# 7.11: Inverse Functions

**At Grade**Created by: CK-12

**Practice**Functions and Inverses

A planet's maximum distance from the sun (in astronomical units) is given by the formula *p* is the period (in years) of the planet's orbit around the sun. What is the inverse of this function?

### Inverse Functions

By now, you are probably familiar with the term “inverse”. Multiplication and division are inverses of each other. More examples are addition and subtraction and the square and square root. We are going to extend this idea to functions. An **inverse relation** maps the output values to the input values to create another relation. In other words, we switch the

Let's find the inverse mapping of

Here, we will find the inverse of this relation by mapping it over the line

If we plot the two relations on the

The blue points are all the points in

If we were to fold the graph on

Domain of

Range of

Domain of

Range of

By looking at the domains and ranges of **one-to-one function**. Each value maps one unique value onto another unique value.

Now, let's find the inverse of

This is a linear function. Let’s solve by doing a little investigation. First, draw the line along with

Notice the points on the function (blue line). Map these points over

The red line in the graph to the right is the inverse of

The equation of the inverse, read “

You may have noticed that the slopes of

**Alternate Method:** There is also an algebraic approach to finding the inverse of any function. Let’s repeat this example using algebra.

Step 1: Change \begin{align*}f(x)\end{align*} to \begin{align*}y\end{align*}.

\begin{align*}y= \frac{2}{3}x-1\end{align*}

Step 2: Switch the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. Change \begin{align*}y\end{align*} to \begin{align*}y^{-1}\end{align*} for the inverse.

\begin{align*}x= \frac{2}{3}y^{-1}-1\end{align*}

Step 3: Solve for \begin{align*}y^\prime\end{align*}.

\begin{align*}x&= \frac{2}{3}y^{-1}-1 \\ \frac{3}{2}(x+1)&= \frac{3}{2} \cdot \left(\frac{2}{3}y^{-1} \right) \\ \frac{3}{2}x+ \frac{3}{2}&=y^{-1}\end{align*}

The algebraic method will work for any type of function.

Finally, let's determine if \begin{align*}g(x)=\sqrt{x-2}\end{align*} and \begin{align*}f(x)=x^2+2\end{align*} are inverses of each other.

There are two different ways to determine if two functions are inverses of each other. The first, is to find \begin{align*}f^{-1}\end{align*} and \begin{align*}g^{-1}\end{align*} and see if \begin{align*}f^{-1}=g\end{align*} and \begin{align*}g^{-1}= f\end{align*}.

\begin{align*}x&= \sqrt{y^{-1}-2}&& && \qquad \quad \ \ x=(y^{-1})^2+2 \\ x^2&=y^{-1}-2 && and && \qquad x-2=(y^{-1})^2 \\ x^2+2&=y^{-1}=g^{-1}(x)&& && \pm \sqrt{x-2}=y^{-1}=f^{-1}(x)\end{align*}

Notice the \begin{align*}\pm\end{align*} sign in front of the square root for \begin{align*}f^{-1}\end{align*}. That means that \begin{align*}g^{-1}\end{align*} is \begin{align*}\sqrt{x-2}\end{align*} and \begin{align*}- \sqrt{x-2}\end{align*}.

Therefore, \begin{align*}f^{-1}\end{align*} is not really a function because it fails the vertical line test. However, if you were to take each part separately, individually, they are functions. You can also think about reflecting \begin{align*}f(x)\end{align*} over \begin{align*}y=x\end{align*}. It would be a parabola on its side, which is not a function.

The inverse of \begin{align*}g\end{align*} would then be only half of the parabola, see below. Despite the restrictions on the domains, \begin{align*}f\end{align*} and \begin{align*}g\end{align*} are inverses of each other.

**Alternate Method:** The second, and easier, way to determine if two functions are inverses of each other is to use composition. If \begin{align*}f \circ g=g \circ f=x\end{align*}, then \begin{align*}f\end{align*} and \begin{align*}g\end{align*} are inverses of each other. Think about it; if everything cancels out and all that remains is \begin{align*}x\end{align*}, each operation within the functions are opposites, making the functions “opposites” or inverses of each other.

\begin{align*}f \circ g&= \sqrt{\left(x^2+2\right)-2} && && g \circ f= \sqrt{x-2}^2+2 \\ &= \sqrt{x^2} && and && \qquad=x-2+2 \\ &=x && && \qquad=x\end{align*}

Because \begin{align*}f \circ g=g \circ f=x\end{align*}, \begin{align*}f\end{align*} and \begin{align*}g\end{align*} are inverses of each other. Both \begin{align*}f \circ g=x\end{align*} and \begin{align*}g \circ f=x\end{align*} in order for \begin{align*}f\end{align*} and \begin{align*}g\end{align*} to be inverses of each other.

### Examples

#### Example 1

Earlier, you were asked to find the inverse of the function \begin{align*}d = p^{\frac {2}{3}}\end{align*}.

In the function \begin{align*}d = p^{\frac {2}{3}}\end{align*}, *d* is the equivalent of *y* and *p* is the equivalent of *x*. So rewrite the equation and follow the step-by-step process illustrated above.

\begin{align*}y = x^{\frac {2}{3}}\end{align*},

Switch the \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. Change \begin{align*}y\end{align*} to \begin{align*}y^{-1}\end{align*} for the inverse.

\begin{align*}x = (y^{-1})^{\frac {2}{3}}\end{align*}

Solve for \begin{align*}y^\prime\end{align*}.

\begin{align*}x = (y^{-1})^{\frac {2}{3}}\\ x^\frac{3}{2} = (y^{-1})^{\frac {2}{3}\cdot \frac{3}{2}}\\ x^\frac{3}{2} = y^{-1}\end{align*}

Now replace *y* and *x* with *d* and *p*. The inverse *d* is \begin{align*}p^\frac{3}{2}\end{align*}.

#### Example 2

Find the inverse of \begin{align*}g(x)=- \frac{3}{4}x+12\end{align*} algebraically.

\begin{align*}y&=- \frac{3}{4}x+12 \\ x&=- \frac{3}{4}y^{-1}+12 \\ x-12&=- \frac{3}{4}y^{-1} \\ - \frac{4}{3}(x-12)&=y^{-1} \\ g^{-1}(x)&=- \frac{4}{3}x+16\end{align*}

#### Example 3

Find the inverse of \begin{align*}f(x)=2x^3+5\end{align*} algebraically. Is the inverse a function?

\begin{align*}y&=2x^3+5 \\ x&=2(y^{-1})^3+5 \\ x-5&=2(y^{-1})^3 \\ \frac{x-5}{2}&=(y^{-1})^3 \\ f^{-1}(x)&= \sqrt[3]{\frac{x-5}{2}}\end{align*}

Yes, \begin{align*}f^{-1}\end{align*} is a function. Plot in your graphing calculator if you are unsure and see if it passes the vertical line test.

#### Example 4

Determine if \begin{align*}h(x)=4x^4-7\end{align*} and \begin{align*}j(x)= \frac{1}{4} \sqrt[4]{x-7}\end{align*} are inverses of each other using compositions.

First, find \begin{align*}h(j(x))\end{align*}.

\begin{align*}h(j(x))&=4 \left(\frac{1}{4} \sqrt[4]{x+7}\right)^4-7 \\ &=4 \cdot \left(\frac{1}{4}\right)^4 x+7-7 \\ &= \frac{1}{64}x\end{align*}

Because \begin{align*}h(j(x)) \ne x\end{align*}, we know that \begin{align*}h\end{align*} and \begin{align*}j\end{align*} are not inverses of each other. Therefore, there is no point to find \begin{align*}j(h(x))\end{align*}.

### Review

Write the inverses of the following functions. State whether or not the inverse is a function.

- \begin{align*}(2, 3), (-4, 8), (-5, 9), (1, 1)\end{align*}
- \begin{align*}(9, -6), (8, -5), (7, 3), (4, 3)\end{align*}

Find the inverses of the following functions algebraically. Note any restrictions to the domain of the inverse functions.

- \begin{align*}f(x)=6x-9\end{align*}
- \begin{align*}f(x)= \frac{1}{4x+3}\end{align*}
- \begin{align*}f(x)= \sqrt{x+7}\end{align*}
- \begin{align*}f(x)=x^2+5\end{align*}
- \begin{align*}f(x)=x^3-11\end{align*}
- \begin{align*}f(x)= \sqrt[5]{x+16}\end{align*}

Determine whether \begin{align*}f\end{align*} and \begin{align*}g\end{align*} are inverses of each other by checking to see whether finding \begin{align*}f \circ g=x\end{align*} or \begin{align*}g \circ f=x\end{align*}. You do not need to show both.

- \begin{align*}f(x)= \frac{2}{3}x-14\end{align*} and \begin{align*}g(x)= \frac{3}{2}x+21\end{align*}
- \begin{align*}f(x)= \frac{x+5}{8}\end{align*} and \begin{align*}g(x)=8x+5\end{align*}
- \begin{align*}f(x)= \sqrt[3]{3x-7}\end{align*} and \begin{align*}g(x)= \frac{x^3}{3}-7\end{align*}
- \begin{align*}f(x)= \frac{x}{x-9},x \ne 9\end{align*} and \begin{align*}g(x)= \frac{9x}{x-1}\end{align*}

Find the inverses of the following functions algebraically. Note any restrictions to the domain of the inverse functions. These problems are a little trickier as you will need to factor out the \begin{align*}y\end{align*} variable to solve. Use the example below as a guide.

\begin{align*}f(x)=\frac{3x+13}{2x-11}\end{align*}

Example:

- \begin{align*}x=\frac{3y+13}{2y-11}\end{align*} First, switch \begin{align*}x\end{align*} and \begin{align*}y\end{align*}
- \begin{align*}2xy-11x=3y+13\end{align*} Multiply both sides by \begin{align*}2y-11\end{align*} to eliminate the fraction
- \begin{align*}2xy-3y=11x+13\end{align*} Now rearrange the terms to get both terms with \begin{align*}y\end{align*} in them on one side and everything else on the other side
- \begin{align*}y(2x-3)=11x+13\end{align*} Factor out the \begin{align*}y\end{align*}
- \begin{align*}y= \frac{11x+13}{2x-3}\end{align*} Finally, Divide both sides by \begin{align*}2x-3\end{align*} to isolate \begin{align*}y\end{align*}.

So, the inverse of \begin{align*}f(x)= \frac{3x+13}{2x-11},x \ne \frac{11}{2}\end{align*} is \begin{align*}f^{-1}(x)= \frac{11x+13}{2x-3},x \ne \frac{3}{2}\end{align*}.

- \begin{align*}f(x)= \frac{x+7}{x},x \ne 0\end{align*}
- \begin{align*}f(x)= \frac{x}{x-8},x \ne 8\end{align*}

**Multi-step problem**

- In many countries, the temperature is measured in degrees Celsius. In the US we typically use degrees Fahrenheit. For travelers, it is helpful to be able to convert from one unit of measure to another. The following problem will help you learn to do this using an inverse function.
- The temperature at which water freezes will give us one point on a line in which \begin{align*}x\end{align*} represents the degrees in Celsius and \begin{align*}y\end{align*} represents the degrees in Fahrenheit. Water freezes at 0 degrees Celsius and 32 degrees Fahrenheit so the first point is (0, 32). The temperature at which water boils gives us the second point (100, 212), because water boils at 100 degrees Celsius or 212 degrees Fahrenheit. Use this information to show that the equation to convert from Celsius to Fahrenheit is \begin{align*}y= \frac{9}{5}x+32\end{align*} or \begin{align*}F= \frac{9}{5}C+32\end{align*}.
- Find the inverse of the equation above by solving for \begin{align*}C\end{align*} to derive a formula that will allow us to convert from Fahrenheit to Celsius.
- Show that your inverse is correct by showing that the composition of the two functions simplifies to either \begin{align*}F\end{align*} or \begin{align*}C\end{align*} (depending on which one you put into the other.)

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.11.

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1-1 function

A function is 1-1 if its inverse is also a function.composite function

A composite function is a function formed by using the output of one function as the input of another function . Composite functions are written in the form or .Function

A function is a relation where there is only one output for every input. In other words, for every value of , there is only one value for .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 functions are functions that 'undo' each other. Formally: and are inverse functions if .inverse function

Inverse functions are functions that 'undo' each other. Formally and are inverse functions if .Inverse Relation

An inverse relation is a relation with output values that are mapped to create input values for a new relation. The input values of the original relation would become the output values for the new relation.One-to-one

A function is one-to-one if its inverse is also a function.Relation

A relation is any set of ordered pairs . A relation can have more than one output for a given input.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.

### Image Attributions

Here you'll learn how to find the inverse of a relation and function.

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