{"id":775,"date":"2019-04-29T17:24:21","date_gmt":"2019-04-29T21:24:21","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=775"},"modified":"2019-12-05T15:55:59","modified_gmt":"2019-12-05T20:55:59","slug":"11-3-inverse-functions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/11-3-inverse-functions\/","title":{"raw":"11.3 Inverse Functions","rendered":"11.3 Inverse Functions"},"content":{"raw":"[latexpage]\r\n\r\nWhen working with mathematical functions, it sometimes becomes useful to undo what the original function does. To do this, you need to find the inverse of the function. This feature is commonly used in exponents and logarithms and in trigonometry.\r\n\r\nIn this topic, you will be looking at functions and seeing if they can be inverses of themselves. The notation used for this procedure is \\(f^{-1}(x)\\) is the inverse of \\(f(x)\\). In practice, this works as follows:\r\n\r\n\\[f^{-1}[f(x)] = x\\]\r\n\r\nThis is a very useful tool used many times over in math. If there are two functions \\(f(x)\\) and \\(g(x)\\) that are inverses of each other (if their composites \u201cundo\u201d each other\u2019s function), their composite functions look like:\r\n\r\n\\[g(f(x)) = x\\text{ and }f(g(x)) = x\\]\r\n
\r\n

Example 11.3.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nAre the functions \\(f(x) = 2x + 20\\) and \\(g(x) = \\dfrac{x}{2} - 10\\) inverses of each other?\r\n\r\nTest if either \\(g(f(x)) = x\\) or \\(f(g(x)) = x\\):\r\n\r\n\\[\\begin{array}{rrl}\r\ng(f(x))&=&\\dfrac{2x+20}{2}-10 \\\\ \\\\\r\n&=&x+10-10 \\\\ \\\\\r\n&=&x\r\n\\end{array}\\]\r\n\r\nThese two functions are inverses of each other. If you had tested \\(f(g(x))\\), you would have gotten the same result, \\(x\\).\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 11.3.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nAre the functions \\(f(x) = (3x + 4)^{\\frac{1}{3}} and \\(g(x) = \\dfrac{x^3 - 4}{3}\\) inverses of each other?\r\n\r\nTest if either \\(g(f(x)) = x\\) or \\(f(g(x)) = x\\).\r\n\r\nFor this problem, it would be easier to work with \\(g(f(x)) = x\\), since \\(x^3\\) will cancel out the radical in the \\(f(x)\\).\r\n\r\n\\[\\begin{array}{rrl}\r\ng(f(x))&=&[(3x+4)^{\\frac{1}{3}}]^3-4 \\\\ \\\\\r\ng(f(x))&=&\\dfrac{3x+4-4}{3} \\\\ \\\\\r\ng(f(x))&=&\\dfrac{3x}{3} \\\\ \\\\\r\ng(f(x))&=&x\r\n\\end{array}\\]\r\n\r\nThese functions are inverses of each other.\r\n\r\n<\/div>\r\n<\/div>\r\nOne of the strategies that is used to find the inverse of another function involves the substitution of the \\(x\\) and \\(y\\) variables of an equation. This is shown in the next few examples.\r\n
\r\n

Example 11.3.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the inverse function of \\(y = x^3 - 8\\).\r\n\r\nThe inverse function is found by substituting \\(y\\) for all \\(x\\) values and \\(x\\) for all \\(y\\) values in the original equation and then isolating for \\(y\\).\r\n\r\nFrom the equation \\(y = x^3 - 8\\), you now get \\(x = y^3 - 8\\).\r\n\r\nIsolating for \\(y\\) yields \\(y^3 = x + 8\\), which simplifies to \\(y = (x + 8)^{\\frac{1}{3}}\\).\r\n\r\nThese equations can be also written as \\(f(x) = x^3 - 8\\) and \\(f^{-1}(x) = (x + 8)^{\\frac{1}{3}}\\).\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 11.3.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the inverse function of \\(f(x) = (x + 4)^3 - 2\\).\r\n\r\n\\[\\begin{array}{rrl}\r\nx&=&(f^{-1}(x) + 4)^3 - 2 \\\\\r\nx+2&=&(f^{-1}(x)+4)^3 \\\\\r\n(x+2)^{\\frac{1}{3}}&=&f^{-1}(x)+4 \\\\\r\nf^{-1}(x)&=&(x+2)^{\\frac{1}{3}}-4\r\n\\end{array}\\]\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nState if the given functions are inverses.\r\n
    \r\n \t
  1. \\(g(x)=-x^5-3\\text{ and }f(x)=\\sqrt[5]{-x-3}\\)<\/li>\r\n \t
  2. \\(g(x)=4-x\\text{ and }f(x)=\\dfrac{4}{x}\\)<\/li>\r\n \t
  3. \\(g(x)=-10x+5\\text{ and }f(x)=\\dfrac{x-5}{10}\\)<\/li>\r\n \t
  4. \\(f(x)=\\dfrac{x-5}{10}\\text{ and }h(x)=10x+5\\)<\/li>\r\n \t
  5. \\(f(x)=\\dfrac{-2}{x+3}\\text{ and }g(x)=\\dfrac{3x+2}{x+2}\\)<\/li>\r\n \t
  6. \\(f(x)=\\dfrac{-x-1}{x-2}\\text{ and }g(x)=\\dfrac{-2x+1}{-x-1}\\)<\/li>\r\n<\/ol>\r\nFor questions 7 to 22, find the inverse of each function.\r\n
      \r\n \t
    1. \\(f(x)=(x-2)^5+3\\)<\/li>\r\n \t
    2. \\(g(x)=\\sqrt[3]{x+1}+2\\)<\/li>\r\n \t
    3. \\(g(x)=\\dfrac{4}{x+2}\\)<\/li>\r\n \t
    4. \\(f(x)=\\dfrac{-3}{x-3}\\)<\/li>\r\n \t
    5. \\(f(x)=\\dfrac{-2x-2}{x+2}\\)<\/li>\r\n \t
    6. \\(g(x)=\\dfrac{9+x}{3}\\)<\/li>\r\n \t
    7. \\(f(x)=\\dfrac{10-x}{5}\\)<\/li>\r\n \t
    8. \\(f(x)=\\dfrac{5x-15}{2}\\)<\/li>\r\n \t
    9. \\(g(x)=-(x-1)^3\\)<\/li>\r\n \t
    10. \\(f(x)=\\dfrac{12-3x}{4}\\)<\/li>\r\n \t
    11. \\(f(x)=(x-3)^3\\)<\/li>\r\n \t
    12. \\(g(x)=\\sqrt[5]{-x}+2\\)<\/li>\r\n \t
    13. \\(g(x)=\\dfrac{x}{x-1}\\)<\/li>\r\n \t
    14. \\(f(x)=\\dfrac{-3-2x}{x+3}\\)<\/li>\r\n \t
    15. \\(f(x)=\\dfrac{x-1}{x+1}\\)<\/li>\r\n \t
    16. \\(h(x)=\\dfrac{x}{x+2}\\)<\/li>\r\n<\/ol>\r\nAnswer Key 11.3<\/a>","rendered":"

      When working with mathematical functions, it sometimes becomes useful to undo what the original function does. To do this, you need to find the inverse of the function. This feature is commonly used in exponents and logarithms and in trigonometry.<\/p>\n

      In this topic, you will be looking at functions and seeing if they can be inverses of themselves. The notation used for this procedure is \"f^{-1}(x)\" is the inverse of \"f(x)\". In practice, this works as follows:<\/p>\n

        <\/span>   <\/span>\"\[f^{-1}[f(x)] = x\]\"<\/p>\n

      This is a very useful tool used many times over in math. If there are two functions \"f(x)\" and \"g(x)\" that are inverses of each other (if their composites \u201cundo\u201d each other\u2019s function), their composite functions look like:<\/p>\n

        <\/span>   <\/span>\"\[g(f(x)) = x\text{ and }f(g(x)) = x\]\"<\/p>\n

      \n
      \n

      Example 11.3.1<\/p>\n<\/header>\n

      \n

      Are the functions \"f(x) = 2x + 20\" and \"g(x) = \dfrac{x}{2} - 10\" inverses of each other?<\/p>\n

      Test if either \"g(f(x)) = x\" or \"f(g(x)) = x\":<\/p>\n

        <\/span>   <\/span>\"\[\begin{array}{rrl}<\/p>\n

      These two functions are inverses of each other. If you had tested \"f(g(x))\", you would have gotten the same result, \"x\".<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      Example 11.3.2<\/p>\n<\/header>\n

      \n

      Are the functions \"f(x) = (3x + 4)^{\frac{1}{3}} and \(g(x) = \dfrac{x^3 - 4}{3}\" inverses of each other?<\/p>\n

      Test if either \"g(f(x)) = x\" or \"f(g(x)) = x\".<\/p>\n

      For this problem, it would be easier to work with \"g(f(x)) = x\", since \"x^3\" will cancel out the radical in the \"f(x)\".<\/p>\n

        <\/span>   <\/span>\"\[\begin{array}{rrl}<\/p>\n

      These functions are inverses of each other.<\/p>\n<\/div>\n<\/div>\n

      One of the strategies that is used to find the inverse of another function involves the substitution of the \"x\" and \"y\" variables of an equation. This is shown in the next few examples.<\/p>\n

      \n
      \n

      Example 11.3.3<\/p>\n<\/header>\n

      \n

      Find the inverse function of \"y = x^3 - 8\".<\/p>\n

      The inverse function is found by substituting \"y\" for all \"x\" values and \"x\" for all \"y\" values in the original equation and then isolating for \"y\".<\/p>\n

      From the equation \"y = x^3 - 8\", you now get \"x = y^3 - 8\".<\/p>\n

      Isolating for \"y\" yields \"y^3 = x + 8\", which simplifies to \"y = (x + 8)^{\frac{1}{3}}\".<\/p>\n

      These equations can be also written as \"f(x) = x^3 - 8\" and \"f^{-1}(x) = (x + 8)^{\frac{1}{3}}\".<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      Example 11.3.4<\/p>\n<\/header>\n

      \n

      Find the inverse function of \"f(x) = (x + 4)^3 - 2\".<\/p>\n

        <\/span>   <\/span>\"\[\begin{array}{rrl}<\/p>\n<\/div>\n<\/div>\n

      Questions<\/h1>\n

      State if the given functions are inverses.<\/p>\n

        \n
      1. \"g(x)=-x^5-3\text{ and }f(x)=\sqrt[5]{-x-3}\"<\/li>\n
      2. \"g(x)=4-x\text{ and }f(x)=\dfrac{4}{x}\"<\/li>\n
      3. \"g(x)=-10x+5\text{ and }f(x)=\dfrac{x-5}{10}\"<\/li>\n
      4. \"f(x)=\dfrac{x-5}{10}\text{ and }h(x)=10x+5\"<\/li>\n
      5. \"f(x)=\dfrac{-2}{x+3}\text{ and }g(x)=\dfrac{3x+2}{x+2}\"<\/li>\n
      6. \"f(x)=\dfrac{-x-1}{x-2}\text{ and }g(x)=\dfrac{-2x+1}{-x-1}\"<\/li>\n<\/ol>\n

        For questions 7 to 22, find the inverse of each function.<\/p>\n

          \n
        1. \"f(x)=(x-2)^5+3\"<\/li>\n
        2. \"g(x)=\sqrt[3]{x+1}+2\"<\/li>\n
        3. \"g(x)=\dfrac{4}{x+2}\"<\/li>\n
        4. \"f(x)=\dfrac{-3}{x-3}\"<\/li>\n
        5. \"f(x)=\dfrac{-2x-2}{x+2}\"<\/li>\n
        6. \"g(x)=\dfrac{9+x}{3}\"<\/li>\n
        7. \"f(x)=\dfrac{10-x}{5}\"<\/li>\n
        8. \"f(x)=\dfrac{5x-15}{2}\"<\/li>\n
        9. \"g(x)=-(x-1)^3\"<\/li>\n
        10. \"f(x)=\dfrac{12-3x}{4}\"<\/li>\n
        11. \"f(x)=(x-3)^3\"<\/li>\n
        12. \"g(x)=\sqrt[5]{-x}+2\"<\/li>\n
        13. \"g(x)=\dfrac{x}{x-1}\"<\/li>\n
        14. \"f(x)=\dfrac{-3-2x}{x+3}\"<\/li>\n
        15. \"f(x)=\dfrac{x-1}{x+1}\"<\/li>\n
        16. \"h(x)=\dfrac{x}{x+2}\"<\/li>\n<\/ol>\n

          Answer Key 11.3<\/a><\/p>\n","protected":false},"author":540,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-775","chapter","type-chapter","status-publish","hentry"],"part":399,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":7,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/775\/revisions"}],"predecessor-version":[{"id":3665,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/775\/revisions\/3665"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/399"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/775\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=775"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=775"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=775"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}