<meta http-equiv="refresh" content="1; url=/nojavascript/"> Functions and Inverses | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Concepts Go to the latest version.

# 3.1: Functions and Inverses

Difficulty Level: At Grade Created by: CK-12

You may not realize it, but you have dealt with inverse functions for most of your life. Inverse functions are functions which sort of 'undo each other'. For instance, $4t = T$ is the function to convert teaspoons to tablespoons. The inverse: $t = T/4$ converts from tablespoons back to teaspoons.

Inverse functions are also used in geometry. Consider the following:

You are asked to design the new box for the iMp3 player that your company makes. You know that the iMp3 is 5.4in tall, 2.3in wide, and .5in thick.

a) What is the volume of a box that will just fit the iMp3?
b) If you knew that the iMp3 2 was soon to be released, and it had 10% less volume without changing height or width, could you prepare a box for it?
c) Can you identify the inverse functions that might be applicable in this scenario?

Embedded Video:

### Guidance

Consider the functions $t(x)=\frac{5}{9}(x-32)$ and $f(x)=\frac{9}{5}x+32$ together. These two functions are inverses.

Informally, if two functions are inverses , then the input of one function is the output of the other.

Formally, the inverse of a function is defined as follows:

Inverse functions
Functions f( x ) and g ( x ) are said to inverses if
f(g(x)) = g(f(x)) = x
Or, using the composite function notation:
f g = g f = x
The following notation is used to indicate inverse functions:
If f ( x ) and g ( x ) are inverse functions, then
f ( x ) = g -1 ( x ) and g ( x ) = f -1 ( x )
The following notation is also used: f = g -1 and g = f -1 .
Note that f -1 (x) does not equal $\frac{1}{f(x)}$ .

Informally, we can identify the inverse of a function as the relation we obtain by switching the domain and range of the function. Because of this definition, you can find an inverse by switching the roles of x and y in an equation. For example, consider the function g ( x ) = 2 x . This is the line y = 2 x. If we switch x and y , we get the equation x = 2 y. Dividing both sides by 2, we get y = 1/2 x . Therefore the functions g(x) = 2 x and y = 1/2 x are inverses. Using function notation, we can write y = 1/2 x as g -1 (x) = 1/2 x .

We can also analyze two functions and determine whether or not they are inverses. Look carefully at the formal definition of inverse functions:

Two functions f ( x ) and g ( x ) are inverses if and only if f(g(x)) = g(f(x)) = x .

That means that just as we can find the inverse of a function by identifying one that fits the definition, we can verify a possible inverse by testing it against the definition.

#### Example A

In the United States, we measure temperature using the Fahrenheit scale. In other countries, people use the Celsius scale. The equation C = 5/9 ( F - 32) can be used to find C, the Celsius temperature, given F , the Fahrenheit temperature. If we write this equation using function notation, we have $t(x)=\frac{5}{9}(x-32)$ . The input of the function is a Fahrenheit temperature, and the output is a Celsius temperature. This function allows us to convert a Fahrenheit temperature into Celsius.

a.) Identify the inverse function of the equation above to get one that will convert Celsius to Fahrenheit.

b.) Use the Celsius to Fahrenheit equation to convert 0deg Celsius into Fahrenheit.

Solution:

Fahrenheit to Celsius: $C = \frac{5} {9} (F - 32)$

Start by isolating F:

$C = \frac{5} {9} (F - 32)$
$\frac{9} {5} C = \frac{9} {5} \times \frac{5} {9} (F - 32)$
$\frac{9} {5} C = F - 32$
$\frac{9} {5} C + 32 = F$

If we write this equation using function notation, we get $f(x)=\frac{9}{5}x+32$ . For this function, the input is the Celsius temperature, and the output is the Fahrenheit temperature.

If it is 0deg Celsius, then we have: $x=0,f(0)=\frac{9}{5}(0)+32=0+32=32$ .

'0deg Celsius = 32deg Fahrenheit '

#### Example B

Find the inverse of each function

a. f ( x ) = 5 x - 8 b. f ( x ) = x 3 .

Solution:

a. First write the function using “ y =” notation, then interchange x and y :

f(x) = 5x - 8 → y = 5x - 8 → x = 5y - 8

Then isolate y :

$x = 5y - 8$
$x + 8 = 5y$
$y = \frac{1} {5}x + \frac{8} {5}$

b. Follow the same process as "a":

First write the function using “ y= ”: f ( x ) = x 3
y = x 3
Now interchange x and y x = y 3
Now isolate y : $y = \sqrt[3]{x}$

Because of the definition of inverse, the graphs of inverses are reflections across the line y = x . The graph below shows $t(x)=\frac{5}{9}(x-32)$ and $f(x)=\frac{9}{5}x+32$ on the same graph, along with the reflection line y = x .

* A note about graphing with software or a graphing calculator: if you look at the graph above, you can see that the lines are reflections over the line y = x . However, if you do not view the graph in a window that shows equal scales of the x - and y -axes, the graph might not look like this.

#### Example C

Use composition of functions to determine if f ( x ) = 2 x + 3 and g(x) = 3 x - 2 are inverses.

Solution

The functions are not inverses.

We only need to check one of the compositions:
$f(g(x)) = f(3x - 2) \to 2(3x - 2) + 3 \to 6x - 4 + 3 \to 6x - 1 \neq x$

Remember the problem at the beginning of the lesson? Were you able to solve it?

You are asked to design the new box for the iMp3 player that your company makes. You know that the iMp3 is 5.4in tall, 2.3in wide, and .5in thick.

a) What is the volume of a box that will just fit the iMp3?
b) If you knew that the iMp3 2 was soon to be released, and it had 10% less volume without changing height or width, could you prepare a box for it?
c) Can you identify the inverse functions that might be applicable in this scenario?

The solution looks like this:

a) Recall that the volume of a rectangular solid is given by: $V = l \cdot w \cdot h$
$\therefore V = 6.21in^{3}$
b) The missing dimension for the new box is height: $.9(6.21) = 5.4 \cdot 2.3 \cdot h$
$\therefore h = .45in$
c) Two inverse functions used here are: $V(h) = (l \cdot w) \cdot h$ and $h(v) = \frac{v}{l \cdot w}$

Verify that the two functions are inverse: Let $h = 2$ and assume a base area of $6$

$V(2) = 6 \cdot 2$ ==> $V(2) = 12$
$h(12) = 12/6$ ==> $h(12) = 2$

This checks out, as $V(h(2)) = 2 = h(V(2))$

### Vocabulary

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

Function composition refers to 'nested functions' or functions within functions.

A function is said to be invertible if its inverse is also a function.

### Guided Practice

1) Find the inverse of the function and determine whether or not it is invertible: $f(x) = x+ 3$

2) If $f(x) = 3x + 10$ find $f^{-1}(x)$

3) If $f(x) = \sqrt{x + 12}$ find $f^{-1}(x)$

4) If $f(x) = {(6, 4), (8, -12), (-2, 22), (10, -10)}$ find $f^{-1}(x)$

5) Determine whether or not problems 3 and 4 are invertible. Explain your answer.

Solutions

1) Follow the steps to find the inverse function:

STEP 1: Replace $f(x)$ with $y$ (if necessary) $f(x) = x+3 \to y = x+3$
STEP 2: Switch $x$ and $y$ in the equation $x = y+3$
STEP 3: Solve for $y$ $x-3 = y$
STEP 4: Replace $y$ with inverse function notation: $f^{-1}(x)$ $f^{-1}(x) = x-3$

Recall the definition of invertibility: If the function $f(x) = x+3$ is invertible, then its inverse $f^{-1}(x) = x + 3$ must also be a function.

$f^{-1}(x) = x + 3$ is a diagonal straight line with a y -intercept 3 units above the origin, it is a function.
$\therefore f(x) = x+3$ is invertible

2) Follow the same steps as problem 1:

Step 1: $f(x) = 3x + 10 \to y = 3x + 10$
Step 2: $x = 3y + 10$
Step 3: $\frac{x - 10}{3} = y$
Step 4: $f^{-1}(x) = \frac{x - 10}{3}$

Answer: $f^{-1}(x) = \frac{x - 10}{3}$

3) Use the same 4 steps:

$f(x) = \sqrt{x + 12} \to y =\sqrt{x + 12}$
$x = \sqrt{y + 12}$
$x^2 = (\sqrt{y+12})^2$ (square both sides)
$x^2 = y + 12$ (simplify)
$x^2 - 12 = y$
$x^2 - 12 = f(x)$

Answer: $f^{-1}(x) =x^2 - 12$

4) Just as when finding the inverse with an equation, exchange x and y.

The ordered pairs (x, y) become (y, x)
$\therefore f(x) = {(6, 4), (8, -12), (-2, 22), (10, -10)} \to$
$\to f^{-1}(x) = {(4, 6), (-12, 8), (22, -2), (-10, 10)}$

5) Both functions are invertible. Explanation: In order for a function to be invertible, the inverse of the function must also be a function.

The equation $f^{-1}(x) =x^2 - 12$ (problem 3) is a parabola with vertex (0, -12), it is indeed a function.
The point set: $f^{-1}(x) = {(4, 6), (-12, 8), (22, -2), (-10, 10)}$ (problem 4) has no x terms with more than 1 associated y value. It is also a function.

### Practice

State if the given functions are inverses of each other:

1. $g(x) =4 - \frac {3}{2}x \to f(x) = \frac{1}{2}x + \frac{3}{2}$
2. $g(n) = \frac {-12 - 2n}{3} \to f(n) = \frac{-5 + 6n}{5}$
3. $f(n) = \frac{-16 + n}{4} \to g(n) = 4n+ 16$
4. $f(n) = 29n - 2)^3 \to g(n) = \frac{4 +\sqrt[3]{4n}}{2}$
5. $g(x) = -\frac {2}{x} - 1 \to f(x) = -\frac{2}{x + 1}$

Find the inverse of each function:

1. $h(x) = \sqrt[3]x - 3$
2. $g(x) =\frac{1}{x} - 2x$
3. $f(x) = 4x$

Find the inverse of each function. Then graph the function and its inverse.

1. $f(x) = -1 - \frac {1}{5}x$
2. $g(x) =\frac {1}{x - 1}$
3. $f(x) = -2x^3 + 1$
4. $g(x) =\frac {-x - 5}{3}x$

Find the inverse of each function and determine invertibility:

1. ${(-5, 1), (2,- 8), (-3, 5), (0, 1)}$
2. ${(6, 1), (3, - 7), (3, -4), (-8, 2)}$

Find the error in the following problem/solution:

1. Given $f(x) = \frac{x - 10}{x}$ find $f^{-1}(x)$ Step 1: $y = \frac{x - 10}{x}$ Step 2: $x = \frac{y -10}{x}$ Step 3: $x (x) = (\frac{y - 10}{x})x$ Step 4: $x^2 = y - 10$ Step 5: $x^2 + 10 = y$ Step 6: $f^{-1}(x) = x^2 + 10$

Nov 01, 2012

May 27, 2014

# We need you!

At the moment, we do not have exercises for Functions and Inverses.