{"id":1465,"date":"2015-11-12T18:35:29","date_gmt":"2015-11-12T18:35:29","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1465"},"modified":"2017-04-03T14:45:41","modified_gmt":"2017-04-03T14:45:41","slug":"find-the-inverse-of-a-polynomial-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/find-the-inverse-of-a-polynomial-function\/","title":{"raw":"Find the inverse of a polynomial function","rendered":"Find the inverse of a polynomial function"},"content":{"raw":"

Two functions f<\/em>\u00a0and g<\/em>\u00a0are inverse functions if for every coordinate pair in f<\/em>, (a<\/em>, b<\/em>), there exists a corresponding coordinate pair in the inverse function, g<\/em>, (b<\/em>, a<\/em>). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.<\/p>\r\n

For a function to have an inverse function<\/strong> the function to create a new function that is one-to-one<\/strong> and would have an inverse function.<\/p>\r\n

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown\u00a0below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"DiagramFigure 2<\/b>[\/caption]\r\n

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x<\/em>\u00a0measured horizontally and y<\/em>\u00a0measured vertically, with the origin at the vertex of the parabola.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 3<\/b>[\/caption]\r\n

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\\left(x\\right)=a{x}^{2}[\/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a<\/em>.<\/p>\r\n\r\n

[latex]\\begin{cases} 18=a{6}^{2}\\hfill \\\\ a=\\frac{18}{36}\\hfill \\\\ \\text{ }=\\frac{1}{2}\\hfill \\end{cases}[\/latex]<\/div>\r\n

Our parabolic cross section has the equation<\/p>\r\n\r\n

[latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/div>\r\n

We are interested in the surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y<\/em>\u00a0the width will be given by 2x<\/em>, so we need to solve the equation above for x<\/em>\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\r\n

To find an inverse, we can restrict our original function to a limited domain on which it is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive x<\/em>\u00a0values. On this domain, we can find an inverse by solving for the input variable:<\/p>\r\n\r\n

[latex]\\begin{cases}y=\\frac{1}{2}{x}^{2}\\hfill \\\\ 2y={x}^{2}\\hfill \\\\ \\text{ }x=\\pm \\sqrt{2y}\\hfill \\end{cases}[\/latex]<\/div>\r\n

This is not a function as written. We are limiting ourselves to positive x<\/em>\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\r\n\r\n

[latex]y=\\frac{{x}^{2}}{2},\\text{ }x>0[\/latex]<\/div>\r\n

Because x<\/em>\u00a0is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x<\/em>. The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\r\n\r\n

[latex]\\begin{cases}\\text{Area} & =l\\cdot w\\hfill \\\\ \\text{ } & =36\\cdot 2x\\hfill \\\\ \\text{ } & =72x\\hfill \\\\ \\text{ } & =72\\sqrt{2y}\\hfill \\end{cases}[\/latex]<\/div>\r\n

This example illustrates two important points:<\/p>\r\n\r\n

  1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\r\n\t
  2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\r\n<\/ol>

    Functions involving roots are often called radical functions<\/strong>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n

    Warning: [latex]{f}^{-1}\\left(x\\right)[\/latex] is not the same as the reciprocal of the function [latex]f\\left(x\\right)[\/latex]. This use of \u20131 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\\left(x\\right)[\/latex], we would need to write [latex]{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}[\/latex].<\/p>\r\n

    An important relationship between inverse functions is that they \"undo\" each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function f<\/em>,\u00a0then f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function f<\/em>\u00a0does to x<\/em>, [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write<\/p>\r\n\r\n

    [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/div>\r\n

    and<\/p>\r\n\r\n

    [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/div>\r\n
    \r\n

    A General Note: Verifying Two Functions Are Inverses of One Another<\/h3>\r\n

    Two functions, f<\/em>\u00a0and g<\/i>, are inverses of one another if for all x<\/em>\u00a0in the domain of f\u00a0<\/em>and g<\/em>.<\/p>\r\n\r\n

    [latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.<\/h3>\r\n
    1. Replace [latex]f\\left(x\\right)[\/latex] with y<\/em>.<\/li>\r\n\t
    2. Interchange x<\/em>\u00a0and y<\/em>.<\/li>\r\n\t
    3. Solve for y<\/em>, and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol><\/div>\r\n
      \r\n
      \r\n
      \r\n

      Example 1: Verifying Inverse Functions<\/h3>\r\n

      Show that [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex] and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses, for [latex]x\\ne 0,-1[\/latex] .<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      Solution<\/h3>\r\n

      We must show that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex].<\/p>\r\n\r\n

      [latex]\\begin{cases}{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{1}{x+1}\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x+1}}-1\\hfill \\\\ \\text{ }=\\left(x+1\\right)-1\\hfill \\\\ \\text{ }=x\\hfill \\\\ f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(\\frac{1}{x}-1\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\left(\\frac{1}{x}-1\\right)+1}\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x}}\\hfill \\\\ \\text{ }=x\\hfill \\end{cases}[\/latex]<\/div>\r\n

      Therefore, [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      \r\n

      Try It 1<\/h3>\r\n

      Show that [latex]f\\left(x\\right)=\\frac{x+5}{3}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=3x - 5[\/latex] are inverses.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n
      \r\n

      Example 2: Finding the Inverse of a Cubic Function<\/h3>\r\n

      Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      Solution<\/h3>\r\n

      This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x<\/em>.<\/p>\r\n\r\n

      [latex]\\begin{cases}\\text{ }y=5{x}^{3}+1\\hfill \\\\ \\text{ }x=5{y}^{3}+1\\hfill \\\\ \\text{ }x - 1=5{y}^{3}\\hfill \\\\ \\text{ }\\frac{x - 1}{5}={y}^{3}\\hfill \\\\ {f}^{-1}\\left(x\\right)=\\sqrt[3]{\\frac{x - 1}{5}}\\hfill \\end{cases}[\/latex]<\/div>\r\n<\/div>\r\n
      \r\n

      Analysis of the Solution<\/h3>\r\n

      Look at the graph of f<\/em>\u00a0and [latex]{f}^{-1}[\/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[\/latex]. This is always the case when graphing a function and its inverse function.<\/p>\r\n

      Also, since the method involved interchanging x<\/em>\u00a0and y<\/em>, notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of f<\/em>, then [latex]\\left(b,a\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Since [latex]\\left(0,1\\right)[\/latex] is on the graph of f<\/em>, then [latex]\\left(1,0\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Similarly, since [latex]\\left(1,6\\right)[\/latex] is on the graph of f<\/em>, then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 4<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      \r\n

      Try It 2<\/h3>\r\n

      Find the inverse function of [latex]f\\left(x\\right)=\\sqrt[3]{x+4}[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

      Two functions f<\/em>\u00a0and g<\/em>\u00a0are inverse functions if for every coordinate pair in f<\/em>, (a<\/em>, b<\/em>), there exists a corresponding coordinate pair in the inverse function, g<\/em>, (b<\/em>, a<\/em>). In other words, the coordinate pairs of the inverse functions have the input and output interchanged.<\/p>\n

      For a function to have an inverse function<\/strong> the function to create a new function that is one-to-one<\/strong> and would have an inverse function.<\/p>\n

      For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown\u00a0below. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.<\/p>\n

      \"Diagram<\/p>\n

      Figure 2<\/b><\/p>\n<\/div>\n

      Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x<\/em>\u00a0measured horizontally and y<\/em>\u00a0measured vertically, with the origin at the vertex of the parabola.<\/p>\n

      \"Graph<\/p>\n

      Figure 3<\/b><\/p>\n<\/div>\n

      From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form [latex]y\\left(x\\right)=a{x}^{2}[\/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a<\/em>.<\/p>\n

      [latex]\\begin{cases} 18=a{6}^{2}\\hfill \\\\ a=\\frac{18}{36}\\hfill \\\\ \\text{ }=\\frac{1}{2}\\hfill \\end{cases}[\/latex]<\/div>\n

      Our parabolic cross section has the equation<\/p>\n

      [latex]y\\left(x\\right)=\\frac{1}{2}{x}^{2}[\/latex]<\/div>\n

      We are interested in the surface area<\/strong> of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y<\/em>\u00a0the width will be given by 2x<\/em>, so we need to solve the equation above for x<\/em>\u00a0and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.<\/p>\n

      To find an inverse, we can restrict our original function to a limited domain on which it is<\/em> one-to-one. In this case, it makes sense to restrict ourselves to positive x<\/em>\u00a0values. On this domain, we can find an inverse by solving for the input variable:<\/p>\n

      [latex]\\begin{cases}y=\\frac{1}{2}{x}^{2}\\hfill \\\\ 2y={x}^{2}\\hfill \\\\ \\text{ }x=\\pm \\sqrt{2y}\\hfill \\end{cases}[\/latex]<\/div>\n

      This is not a function as written. We are limiting ourselves to positive x<\/em>\u00a0values, so we eliminate the negative solution, giving us the inverse function we\u2019re looking for.<\/p>\n

      [latex]y=\\frac{{x}^{2}}{2},\\text{ }x>0[\/latex]<\/div>\n

      Because x<\/em>\u00a0is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x<\/em>. The trough is 3 feet (36 inches) long, so the surface area will then be:<\/p>\n

      [latex]\\begin{cases}\\text{Area} & =l\\cdot w\\hfill \\\\ \\text{ } & =36\\cdot 2x\\hfill \\\\ \\text{ } & =72x\\hfill \\\\ \\text{ } & =72\\sqrt{2y}\\hfill \\end{cases}[\/latex]<\/div>\n

      This example illustrates two important points:<\/p>\n

        \n
      1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.<\/li>\n
      2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.<\/li>\n<\/ol>\n

        Functions involving roots are often called radical functions<\/strong>. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n

        Warning: [latex]{f}^{-1}\\left(x\\right)[\/latex] is not the same as the reciprocal of the function [latex]f\\left(x\\right)[\/latex]. This use of \u20131 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\\left(x\\right)[\/latex], we would need to write [latex]{\\left(f\\left(x\\right)\\right)}^{-1}=\\frac{1}{f\\left(x\\right)}[\/latex].<\/p>\n

        An important relationship between inverse functions is that they “undo” each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function f<\/em>,\u00a0then f<\/em>\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function f<\/em>\u00a0does to x<\/em>, [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write<\/p>\n

        [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }f[\/latex]<\/div>\n

        and<\/p>\n

        [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x,\\text{for all }x\\text{ in the domain of }{f}^{-1}[\/latex]<\/div>\n
        \n

        A General Note: Verifying Two Functions Are Inverses of One Another<\/h3>\n

        Two functions, f<\/em>\u00a0and g<\/i>, are inverses of one another if for all x<\/em>\u00a0in the domain of f\u00a0<\/em>and g<\/em>.<\/p>\n

        [latex]g\\left(f\\left(x\\right)\\right)=f\\left(g\\left(x\\right)\\right)=x[\/latex]<\/div>\n<\/div>\n
        \n

        How To: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one.<\/h3>\n
          \n
        1. Replace [latex]f\\left(x\\right)[\/latex] with y<\/em>.<\/li>\n
        2. Interchange x<\/em>\u00a0and y<\/em>.<\/li>\n
        3. Solve for y<\/em>, and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n
          \n
          \n
          \n

          Example 1: Verifying Inverse Functions<\/h3>\n

          Show that [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex] and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses, for [latex]x\\ne 0,-1[\/latex] .<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

          We must show that [latex]{f}^{-1}\\left(f\\left(x\\right)\\right)=x[\/latex] and [latex]f\\left({f}^{-1}\\left(x\\right)\\right)=x[\/latex].<\/p>\n

          [latex]\\begin{cases}{f}^{-1}\\left(f\\left(x\\right)\\right)={f}^{-1}\\left(\\frac{1}{x+1}\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x+1}}-1\\hfill \\\\ \\text{ }=\\left(x+1\\right)-1\\hfill \\\\ \\text{ }=x\\hfill \\\\ f\\left({f}^{-1}\\left(x\\right)\\right)=f\\left(\\frac{1}{x}-1\\right)\\hfill \\\\ \\text{ }=\\frac{1}{\\left(\\frac{1}{x}-1\\right)+1}\\hfill \\\\ \\text{ }=\\frac{1}{\\frac{1}{x}}\\hfill \\\\ \\text{ }=x\\hfill \\end{cases}[\/latex]<\/div>\n

          Therefore, [latex]f\\left(x\\right)=\\frac{1}{x+1}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=\\frac{1}{x}-1[\/latex] are inverses.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          \n

          Try It 1<\/h3>\n

          Show that [latex]f\\left(x\\right)=\\frac{x+5}{3}[\/latex]\u00a0and [latex]{f}^{-1}\\left(x\\right)=3x - 5[\/latex] are inverses.<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n

          \n
          \n
          \n

          Example 2: Finding the Inverse of a Cubic Function<\/h3>\n

          Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

          This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x<\/em>.<\/p>\n

          [latex]\\begin{cases}\\text{ }y=5{x}^{3}+1\\hfill \\\\ \\text{ }x=5{y}^{3}+1\\hfill \\\\ \\text{ }x - 1=5{y}^{3}\\hfill \\\\ \\text{ }\\frac{x - 1}{5}={y}^{3}\\hfill \\\\ {f}^{-1}\\left(x\\right)=\\sqrt[3]{\\frac{x - 1}{5}}\\hfill \\end{cases}[\/latex]<\/div>\n<\/div>\n
          \n

          Analysis of the Solution<\/h3>\n

          Look at the graph of f<\/em>\u00a0and [latex]{f}^{-1}[\/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[\/latex]. This is always the case when graphing a function and its inverse function.<\/p>\n

          Also, since the method involved interchanging x<\/em>\u00a0and y<\/em>, notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of f<\/em>, then [latex]\\left(b,a\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Since [latex]\\left(0,1\\right)[\/latex] is on the graph of f<\/em>, then [latex]\\left(1,0\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex]. Similarly, since [latex]\\left(1,6\\right)[\/latex] is on the graph of f<\/em>, then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].<\/p>\n

          \"Graph<\/p>\n

          Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          \n

          Try It 2<\/h3>\n

          Find the inverse function of [latex]f\\left(x\\right)=\\sqrt[3]{x+4}[\/latex].<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

          \n\t\t\t

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