# Chapter 6: Inverse Functions

#### Definition

A function ﻿$f(x)$﻿ is ONE-TO-ONE on a domain ﻿$D$﻿ if ﻿$f(x_1) \neq f(x_2)$﻿ Whenever ﻿$x_1 \neq x_2$﻿ in ﻿$D$﻿

Plainly ﻿$f(x)$﻿ is one-to-one if every ﻿$x$﻿ gets sent to a different value

#### Definition

The Horizontal Line Test - A function is one-to-one if and only if each horizontal line meets it graph at most once.

#### Discuss

Which of these functions is one-to-one ?

1. ﻿$y = |x|$﻿
2. ﻿$y= x^3$﻿
3. ﻿$y=3x+2$﻿
4. ﻿$y= \sin (x)$﻿
5. ﻿$y =e^x$﻿
6. ﻿$y= x^2$﻿

#### Definition

If ﻿$f(x)$﻿ is one-to-one on ﻿$D$﻿ and has range ﻿$R$﻿ then the INVERSE FUNCTION is defined by:

﻿$f^{-1}(b) = a$﻿ if ﻿$f(a) =b$﻿

Why do we need one-to-one?

If ﻿$f (1) = f(2) =3$﻿ then what is ﻿$f^{-1}(3)$﻿?

It could be ﻿$f^{-1}(3) = 1$﻿ or ﻿$f^{-1}(3) =2$﻿

For ﻿$f^{-1}(x)$﻿ to be a function we need one output

### The Inversion Procedure

To invert ﻿$f(x)$﻿

Solve ﻿$y= f(x)$﻿ for ﻿$x$﻿. Obtain ﻿$x = f^{ -1}(y)$﻿

Switch ﻿$x$﻿ and ﻿$y$﻿ to obtain ﻿$y = f^{-1}(x)$﻿.

#### Example

Invert the function ﻿$y = \frac{1}{2} x+ 1$﻿

﻿$y = \frac{1}{2} x +1 \Leftrightarrow 2y = x+2$﻿

﻿$\Leftrightarrow 2y -2 =x$﻿

Thus, ﻿$x= f^{-1}(y) = 2y-2$﻿

We obtain ﻿$f^{-1}(x) = 2x-2$﻿

﻿$*$﻿ Check your work

﻿$f(f^{-1}(x) = f(2x-2) = \frac{1}{2}(2x-2)+1$﻿

﻿$=$﻿ ﻿$x-1+1 =x$﻿

﻿$f^{-1}(f(x))=2 (\frac{1}{2} x +1) -2 = x +2 -2$﻿

﻿$=x$﻿

#### Example

Invert the function ﻿$y=x^2$﻿ on the domain ﻿$[0, \infty)$﻿.

﻿$*$﻿ ﻿$y=x^2$﻿ is not one-to-one on ﻿$(- \infty , \infty)$﻿ but it is on ﻿$[0, \infty)$﻿.

﻿$y = x ^2 \Rightarrow \sqrt{y} = x \Rightarrow x = f ^{-1} (y) = \sqrt{y}$﻿

We obtain ﻿$f ^{-1} (x) = \sqrt{x}$﻿

Check:

﻿$f(f^{-1}(x)) = (\sqrt{x})^2 = x$﻿

﻿$f^{-1}(f(x)) = \sqrt{x^2} = |x|=x$﻿

### Inverse Trigonometric Functions

Idea: The trig functions are NOT one-to-one.

﻿$0 = \sin (0) = \sin (2 \pi) = \sin (4 \pi) = ...$﻿

However, we want to invest them

### Domain Restrictions We get the following inverses:

﻿$y = \arcsin (x) \rightarrow x$﻿ in ﻿$[-1,1]$﻿

﻿$y = \arccos (x) \rightarrow x$﻿ in ﻿$[-1,1]$﻿

﻿$y = \arctan (x) \rightarrow x$﻿ in ﻿$( - \infty, \infty)$﻿

#### Discuss

What are the ranges of:

﻿$\arcsin x, \arccos x$﻿ and ﻿$\arctan x$﻿

#### Fun Fact #### Example

Calculate ﻿$\arcsin (\frac{\sqrt{3}}{2})$﻿

Find ﻿$\theta$﻿ in ﻿$[- \frac {\pi}{2}, \ \frac {\pi}{2}]$﻿ such that ﻿$\sin \theta = \frac {\sqrt{3}}{2}$﻿

We know ﻿$\sin \frac {\pi}{6} = \frac {\sqrt{3}}{2}$﻿ since Thus ﻿$\arcsin (\frac{\sqrt{3}}{2})= \frac{\pi}{6}$﻿

#### Example

Calculate ﻿$\arccos (-\frac{1}{2})$﻿

Find ﻿$\theta$﻿ in ﻿$[0,\pi]$﻿ such that ﻿$\cos \theta = - \frac{1}{2}$﻿

We know ﻿$\cos (\frac{2 \pi}{3}) = -\frac{1}{2}$﻿ since Thus , ﻿$\arccos(- \frac {1}{2}) = \frac {2 \pi}{3}$﻿

#### Example

Sketch ﻿$y = \arcsin (x)$﻿ 