Your instructor gives you a function,

Are you able to do this?

Keep reading, and by the end of this Concept, you'll be able to find the inverse of the function through graphing instead of algebra.

### Watch This

James Sousa: Animation Inverse Function

### Guidance

Determining an inverse function algebraically can be both involved and difficult, so it is useful to know how to map

The points

The points

#### Example A

Find the inverse of

**Solution:** From the last section, we know that the inverse of this function is

A: (4, -1)

B: (4.8, -5)

C: (2, -0.3)

D: (0, -0.2)

E: (5.3, 3.3)

F: (6, 1)

G: (8, 0.3)

H: (11, 0.2)

Now, take these eight points, switch the

Not all functions have inverses that are one-to-one. However, the inverse can be modified to a one-to-one function if a “restricted domain” is applied to the inverse function.

#### Example B

Find the inverse of

**Solution**: Let’s use the graphic approach for this one. The function is graphed in blue and its inverse is red.

Clearly, the inverse relation is not a function because it does not pass the vertical line test. This is because all parabolas fail the horizontal line test. To “make” the inverse a function, we restrict the domain of the original function. For parabolas, this is fairly simple. To find the inverse of this function algebraically, we get

This technique of sectioning the inverse is applied to finding the inverse of trigonometric functions because it is periodic.

#### Example C

Find the inverse of

**Solution:** To find the inverse by mapping, pick several points on

A: (0, -.5)

B: (-1, 2)

C: (1, 0)

D: (-2, .75)

E: (2, .125)

F: (-3, .57)

G: (3, .18)

Now, take these seven points, switch the

Not all functions have inverses that are one-to-one. However, the inverse can be modified to a one-to-one function if a “restricted domain” is applied to the inverse function.

### Guided Practice

1. Study the following graph and answer these questions:

(a) Is the graphed relation a function?

(b) Does the relation have an inverse that is a function?

2. Find the inverse of

3. Find the inverse of

**Solutions:**

1. The graph represents a one-to-one function. It passes both a vertical and a horizontal line test. The inverse would be a function.

2. By selecting 4-5 points and switching the

3. By selecting 4-5 points and switching the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values, you will get the red graph below.

### Concept Problem Solution

The original equation is \begin{align*}f(x) = (x - 1)^2 + 3\end{align*}.

Here is a plot of the function:

Notice that the domain of the function under examination has to be restricted to 1 or greater, so that the function will pass the horizontal line test. Some points that are on this graph are:

A: (1, 3)

B: (2, 4)

C: (3, 7)

To map the inverse function, first take each point and switch the "x" and "y" values:

\begin{align*}A^{-1}: \ (3, 1)\end{align*}

\begin{align*}B^{-1}: \ (4, 2)\end{align*}

\begin{align*}C^{-1}: \ (7, 3)\end{align*}

Then connect these dots, and you can see the graph of the inverse function. The inverse function graph looks like this:

In this case, the range of the function has to be restricted to be 1 or greater, so that the inverse function will pass the vertical line test.

### Explore More

For each of the following graphs answer these questions:

(a) Is the graphed relation a function?

(b) Does the relation have an inverse that is a function?

Find the inverse of each function using the mapping principle.

- \begin{align*}y=x^2+x-2\end{align*}
- \begin{align*}y=x^3\end{align*}
- \begin{align*}y=\sin(x-\frac{\pi}{2})\end{align*}
- \begin{align*}y=\cos(2x)\end{align*}
- \begin{align*}y=\frac{1}{x}\end{align*}
- \begin{align*}y=x^2-9\end{align*}
- \begin{align*}y=-2+\sin(\frac{1}{2}x)\end{align*}
- What type of points will be in the same place in both a function and its inverse?
- When you graph a function and its inverse on the same set of axes, where is the line of reflection? Why?