{"id":1973,"date":"2021-08-19T16:03:43","date_gmt":"2021-08-19T16:03:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/review-for-substitution\/"},"modified":"2021-11-17T01:32:32","modified_gmt":"2021-11-17T01:32:32","slug":"review-for-substitution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/review-for-substitution\/","title":{"raw":"Skills Review for Substitution and Integrals Involving Exponential and Logarithmic Functions","rendered":"Skills Review for Substitution and Integrals Involving Exponential and Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Decompose a composite function into its component functions<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Substitution and Integrals Involving Exponential and Logarithmic Functions sections, we will learn all about using substitution as an integration method. Substitution is basically the process used to find the antiderivative of a function that was differentiated using the chain rule. That being said, it is important to be able to look at a composite function and identify the inside function and outside function. Usually, the inside function is what we set our substitution variable equal to.\r\n<h2>Identify Components of Composite Functions<\/h2>\r\nIn some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Components of Composite Functions<\/h3>\r\nWrite [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.\r\n\r\n[reveal-answer q=\"702975\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"702975\"]\r\n\r\nWe are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex]. As one possibility, we might notice that the expression [latex]5-{x}^{2}[\/latex] is the inside of the square root. We could then decompose the function as\r\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=5-{x}^{2}\\hspace{2mm}\\text{and}\\hspace{2mm}g\\left(x\\right)=\\sqrt{x}[\/latex]<\/p>\r\nWe can check our answer by recomposing the functions.\r\n<p style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g\\left(5-{x}^{2}\\right)=\\sqrt{5-{x}^{2}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Components of Composite Functions<\/h3>\r\nWrite [latex]f\\left(x\\right)=e^{4x-3}[\/latex] as the composition of two functions.\r\n\r\n[reveal-answer q=\"702977\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"702977\"]\r\n\r\nWe are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex]. As one possibility, we might notice that the expression [latex]4x-3[\/latex] is within the exponent of the exponential function. We could then decompose the function as\r\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=4x-3\\hspace{2mm}\\text{and}\\hspace{2mm}g\\left(x\\right)=e^{x}[\/latex]<\/p>\r\nWe can check our answer by recomposing the functions.\r\n<p style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g(4x-3)=e^{4x-3}[\/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\nWrite [latex]f\\left(x\\right)=\\dfrac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.\r\n\r\n[reveal-answer q=\"489928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"489928\"]\r\n\r\nPossible answer:\r\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\r\n[latex]h\\left(x\\right)=\\dfrac{4}{3-x}[\/latex]\r\n\r\n[latex]f=h\\circ g[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]32917[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Decompose a composite function into its component functions<\/li>\n<\/ul>\n<\/div>\n<p>In the Substitution and Integrals Involving Exponential and Logarithmic Functions sections, we will learn all about using substitution as an integration method. Substitution is basically the process used to find the antiderivative of a function that was differentiated using the chain rule. That being said, it is important to be able to look at a composite function and identify the inside function and outside function. Usually, the inside function is what we set our substitution variable equal to.<\/p>\n<h2>Identify Components of Composite Functions<\/h2>\n<p>In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Components of Composite Functions<\/h3>\n<p>Write [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex] as the composition of two functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q702975\">Show Solution<\/span><\/p>\n<div id=\"q702975\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex]. As one possibility, we might notice that the expression [latex]5-{x}^{2}[\/latex] is the inside of the square root. We could then decompose the function as<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=5-{x}^{2}\\hspace{2mm}\\text{and}\\hspace{2mm}g\\left(x\\right)=\\sqrt{x}[\/latex]<\/p>\n<p>We can check our answer by recomposing the functions.<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g\\left(5-{x}^{2}\\right)=\\sqrt{5-{x}^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Components of Composite Functions<\/h3>\n<p>Write [latex]f\\left(x\\right)=e^{4x-3}[\/latex] as the composition of two functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q702977\">Show Solution<\/span><\/p>\n<div id=\"q702977\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=g\\left(h\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex]. As one possibility, we might notice that the expression [latex]4x-3[\/latex] is within the exponent of the exponential function. We could then decompose the function as<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(x\\right)=4x-3\\hspace{2mm}\\text{and}\\hspace{2mm}g\\left(x\\right)=e^{x}[\/latex]<\/p>\n<p>We can check our answer by recomposing the functions.<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(h\\left(x\\right)\\right)=g(4x-3)=e^{4x-3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write [latex]f\\left(x\\right)=\\dfrac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q489928\">Show Solution<\/span><\/p>\n<div id=\"q489928\" class=\"hidden-answer\" style=\"display: none\">\n<p>Possible answer:<\/p>\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\n<p>[latex]h\\left(x\\right)=\\dfrac{4}{3-x}[\/latex]<\/p>\n<p>[latex]f=h\\circ g[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm32917\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32917&theme=oea&iframe_resize_id=ohm32917&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-1973\">\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 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