Inverse Functions - Inverse of Linear Functions

(How to find an Inverse Function Part 1)


Given a linear function: \[f(x) = ax+b\] we now learn about how to find its inverse function: \[f^{-1}(x)\]

We've see what an inverse function is and we've seen that a function \(f(x)\) only has an inverse if it is a one-to-one mapping.
A linear functions, \(f(x) = ax+b\) is represented by a line with equation \(y = ax+b\), which passes the horizontal line test and is definitely a one-to-one mapping; linear functions therefore have an inverse.


What we'll learn here

The content of this page, and what we'll learn here, can be summarized as follows:

  • We start by making a note of the two-step method for finding an inverse function.
  • We then watch a couple of detailed tutorials for finding any linear function's inverse function.
  • We work through an exercise that consists of finding the inverse of several linear functions.

Method

The method for finding a function's inverse can be summarized in two steps:

  • Step 1: rearrange the expression \(y=f(x)\) to make \(x\) the subject.
    By the end of this you sould have an expression looking like \(x = f(y)\).
  • Step 2: swap \(x\) and \(y\) in the expression obtained at the end of Step 1, the expression obtained is \(y=f^{-1}(x)\).

Tutorial 1

In this tutorial, we show how to use our two-step method for finding the inverse function of a linear function, \(f(x) = ax+b\).

In particular we find the inverse function of the following two functions: \[f(x) = 3x \quad \text{and} \quad f(x) = 2x+8\]





Example 1

Given the function defined by: \[f(x) = 2x+4\] find an expression for its inverse function \(f^{-1}(x)\).

Solution

We follow our two-step method:

  • Step 1: We rearrange \(y=f(x)\) to make \(x\) the subject.
    In other words we rearrange \(y = 2x+4\).
    This is done here: \[\begin{aligned} & y = 2x + 4 \\ & y - 4 = 2x \\ & \frac{y-4}{2} = x \\ & \frac{y}{2} - 2 = x \\ & x = \frac{y}{2} - 2 \end{aligned}\]
  • Step 2: swap \(x\) and \(y\). The expression obtained will then be \(y = f^{-1}(x)\).

    Swapping \(x\) and \(y\) in \(x = \frac{y}{2} - 2\) leads to: \[y = \frac{x}{2} - 2\] Since this corresponds to \(y = f^{-1}(x)\), we can state that the inverse function is: \[f^{-1}(x) = \frac{x}{2} - 2 \]

Tutorial 2

In this tutorial, we show how to use our two-step method for finding the inverse function of a linear function, \(f(x) = \frac{x}{a}+b\).

In particular we find the inverse function of the following two functions: \[f(x) = \frac{x}{3} \quad \text{and} \quad f(x) = \frac{x}{5} + 1 \]





Exercise 1

Find the inverse function for each of the following functions:

  1. \(f(x) = 2x-8\)
  2. \(f(x) = \frac{x}{2}+3\)
  3. \(f(x) = \frac{3}{x}\)
  4. \(f(x) = \sqrt[3]{x}+2\)
  5. \(f(x) = \sqrt{x}-2\)
  6. \(f(x) = -3x+9\)
  7. \(f(x) = 6 - \frac{x}{3}\)
  8. \(f(x) = 4-2\sqrt{x}\)

Answers Without Working

  1. For \(f(x) = 2x-8\) we find: \[f^{-1}(x) = \frac{x+8}{2}\]
  2. For \(f(x) = \frac{x}{2}+3\) we find: \[f^{-1}(x) = 2x-6\]
  3. For \(f(x) = \frac{3}{x}\) we find: \[f^{-1}(x) = \frac{3}{x}\]
  4. For \(f(x) = \sqrt[3]{x}+2\) we find: \[f^{-1}(x) = \begin{pmatrix}x - 2 \end{pmatrix}^3\]
    we can also write:
    \[f^{-1}(x) = x^3-6x^2+12x-8\]
  5. For \(f(x) = \sqrt{x}-2\) we find: \[f^{-1}(x) = x^2+2x+1\]
  6. For \(f(x) = -3x+9\) we find: \[f^{-1}(x) = \frac{9-x}{3}\]
  7. For \(f(x) = 6 - \frac{x}{3}\) we find: \[f^{-1}(x) = 18-3x\]
  8. For \(f(x) = 4-2\sqrt{x}\) we find: \[f^{-1}(x) = \frac{x^2-8x+16}{4}\]


Finding the Inverse of Rational Functions

We now learn how to find the inverse of a rational function, of the form: \[f(x) = \frac{ax+b}{cx+d}\] For example, we'll know how to find the inverse function of: \[f(x) = \frac{2x+5}{x-3}\]

Exercise 2

Find the inverse function for each of the following functions:

  1. \(f(x) = \frac{3}{x} - 2\)
  2. \(f(x) = \frac{5}{x+2}\)
  3. \(f(x) = \frac{1}{2x-4}+2\)
  4. \(f(x) = \frac{3x}{2x-1}\)
  5. \(f(x) = \frac{2x+5}{x-3}+1\)
  6. \(f(x) = \frac{5}{2x-4}- 3\)
  7. \(f(x) = \frac{x-3}{x+1}\)
  8. \(f(x) = \frac{2x}{3x+6}\)

Answers Without Working

  1. For \(f(x) = \frac{3}{x} - 2\) we find: \[f^{-1}(x) = \frac{3}{x+2}\]
  2. For \(f(x) = \frac{5}{x+2}\) we find: \[f^{-1}(x) = \frac{5-2x}{x}\]
  3. For \(f(x) = \frac{1}{2x-4}+2\) we find: \[f^{-1}(x) = \frac{4x-7}{2x-4}\]
  4. For \(f(x) = \frac{3x}{2x-1}\) we find: \[f^{-1}(x) = \frac{x}{2x-3}\]
  5. For \(f(x) = \frac{2x+5}{x-3}+1\) we find: \[f^{-1}(x) = \frac{2+3x}{x-3}\]
  6. For \(f(x) = \frac{5}{2x-4}- 3\) we find: \[f^{-1}(x) = \frac{4x+17}{2x-6}\]
  7. For \(f(x) = \frac{x-3}{x+1}\) we find: \[f^{-1}(x) = \frac{-3-x}{x-1}\]
    better written as:
    \[f^{-1}(x) = \frac{3+x}{1-x}\]
  8. For \(f(x) = \frac{2x}{3x+6}\) we find: \[f^{-1}(x) = \frac{-6x}{3x-2}\]
    better written as:
    \[f^{-1}(x) = \frac{6x}{2-3x}\]

A Must-Know Example: Inverse of Quadratic Functions

(when we have to choose between two inverses)


At times finding the inverse function, \(f^{-1}(x)\), won't be quite as obvious.

In exams we'll often be asked to find the inverse function of a quadratic function, for which we're told \(x\geq p\) or \(x\leq q\), where \(p\) and \(q\) could be any two real numbers.

Say we're given the function \(f(x)=x^2\), for \(x\geq 0\), and we're asked to find \(f^{-1}(x)\).

Following our twp-step method for finding the inverse leads to:

  • Step 1: starting from \(y=x^2\), we make \(x\) the subject: \[\begin{aligned} & y = x^2 \\ & \pm \sqrt{y} = x \\ & x = \pm \sqrt{y} \end{aligned}\] We're faced with two options, either:
    • \(x = -\sqrt{y}\), or
    • \(x = \sqrt{y}\).
    To know which of the two, we need to remember that we were told that \(f(x)\) was to be considered for \(x\geq 0\); in other words we have to look back at the domain of \(f(x)\).

    This allows us to eliminate the option \(x = -\sqrt{y}\) since that will always be negative.
    Conseuqnetly we choose: \[x = \sqrt{y}\]
  • Step 2: swap \(x\) and \(y\), the expression obtained is \(y=f^{-1}(x)\): \[y = \sqrt{x}\] Since this is \(y=f^{-1}(x)\), we can state that the function's inverse function is: \[f^{-1}(x) = \sqrt{x}\]

Method: Inverse Function of a Quadratic

  • Step 1: rearrange the equation to make \(x\) the subject.
    By the end of this step you should have an expression looking like:
  • Step 2: use the domain of \(f(x)\) to choose wether to replace the \(\pm \) by \(+\) or \(-\).
  • Step 3: swap \(x\) and \(y\).
    The expression obtained at the end of this step is: \[y = f^{-1}(x)\]

Exercise 3

Find the inverse function

  1. \(f(x)=x^2 - 9\) for \(x \geq 0\).
  2. \(f(x) = 2x^2 - 8\) for \( x \leq 0\).
  3. \(f(x) = x^2 - 4x+3\) for \(x \geq 2\).
  4. \(f(x) = 2x^2+4x - 2\) for \(x \geq -1\).
  5. \(f(x) = -2x^2+12x - 16\) for \(x\leq 3\).
  6. \(f(x) = 3x^2-6x-9\) for \(x \geq 1\).
  7. \(f(x) = -4x^2-16x-8\) for \(x \leq -2\).
  8. \(f(x) = 2x^2 - 16x+24\) for \(x \geq 4\).

Answers Without Working

  1. For \(f(x)=x^2 - 9\), defined for \(x \geq 0\), we find: \[f^{-1}(x) = \sqrt{x+9}\]
  2. For \(f(x) = 2x^2 - 8\), defined for \( x \leq 0\), we find: \[f^{-1}(x) = - \sqrt{\frac{x+8}{2}}\]
  3. For \(f(x) = x^2 - 4x+3\), defined for \(x \geq 2\), we find: \[f^{-1}(x) = 2 + \sqrt{x+1}\]
  4. For \(f(x) = 2x^2+4x - 2\), defined for \(x \geq -1\), we find: \[f^{-1}(x) = -1 +\sqrt{\frac{x+4}{2}}\]
  5. For \(f(x) = -2x^2+12x - 16\), defined for \(x\leq 3\), we find: \[f^{-1}(x) = 3 - \sqrt{\frac{2-x}{2}}\]
  6. For \(f(x) = 3x^2 - 6x - 9\), defined for \(x \geq 1\), we find: \[f^{-1}(x) = 1+\sqrt{\frac{x+12}{3}}\]
  7. For \(f(x) = -4x^2 - 16x - 8\), defined for \(x \leq -2\), we find: \[f^{-1}(x) = -2 - \sqrt{\frac{8-x}{4}}\]
  8. For \(f(x) = 2x^2 - 16x + 24\), defined for \(x \geq 4\), we find: \[f^{-1}(x) = 4 + \sqrt{\frac{x+8}{2}}\]