Lesson Objectives
• Learn how to find the inverse of a domain restricted function

## How to Find the Inverse of a Domain Restricted Function

In this lesson, we want to learn how to find the inverse of a domain restricted function. Let's begin with an example.
Example #1: Find the inverse. $$f(x)=\sqrt{x - 9}$$ We know this function is a one-to-one function, therefore, it will have an inverse. What happens when we apply our normal procedure?
Step 1) Write f(x) as y: $$y=\sqrt{x - 9}$$ Step 2) Swap x and y: $$x=\sqrt{y - 9}$$ Step 3) Now solve for y: $$x=\sqrt{y - 9}$$ $$x^2=(\sqrt{y - 9})^2$$ $$x^2=y - 9$$ $$y=x^2 + 9$$ We know that y = x2 + 9 is not a one-to-one function. So what is the actual inverse? We are on the right track, however, we need to include a domain restriction. Let's go back to our original function f(x). $$f(x)=\sqrt{x - 9}$$ $$domain: [9, \infty)$$ $$range: [0, \infty)$$ Recall that for an inverse, we swap the domain and range. So the range will be the domain for the inverse. $$f^{-1}(x)=x^2 + 9, x ≥ 0$$

#### Skills Check:

Example #1

Find the inverse. $$f(x)=\sqrt{x - 11}$$

A
No Inverse
B
$$f^{-1}(x)=2x^2 - 11$$
C
$$f^{-1}(x)=x^2 - 11, x ≤ 0$$
D
$$f^{-1}(x)=x^2 + 11, x ≤ 0$$
E
$$f^{-1}(x)=x^2 + 11, x ≥ 0$$

Example #2

Find the inverse. $$f(x)=\frac{1}{\sqrt{x + 2}}$$

A
No Inverse
B
$$f^{-1}(x)=\frac{1}{x^2}- 2, x ≤ 0$$
C
$$f^{-1}(x)=\frac{1}{x^2}- 2, x < 0$$
D
$$f^{-1}(x)=\frac{1}{x^2}- 2, x > 0$$
E
$$f^{-1}(x)=\frac{1}{x^2}- 2, x ≥ 0$$

Example #3

Find the inverse. $$f(x)=x^2 + 4, x ≥ 0$$

A
No Inverse
B
$$f^{-1}(x)=\sqrt{x - 4}, x ≥ 0$$
C
$$f^{-1}(x)=-\sqrt{x - 4}, x ≤ 4$$
D
$$f^{-1}(x)=\sqrt{x - 4}, x ≥ 4$$
E
$$f^{-1}(x)=-\sqrt{x - 4}, x ≥ 4$$         