{"id":152,"date":"2021-02-03T22:01:09","date_gmt":"2021-02-03T22:01:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=152"},"modified":"2021-03-18T02:24:04","modified_gmt":"2021-03-18T02:24:04","slug":"summary-of-inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/summary-of-inverse-functions\/","title":{"raw":"Summary of Inverse Functions","rendered":"Summary of Inverse Functions"},"content":{"raw":"<div id=\"fs-id1170572470426\" class=\"learning-objectives\">\r\n<h3>EsSential Concepts<\/h3>\r\n<ul id=\"fs-id1170572470433\">\r\n \t<li>For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.<\/li>\r\n \t<li>If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.<\/li>\r\n \t<li>For a function [latex]f[\/latex] and its inverse [latex]f^{-1}, \\, f(f^{-1}(x))=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f^{-1}[\/latex] and [latex]f^{-1}(f(x))=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/li>\r\n \t<li>Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.<\/li>\r\n \t<li>The graph of a function [latex]f[\/latex] and its inverse [latex]f^{-1}[\/latex] are symmetric about the line [latex]y=x[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<ul id=\"fs-id1170572453126\">\r\n \t<li><strong>Inverse functions<\/strong>\r\n[latex]f^{-1}(f(x))=x[\/latex]\u00a0 for all\u00a0 [latex]x[\/latex]\u00a0 in\u00a0 [latex]D[\/latex], and\u00a0 [latex]f(f^{-1}(y))=y[\/latex]\u00a0 for all\u00a0 [latex]y[\/latex]\u00a0 in\u00a0 [latex]R[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1170572229168\" class=\"definition\">\r\n \t<dt>horizontal line test<\/dt>\r\n \t<dd id=\"fs-id1170572229174\">a function [latex]f[\/latex] is one-to-one if and only if every horizontal line intersects the graph of [latex]f[\/latex], at most, once<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170572229190\" class=\"definition\">\r\n \t<dt>inverse function<\/dt>\r\n \t<dd id=\"fs-id1170572229195\">for a function [latex]f[\/latex], the inverse function [latex]f^{-1}[\/latex] satisfies [latex]f^{-1}(y)=x[\/latex] if [latex]f(x)=y[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170572482608\" class=\"definition\">\r\n \t<dt>inverse trigonometric functions<\/dt>\r\n \t<dd id=\"fs-id1170572482614\">the inverses of the trigonometric functions are defined on restricted domains where they are one-to-one functions<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170572482619\" class=\"definition\">\r\n \t<dt>one-to-one function<\/dt>\r\n \t<dd id=\"fs-id1170572482624\">a function [latex]f[\/latex] is one-to-one if [latex]f(x_1) \\ne f(x_2)[\/latex] if [latex]x_1 \\ne x_2[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1170572482683\" class=\"definition\">\r\n \t<dt>restricted domain<\/dt>\r\n \t<dd id=\"fs-id1170572482689\">a subset of the domain of a function [latex]f[\/latex]<\/dd>\r\n<\/dl>","rendered":"<div id=\"fs-id1170572470426\" class=\"learning-objectives\">\n<h3>EsSential Concepts<\/h3>\n<ul id=\"fs-id1170572470433\">\n<li>For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.<\/li>\n<li>If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.<\/li>\n<li>For a function [latex]f[\/latex] and its inverse [latex]f^{-1}, \\, f(f^{-1}(x))=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f^{-1}[\/latex] and [latex]f^{-1}(f(x))=x[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/li>\n<li>Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.<\/li>\n<li>The graph of a function [latex]f[\/latex] and its inverse [latex]f^{-1}[\/latex] are symmetric about the line [latex]y=x[\/latex].<\/li>\n<\/ul>\n<\/div>\n<h2>Key Equations<\/h2>\n<ul id=\"fs-id1170572453126\">\n<li><strong>Inverse functions<\/strong><br \/>\n[latex]f^{-1}(f(x))=x[\/latex]\u00a0 for all\u00a0 [latex]x[\/latex]\u00a0 in\u00a0 [latex]D[\/latex], and\u00a0 [latex]f(f^{-1}(y))=y[\/latex]\u00a0 for all\u00a0 [latex]y[\/latex]\u00a0 in\u00a0 [latex]R[\/latex].<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1170572229168\" class=\"definition\">\n<dt>horizontal line test<\/dt>\n<dd id=\"fs-id1170572229174\">a function [latex]f[\/latex] is one-to-one if and only if every horizontal line intersects the graph of [latex]f[\/latex], at most, once<\/dd>\n<\/dl>\n<dl id=\"fs-id1170572229190\" class=\"definition\">\n<dt>inverse function<\/dt>\n<dd id=\"fs-id1170572229195\">for a function [latex]f[\/latex], the inverse function [latex]f^{-1}[\/latex] satisfies [latex]f^{-1}(y)=x[\/latex] if [latex]f(x)=y[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170572482608\" class=\"definition\">\n<dt>inverse trigonometric functions<\/dt>\n<dd id=\"fs-id1170572482614\">the inverses of the trigonometric functions are defined on restricted domains where they are one-to-one functions<\/dd>\n<\/dl>\n<dl id=\"fs-id1170572482619\" class=\"definition\">\n<dt>one-to-one function<\/dt>\n<dd id=\"fs-id1170572482624\">a function [latex]f[\/latex] is one-to-one if [latex]f(x_1) \\ne f(x_2)[\/latex] if [latex]x_1 \\ne x_2[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1170572482683\" class=\"definition\">\n<dt>restricted domain<\/dt>\n<dd id=\"fs-id1170572482689\">a subset of the domain of a function [latex]f[\/latex]<\/dd>\n<\/dl>\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-152\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><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":20,"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\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-152","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/152","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":8,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/152\/revisions"}],"predecessor-version":[{"id":1380,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/152\/revisions\/1380"}],"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\/152\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=152"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=152"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=152"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=152"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}