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 $y = \arcsin (x)$