{"id":926,"date":"2025-07-16T18:19:28","date_gmt":"2025-07-16T18:19:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=926"},"modified":"2026-01-13T21:00:33","modified_gmt":"2026-01-13T21:00:33","slug":"polynomial-equations-and-inequalities","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/polynomial-equations-and-inequalities\/","title":{"raw":"Polynomial Equations and Inequalities: Learn It 2","rendered":"Polynomial Equations and Inequalities: Learn It 2"},"content":{"raw":"

Inverse of Polynomial Functions<\/h2>\r\n

\"GravelA mound of gravel is in the shape of a cone with the height equal to twice the radius.<\/span><\/p>\r\n

The volume is found using a formula from elementary geometry.<\/p>\r\n\r\n

[latex]\\begin{align}V&=\\frac{1}{3}\\pi {r}^{2}h \\\\ &=\\frac{1}{3}\\pi {r}^{2}\\left(2r\\right) \\\\ &=\\frac{2}{3}\\pi {r}^{3} \\end{align}[\/latex]<\/div>\r\n \r\n

We have written the volume V<\/em>\u00a0in terms of the radius r<\/em>. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What is the radius of the new cone? To answer this question, we use the formula<\/p>\r\n\r\n

[latex]r=\\sqrt[3]{\\dfrac{3V}{2\\pi }\\\\}[\/latex]<\/div>\r\n

This function is the inverse of the formula for V<\/em>\u00a0in terms of r<\/em>.<\/p>\r\n\r\n

Find the inverse of a polynomial function<\/h3>\r\n
Recall that two functions [latex]f[\/latex] and [latex]g[\/latex] are inverse functions if for every coordinate pair in [latex]f[\/latex], [latex](a, b)[\/latex], there exists a corresponding coordinate pair in the inverse function, [latex]g[\/latex], [latex](b, a)[\/latex]. In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses that are also functions. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.<\/section>While it is not possible to find an inverse function of most polynomial functions, some basic polynomials do have inverses that are functions. Such functions are called invertible functions<\/strong>, and we use the notation [latex]{f}^{-1}\\left(x\\right)[\/latex].\r\n\r\n
Warning<\/strong>: [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 [latex]\u20131[\/latex] 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].<\/section>An important relationship between inverse functions is that they \"undo\" each other. If [latex]{f}^{-1}[\/latex] is the inverse of a function [latex]f[\/latex],\u00a0then [latex]f[\/latex]\u00a0is the inverse of the function [latex]{f}^{-1}[\/latex]. In other words, whatever the function [latex]f[\/latex]\u00a0does to [latex]x[\/latex], [latex]{f}^{-1}[\/latex] undoes it\u2014and vice-versa. More formally, we write\r\n

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

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

How To: Given a polynomial function, find the inverse of the function\u00a0<\/strong>\r\n
    \r\n \t
  1. Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\r\n \t
  2. Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\r\n \t
  3. Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\r\n \t
  4. Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].[reveal-answer q=\"289537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"289537\"]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 [latex]x[\/latex].\r\n

    [latex]\\begin{align}y&=5{x}^{3}+1 \\\\[1mm] x&=5{y}^{3}+1 \\\\[1mm] x - 1&=5{y}^{3} \\\\[1mm] \\dfrac{x - 1}{5}&={y}^{3} \\\\[4mm] {f}^{-1}\\left(x\\right)&=\\sqrt[3]{\\dfrac{x - 1}{5}} \\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nLook at the graph of [latex]f[\/latex] and [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.\r\n\r\nAlso, since the method involved interchanging [latex]x[\/latex]\u00a0and [latex]y[\/latex], notice corresponding points. If [latex]\\left(a,b\\right)[\/latex] is on the graph of [latex]f[\/latex], 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 [latex]f[\/latex], 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 [latex]f[\/latex], then [latex]\\left(6,1\\right)[\/latex] is on the graph of [latex]{f}^{-1}[\/latex].\r\n\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm_question hide_question_numbers=1]318905[\/ohm_question]<\/section>","rendered":"

    Inverse of Polynomial Functions<\/h2>\n

    \"GravelA mound of gravel is in the shape of a cone with the height equal to twice the radius.<\/span><\/p>\n

    The volume is found using a formula from elementary geometry.<\/p>\n

    [latex]\\begin{align}V&=\\frac{1}{3}\\pi {r}^{2}h \\\\ &=\\frac{1}{3}\\pi {r}^{2}\\left(2r\\right) \\\\ &=\\frac{2}{3}\\pi {r}^{3} \\end{align}[\/latex]<\/div>\n

     <\/p>\n

    We have written the volume V<\/em>\u00a0in terms of the radius r<\/em>. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What is the radius of the new cone? To answer this question, we use the formula<\/p>\n

    [latex]r=\\sqrt[3]{\\dfrac{3V}{2\\pi }\\\\}[\/latex]<\/div>\n

    This function is the inverse of the formula for V<\/em>\u00a0in terms of r<\/em>.<\/p>\n

    Find the inverse of a polynomial function<\/h3>\n
    Recall that two functions [latex]f[\/latex] and [latex]g[\/latex] are inverse functions if for every coordinate pair in [latex]f[\/latex], [latex](a, b)[\/latex], there exists a corresponding coordinate pair in the inverse function, [latex]g[\/latex], [latex](b, a)[\/latex]. In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses that are also functions. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.<\/section>\n

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

    Warning<\/strong>: [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 [latex]\u20131[\/latex] 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].<\/section>\n

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

    How To: Given a polynomial function, find the inverse of the function\u00a0<\/strong><\/p>\n
      \n
    1. Verify that\u00a0[latex]f[\/latex] is a one-to-one function.<\/li>\n
    2. Replace [latex]f\\left(x\\right)[\/latex] with [latex]y[\/latex].<\/li>\n
    3. Interchange [latex]x[\/latex]\u00a0and [latex]y[\/latex].<\/li>\n
    4. Solve for [latex]y[\/latex], and rename the function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\n
      Find the inverse of the function [latex]f\\left(x\\right)=5{x}^{3}+1[\/latex].<\/p>\n