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# Functions and Inverses

## Undo the original function; reflections across y = x.

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Practice Functions and Inverses
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Inverses of Functions

Functions are commonly known as rules that take inputs and produce outputs. An inverse function does exactly the reverse, undoing what the original function does. How can you tell if two functions are inverses?

#### Guidance

A function is written as $f(x)$  and its inverse is written as $f^{-1} (x)$ . A common misconception is to see the -1 and interpret it as an exponent and write $\frac{1}{f(x)}$ , but this is not correct. Instead, $f^{-1} (x)$ should be viewed as a new function from the range of $f(x)$  back to the domain.

It is important to see the cycle that starts with $x$ , becomes  $y$ and then goes back to $x$ . In order for two functions to truly be inverses of each other, this cycle must hold algebraically.

$f(f^{-1} (x) )=x$ and $f^{-1} (f(x) )=x$

When given a function there are two steps to take to find its inverse. In the original function, first switch the variables  $x$ and $y$ . Next, solve the function for $y$ . This will give you the inverse function. After finding the inverse, it is important to check both directions of compositions to make sure that together the function and its inverse produce the value $x$ . In other words, verify that $f(f^{-1} (x) )=x$  and $f^{-1} (f(x) )=x$ .

Graphically, inverses are reflections across the line $y=x$ . Below you see inverses $y =e^x$  and $y= \ln x$ .  Notice how the  $(x, y)$ coordinates in one graph become  $(y, x)$ coordinates in the other graph.

In order to decide whether an inverse function is also actually a function you can use the vertical line test on the inverse function like usual. You can also use the horizontal line test on the original function. The horizontal line test is exactly like the vertical line test except the lines simply travel horizontally.

Example A

Find the inverse, then verify the inverse algebraically. $f(x)=y=(x+1)^2+4$

Solution: To find the inverse, switch  $x$ and  $y$ then solve for $y$

$x&=(y+1)^2+4\\x-4&=(y+1)^2\\\pm \sqrt{x-4}&=y+1\\-1 \pm \sqrt{x-4}&=y=f^{-1} (x)$

To verify algebraically, you must show $x= f(f^{-1} (x) )=f^{-1} (f(x) )$ :

$f(f^{-1} (x) ) & = f\left(-1 \pm \sqrt{x-4} ~ \right) \\ & = \big( (-1 \pm \sqrt{x-4} ~ )+1 \big)^2+4 \\ & = \left( \pm \sqrt{x-4} ~ \right)^2+4 \\ & = x-4+4 = x$

$f^{-1} \big( f(x) \big) & = f^{-1} \big( (x+1)^2+4 \big) \\ & =-1 \pm \sqrt{ \big( (x+1)^2+4 \big) - 4} \\ & =-1 \pm \sqrt{(x+1)^2} \\ & =-1+x+1=x$

As you can see from the graph, the $\pm$  causes the inverse to be a relation instead of a function. This can be observed in the graph because the original function does not pass the horizontal line test and the inverse does not pass the vertical line test

Example B

Find the inverse of the function and then verify that $x= f(f^{-1} (x) )=f^{-1} (f(x) )$ .

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

Solution: Sometimes it is quite challenging to switch  $x$ and  $y$ and then solve for $y$ . You must be careful with your algebra.

$x &= \frac{y+1}{y-1}\\x(y-1)&=y+1\\xy-x&=y+1\\xy-y&=x+1\\y(x-1)&=x+1\\y&= \frac{x+1}{x-1}$

This function turns out to be its own inverse. Since they are identical, you only need to show that $x= f(f^{-1} (x) )$ .

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

Example C

What is the inverse of $f(x)=y=\sin x$ ?

Solution: The sine function does not pass the horizontal line test and so its true inverse is not a function.

However, if you restrict the domain to just the part of the $x$ -axis between $- \frac{\pi}{2}$  and $\frac{\pi}{2}$  then it will pass the horizontal line test and the inverse will be a function.

The inverse of the sine function is called the arcsine function, $f(x)=\sin^{-1}(x)$ , and is shown in black. It is truncated so that it only inverts a part of the whole sine wave. You will study periodic functions and their inverses in more detail later.

Concept Problem Revisited

You can tell that two functions are inverses if each undoes the other, always leaving the original $x$

#### Vocabulary

A relation is a general term for functions and non-functions that relate two variables and may or may not pass the vertical line test

Two relations are inverses of each other if they are reflections across the line  $y=x$ .

The horizontal line test is a means of discovering whether the inverse of a function is also a function.

#### Guided Practice

1. Determine the inverse of $f(x)=5+\frac{x}{2}$ . Verify that the inverse is actually the inverse.

2. Determine if $f(x)=\frac{3}{7}x-21$  and $g(x)=\frac{7}{3}x+21$ are inverses of one another.

3. Determine the inverse of $f(x)=\frac{x}{x+4}$ .

1. To find the inverse,

$y&=5+ \frac{x}{2}\\x&=5+ \frac{y}{2}\\x-5&=\frac{y}{2}\\2(x-5)&=y=f^{-1}(x)$

Verification:

$2\left(5+\frac{x}{2}-5\right)=2\left(\frac{x}{2}\right)=x$

$5+\frac{2(x-5)}{2}=5+x-5=x$

They are truly inverses of each other.

2. Even though $f(x)=\frac{3}{7}x-21$  and $g(x)=\frac{7}{3}x+21$  have some inverted pieces, they are not inverses of each other. In order to show this, you must show that the composition does not simplify to $x$ . $\frac{3}{7}\left(\frac{7}{3}x+21\right) -21=x+9-21=x-12 \ne x$

3. To find the inverse, switch x and y.

$f(x)&=y=\frac{x}{x+4}\\x&= \frac{y}{y+4}\\x(y+4)&=y\\xy+4x&=y\\xy-y&=-4x\\y(x-1)&=-4x\\f^{-1}(x)&=y=-\frac{4x}{x-1}$

#### Practice

Consider $f(x)=x^3$ .

1. Sketch $f(x)$  and $f^{-1}(x)$ .

2. Find  $f^{-1}(x)$ algebraically. It is actually a function?

3. Verify algebraically that $f(x)$  and $f^{-1}(x)$  are inverses.

Consider $g(x)=\sqrt{x}$ .

4. Sketch $g(x)$  and $g^{-1}(x)$ .

5. Find $g^{-1}(x)$ algebraically. It is actually a function?

6. Verify algebraically that $g(x)$  and $g^{-1}(x)$  are inverses.

Consider $h(x)=|x|$ .

7. Sketch $h(x)$  and $h^{-1}(x)$ .

8. Find $h^{-1}(x)$ algebraically. It is actually a function?

9. Verify graphically that $h(x)$  and $h^{-1}(x)$  are inverses.

Consider $j(x)=2x-5$ .

10. Sketch $j(x)$  and $j^{-1}(x)$ .

11. Find  $j^{-1}(x)$ algebraically. It is actually a function?

12. Verify algebraically that $j(x)$  and $j^{-1}(x)$  are inverses.

13. Use the horizontal line test to determine whether or not the inverse of $f(x)=x^3-2x^2+1$ is also a function.

14. Are $g(x)=\ln (x+1)$ and $h(x)=e^{x-1}$ inverses? Explain.

15. If you were given a table of values for a function, how could you create a table of values for the inverse of the function?

### Vocabulary Language: English

composite function

composite function

A composite function is a function $h(x)$ formed by using the output of one function $g(x)$ as the input of another function $f(x)$. Composite functions are written in the form $h(x)=f(g(x))$ or $h=f \circ g$.
Function

Function

A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Horizontal Line Test

Horizontal Line Test

The horizontal line test says that if a horizontal line drawn anywhere through the graph of a function intersects the function in more than one location, then the function is not one-to-one and not invertible.
inverse

inverse

Inverse functions are functions that 'undo' each other. Formally: $f(x)$ and $g(x)$ are inverse functions if $f(g(x)) = g(f(x)) = x$.
Relation

Relation

A relation is any set of ordered pairs $(x, y)$. A relation can have more than one output for a given input.
Vertical Line Test

Vertical Line Test

The vertical line test says that if a vertical line drawn anywhere through the graph of a relation intersects the relation in more than one location, then the relation is not a function.