# 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: $f^{-1}(x) = 4 + \sqrt{\frac{x+8}{2}}$