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7.11: Inverse Functions

Difficulty Level: At Grade Created by: CK-12

A planet's maximum distance from the sun (in astronomical units) is given by the formula $d = p^{\frac {2}{3}}$ , were p is the period (in years) of the planet's orbit around the sun. What is the inverse of this function?

Guidance

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 $x$ and $y$ values. The domain of the original relation becomes the range of the inverse relation and the range of the original relation becomes the domain of the inverse relation.

Example A

Find the inverse mapping of $S=\left \{(6, -1), (-2, -5), (-3, 4), (0, 3), (2, 2)\right \}$ .

Solution: Here, we will find the inverse of this relation by mapping it over the line $y=x$ . As was stated above in the definition, the inverse relation switched the domain and range of the original function. So, the inverse of this relation, $S$ , is $S^{-1}$ (said “ $s$ inverse”) and will flip all the $x$ and $y$ values.

$S^{-1}=\left \{(-1, 6), (-5, -2), (4, -3), (3, 0), (2, 2)\right \}$

If we plot the two relations on the $x-y$ plane, we have:

The blue points are all the points in $S$ and the red points are all the points in $S^{-1}$ . Notice that the points in $S^{-1}$ are a reflection of the points in $S$ over the line, $y=x$ . All inverses have this property.

If we were to fold the graph on $y=x$ , each inverse point $S^{-1}$ should lie on the original point from $S$ . The point $(2, 2)$ lies on this line, so it has no reflection. Any value on this line will remain the same.

Domain of $S$ : $x \in \left \{6, -2, -3, 0, 2\right \}$

Range of $S$ : $y \in \left \{-1, -5, 4, 3, 2\right \}$

Domain of $S^\prime$ : $x \in \left \{-1, -5, 4, 3, 2\right \}$

Range of $S^\prime$ : $y \in \left \{6, -2, -3, 0, 2\right \}$

By looking at the domains and ranges of $S$ and $S^{-1}$ , we see that they are both functions (no $x$ -values repeat). When the inverse of a function is also a function, we say that the original function is a one-to-one function . Each value maps one unique value onto another unique value.

Example B

Find the inverse of $f(x)= \frac{2}{3}x-1$ .

Solution: This is a linear function. Let’s solve by doing a little investigation. First, draw the line along with $y=x$ on the same set of axes.

Notice the points on the function (blue line). Map these points over $y=x$ by switching their $x$ and $y$ values. You could also fold the graph along $y=x$ and trace the reflection.

The red line in the graph to the right is the inverse of $f(x)= \frac{2}{3}x-1$ . Using slope triangles between (-1, 0) and (1, 3), we see that the slope is $\frac{3}{2}$ . Use (-1, 0) to find the $y$ -intercept.

$f^{-1}(x)&= \frac{3}{2}x+b \\0&= \frac{3}{2}(-1)+b \\\frac{3}{2}&=b$

The equation of the inverse, read “ $f$ inverse”, is $f^{-1}(x)= \frac{3}{2}x+ \frac{3}{2}$ .

You may have noticed that the slopes of $f$ and $f^{-1}$ are reciprocals of each other. This will always be the case for linear functions. Also, the $x$ -intercept of $f$ becomes the $y$ -intercept of $f^{-1}$ and vice versa.

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

1. Change $f(x)$ to $y$ .

$y= \frac{2}{3}x-1$

2. Switch the $x$ and $y$ . Change $y$ to $y^{-1}$ for the inverse.

$x= \frac{2}{3}y^{-1}-1$

3. Solve for $y^\prime$ .

$x&= \frac{2}{3}y^{-1}-1 \\\frac{3}{2}(x+1)&= \frac{3}{2} \cdot \left(\frac{2}{3}y^{-1} \right) \\\frac{3}{2}x+ \frac{3}{2}&=y^{-1}$

The algebraic method will work for any type of function.

Example C

Determine if $g(x)=\sqrt{x-2}$ and $f(x)=x^2+2$ are inverses of each other.

Solution: There are two different ways to determine if two functions are inverses of each other. The first, is to find $f^{-1}$ and $g^{-1}$ and see if $f^{-1}=g$ and $g^{-1}= f$ .

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

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

Therefore, $f^{-1}$ 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 $f(x)$ over $y=x$ . It would be a parabola on its side, which is not a function.

The inverse of $g$ would then be only half of the parabola, see below. Despite the restrictions on the domains, $f$ and $g$ 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 $f \circ g=g \circ f=x$ , then $f$ and $g$ are inverses of each other. Think about it; if everything cancels out and all that remains is $x$ , each operation within the functions are opposites, making the functions “opposites” or inverses of each other.

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

Because $f \circ g=g \circ f=x$ , $f$ and $g$ are inverses of each other. Both $f \circ g=x$ and $g \circ f=x$ in order for $f$ and $g$ to be inverses of each other.

Intro Problem Revisit In the function $d = p^{\frac {2}{3}}$ , d is the equivalent of y and p is the equivalent of x . So rewrite the equation and follow the step-by-step process.

$y = x^{\frac {2}{3}}$ ,

Switch the $x$ and $y$ . Change $y$ to $y^{-1}$ for the inverse.

$x = (y^{-1})^{\frac {2}{3}}$

3. Solve for $y^\prime$ .

$x = (y^{-1})^{\frac {2}{3}}\\x^\frac{3}{2} = (y^{-1})^{\frac {2}{3}\cdot \frac{3}{2}}\\x^\frac{3}{2} = y^{-1}$

Now replace y and x with d and p . The inverse d is $p^\frac{3}{2}$ .

Guided Practice

1. Find the inverse of $g(x)=- \frac{3}{4}x+12$ algebraically.

2. Find the inverse of $f(x)=2x^3+5$ algebraically. Is the inverse a function?

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

1. Use the steps given in the Alternate Method for Example B.

$y&=- \frac{3}{4}x+12 \\x&=- \frac{3}{4}y^{-1}+12 \\x-12&=- \frac{3}{4}y^{-1} \\- \frac{4}{3}(x-12)&=y^{-1} \\g^{-1}(x)&=- \frac{4}{3}x+16$

2. Again, use the steps from Example B.

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

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

3. First, find $h(j(x))$ .

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

Because $h(j(x)) \ne x$ , we know that $h$ and $j$ are not inverses of each other. Therefore, there is no point to find $j(h(x))$ .

Vocabulary

Inverse Relation/Function
When a relation or function’s output values are mapped to create input values for a new relation (or function). The input values of the original function would become the output values for the new relation (or function).
One-to-one Function
When the inverse of a function is also a function.

Explore More

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

1. $(2, 3), (-4, 8), (-5, 9), (1, 1)$
2. $(9, -6), (8, -5), (7, 3), (4, 3)$

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

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

Determine whether $f$ and $g$ are inverses of each other by checking to see whether finding $f \circ g=x$ or $g \circ f=x$ . You do not need to show both.

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

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 $y$ variable to solve. Use the example below as a guide.

$f(x)=\frac{3x+13}{2x-11}$

Example:

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

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

1. $f(x)= \frac{x+7}{x},x \ne 0$
2. $f(x)= \frac{x}{x-8},x \ne 8$

Multi-step problem

1. 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.
1. The temperature at which water freezes will give us one point on a line in which $x$ represents the degrees in Celsius and $y$ 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 $y= \frac{9}{5}x+32$ or $F= \frac{9}{5}C+32$ .
2. Find the inverse of the equation above by solving for $C$ to derive a formula that will allow us to convert from Fahrenheit to Celsius.
3. Show that your inverse is correct by showing that the composition of the two functions simplifies to either $F$ or $C$ (depending on which one you put into the other.)

Mar 12, 2013

Jan 06, 2015

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