(x+2\\right)}^{\\frac{2}{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164967\">Show Solution<\/span><\/p>\n<div id=\"q164967\" class=\"hidden-answer\" style=\"display: none\">\n<p>Factor out the term with the lowest value of the exponent. In this case, that would be [latex]{\\left(x+2\\right)}^{-\\frac{1}{3}}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4\\left(x+2\\right)\\right)\\hfill & \\text{Factor out the GCF}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4x+8\\right)\\hfill & \\text{Simplify}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(7x+8\\right)\\hfill & \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Factor [latex]2{\\left(5a - 1\\right)}^{\\frac{3}{4}}+7a{\\left(5a - 1\\right)}^{-\\frac{1}{4}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q989124\">Show Solution<\/span><\/p>\n<div id=\"q989124\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]{\\left(5a - 1\\right)}^{-\\frac{1}{4}}\\left(17a - 2\\right)[\/latex]<\/p><\/div>\n<\/div>\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-3889\">\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>Modification and Revision . <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 Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen 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