Lesson Objectives

- Demonstrate an understanding of inverse functions
- Learn how to determine if two functions are inverses

## How to Determine if Two Functions are Inverses

How can we determine if two functions are inverses? We previously stated that the domain of a function becomes the range of its inverse and the range of a function becomes the domain of its inverse. Let's take a simple function such as: $$f(x)=2x$$ This function maps an x-value of 5 to a y-value of 10. $$f(5)=2(5)=10$$ If we find the inverse of the function, it will map an x-value of 10 to a y-value of 5. $$f^{-1}(x)=\frac{x}{2}$$ $$f^{-1}(10)=\frac{10}{2}=5$$ We can think about the inverse as reversing the procedure of the original function. In the original function, we multiply 2 by an unknown number (x). In the inverse, we take the unknown number (x) and divide by 2. We know that multiplying by 2 and dividing by 2 are opposite operations and will cancel each other out. What happens if we plug f

For inverses f and f

We can determine if two functions are inverses by checking the composition of the functions and verifying that their domains and ranges match appropriately. In other words, we can prove that two functions are inverses by plugging the inverse function in for x in the original function. If they are inverses, the result should just be x. We also need to check the other scenario. We will plug in the original function for x in the inverse function. Again, if they are inverses, the result should just be x.

We can also write this with different notation. We can use f and g, instead of f and f

Let's look at a few examples.

Example 1: Determine if the functions f and g are inverses. $$f(x)=\sqrt[5]{x + 1}$$ $$g(x)=x^5 - 1$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=\sqrt[5]{(x^5 - 1) + 1}$$ $$f(g(x))=\sqrt[5]{x^5}$$ $$f(g(x))=x$$ Now we will plug f(x) in for x in g(x). $$g(f(x))=(\sqrt[5]{x + 1})^5 - 1$$ $$g(f(x))=x + 1 - 1$$ $$g(f(x))=x$$ In each case, we get x as a result. We can state that these two functions are inverses.

Example 2: Determine if the functions f and g are inverses. $$f(x)=3x - 7$$ $$g(x)=\frac{x + 3}{7}$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=3\left(\frac{x + 3}{7}\right) - 7$$ $$f(g(x))=\frac{3x + 9}{7}- 7$$ $$f(g(x))=\frac{3x + 9}{7}- \frac{49}{7}$$ $$f(g(x))=\frac{3x + 9 - 49}{7}$$ $$f(g(x))=\frac{3x - 40}{7}$$ Notice that the result here is not x. We can stop and state that these two functions are not inverses.

Example 3: Determine if the functions f and g are inverses. $$f(x)=\frac{4}{x}- 1$$ $$g(x)=\frac{4}{x + 1}$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=\large{\frac{\frac{4}{1}}{\frac{4}{x + 1}}}- 1$$ $$f(g(x))=\frac{4}{1}\cdot \frac{x + 1}{4}- 1$$ $$f(g(x))=\frac{\cancel{4}}{1}\cdot \frac{x + 1}{\cancel{4}}- 1$$ $$f(g(x))=x + 1 - 1$$ $$f(g(x))=x$$ Now we will plug f(x) in for x in g(x). $$g(f(x))=\frac{4}{\large{\frac{4}{x}}- 1 + 1}$$ $$g(f(x))=\large{\frac{\frac{4}{1}}{\frac{4}{x}}}$$ $$g(f(x))=\frac{\cancel{4}}{1}\cdot \frac{x}{\cancel{4}}$$ $$g(f(x))=x$$ In each case, we get x as a result. We can state that these two functions are inverses.

^{-1}(x) in for x in the original function? $$f(f^{-1}(x))=2\left(\frac{x}{2}\right)$$ $$\require{cancel}f(f^{-1}(x))=\cancel{2}\left(\frac{x}{\cancel{2}}\right)=x$$ Additionally, let's plug f(x) in for x in f^{-1}(x). $$f^{-1}(f(x))=\frac{2x}{2}$$ $$f^{-1}(f(x))=\frac{\cancel{2}x}{\cancel{2}}=x$$ Based on our example, we can state the following rule:For inverses f and f

^{-1}: $$f(f^{-1}(x))=x$$ For every x in the domain of f^{-1}. $$f^{-1}(f(x))=x$$ For every x in the domain of f.We can determine if two functions are inverses by checking the composition of the functions and verifying that their domains and ranges match appropriately. In other words, we can prove that two functions are inverses by plugging the inverse function in for x in the original function. If they are inverses, the result should just be x. We also need to check the other scenario. We will plug in the original function for x in the inverse function. Again, if they are inverses, the result should just be x.

We can also write this with different notation. We can use f and g, instead of f and f

^{-1}. If f(x) and g(x) are inverses: $$f(g(x)) = x$$ For every x in the domain of g. $$g(f(x)) = x$$ For every x in the domain of f.Let's look at a few examples.

Example 1: Determine if the functions f and g are inverses. $$f(x)=\sqrt[5]{x + 1}$$ $$g(x)=x^5 - 1$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=\sqrt[5]{(x^5 - 1) + 1}$$ $$f(g(x))=\sqrt[5]{x^5}$$ $$f(g(x))=x$$ Now we will plug f(x) in for x in g(x). $$g(f(x))=(\sqrt[5]{x + 1})^5 - 1$$ $$g(f(x))=x + 1 - 1$$ $$g(f(x))=x$$ In each case, we get x as a result. We can state that these two functions are inverses.

Example 2: Determine if the functions f and g are inverses. $$f(x)=3x - 7$$ $$g(x)=\frac{x + 3}{7}$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=3\left(\frac{x + 3}{7}\right) - 7$$ $$f(g(x))=\frac{3x + 9}{7}- 7$$ $$f(g(x))=\frac{3x + 9}{7}- \frac{49}{7}$$ $$f(g(x))=\frac{3x + 9 - 49}{7}$$ $$f(g(x))=\frac{3x - 40}{7}$$ Notice that the result here is not x. We can stop and state that these two functions are not inverses.

Example 3: Determine if the functions f and g are inverses. $$f(x)=\frac{4}{x}- 1$$ $$g(x)=\frac{4}{x + 1}$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=\large{\frac{\frac{4}{1}}{\frac{4}{x + 1}}}- 1$$ $$f(g(x))=\frac{4}{1}\cdot \frac{x + 1}{4}- 1$$ $$f(g(x))=\frac{\cancel{4}}{1}\cdot \frac{x + 1}{\cancel{4}}- 1$$ $$f(g(x))=x + 1 - 1$$ $$f(g(x))=x$$ Now we will plug f(x) in for x in g(x). $$g(f(x))=\frac{4}{\large{\frac{4}{x}}- 1 + 1}$$ $$g(f(x))=\large{\frac{\frac{4}{1}}{\frac{4}{x}}}$$ $$g(f(x))=\frac{\cancel{4}}{1}\cdot \frac{x}{\cancel{4}}$$ $$g(f(x))=x$$ In each case, we get x as a result. We can state that these two functions are inverses.

#### Skills Check:

Example #1

Determine if the two functions are inverses. $$f(x)=\sqrt[5]{x}- 2$$ $$g(x)=(x + 2)^5$$

Please choose the best answer.

A

Inverses

B

Not Inverses

Example #2

Determine if the two functions are inverses. $$f(x)=\frac{2}{3}x + \frac{2}{3}$$ $$g(x)=5x - 5$$

Please choose the best answer.

A

Inverses

B

Not Inverses

Example #3

Determine if the two functions are inverses. $$f(x)=\frac{1}{x + 1}$$ $$g(x)=\frac{1 - x}{x}$$

Please choose the best answer.

A

Inverses

B

Not Inverses

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test