﻿ Inverse of a function in math. Tutorial explaining inverses step by step, several practice problems, plus a free worksheet with answer key

# Inverse of a Function

#### What is the Inverse of a function?

A conceptual analogy

If the function itself is considered a "DO" action, then the inverse is the "UNDO".

#### What about the domain and Range?

Definition: The inverse of a function is when the domain and the range trade places. All elements of the domain become the range, and all elements of the range become the domain.

Inverse of a Function Demonstration

Therefore, the inverse of a function is equivalent to what kind of transformation?

Reflection over the line y = x.

Another Example
Original function f(x) Inverse of function or f-1(x)

{(0, 3) , (1, 4) , (2, 5)}

{(3, 0) , (4, 1) , (5, 2)}

Explore the relationship of the domain and range more, with our interactive function applet.

#### Is the inverse of a function always a function?

To answer that question let's look at the function in Diagram 1. In that graph, you can see
the original function and its inverse .

Is the inverse of that function , also a function?

No

As you can see, the inverse of that function does not pass the vertical line test and, therefore , is not a function.

Diagram I

Conclusion: The inverse of a function is not necessarily a function.

#### So...when is the inverse of a function also a function?

f-1(x), the inverse, is itself a function only when f(x), the original function, is a one-to-one function.

Remember: 1 to 1 functions must pass the horizontal line test!

Example 1

In this case, f(x) is a function, but f-1(x) is nota function.

Function #1, f(x) is not a one to one function because it does not pass the horizontal line test .

. f-1(x) is not a function.

Example 2

In this case, f(x) is a function and f-1(x) is also a function.

This function passes the horizontal line test so, its inverse will also be a function.

### Practice Problems Part I

##### Problem 1

Remember: $$f^{-1} (x)$$ means 'the inverse of f(x) '.

$$f^{-1} (x)$$ = {(2, 1) , (4, 3) , (6, 5)}

##### Problem 2

Remember: $$f^{-1} (x)$$ means 'the inverse of f(x) '.

$$f^{-1} (x)$$ = {(14, 33) , (15, 23) , (12, 11) , (14, 13)}}

##### Problem 3

Remember: $$f^{-1} (x)$$ means 'the inverse of f(x) '.

$$f^{-1} (x)$$ = {(12, -1) , (114, 13) , (61, 15) , (12, 1)}

##### Problem 4

y = 3 is a horizontal line. In a horizontal line all elements of the range are 3.

So, it looks something $$f^{-1} (x)$$ ={(-1,3) , (0, 3) , (1, 3)}.

Therefore the inverse of this function will be whatever line has 3 for all elements in its domain.

Therefore the inverse of y = 3 is the line x = 3.

### Solving Algebraically for the Inverse

#### What is the inverse of f(x) = x + 1?

Just like in our prior examples, we need to switch the domain and range. I n an equation, the domain is represented by the x variable and the range by the y variable.

To Summarize

f(x): took an element from the domain and added 1 to arrive at the corresponding element in the range.

f-1(x) : took an element from the domain and subtracted 1 to arrive at the corresponding element in the range.

### Practice Problems Part II

##### Problem 1
Step 1

Replace f(x) with y.

$$y = x + 22$$

Step 2

Switch x and y.

$$x = y + 22$$

Step 3

Solve for new 'y'.

Step 4

Replace 'y' with f-1(x).

$$f^{-1} (x) = x - 22$$

##### Problem 2
Step 1

Replace f(x) with y.

$$y = 2x$$

Step 2

Switch x and y.

$$x = 2y$$

Step 3

Solve for new 'y'.

$$\frac{1}{2} x = \frac{1}{2} \cdot 2y \\ \frac{1}{2} x = y$$

Step 4

Replace 'y' with f-1(x).

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

##### Problem 3
Step 1

Replace f(x) with y.

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

Step 2

Switch x and y.

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

Step 3

Solve for new 'y'.

$$\color{Red}{2 \cdot} x = \color{Red}{2 \cdot} \big (\frac{1}{2}y + 3 \big) \\ 2 \cdot x = \frac{2}{2}\cdot y + 2 \cdot 3 \\ 2x = y + 6 \\ 2x -6 = y$$

Step 4

Replace 'y' with f-1(x).

$$f^{-1} (x) = 2x -6$$

##### Problem 4
Step 1

Replace f(x) with y.

$$y = x^2 + 3$$

Step 2

Switch x and y.

$$x = y^2 + 3$$

Step 3

Solve for new 'y'.

$$x \color{Red}{-3} = y^2 + 3 \color{Red}{-3} \\ x -3 = y^2 \\ \sqrt{x-3} = \sqrt{y^2} \\ \sqrt{x-3} = y$$

Step 4

Replace 'y' with f-1(x).

$$f^{-1} (x) = \sqrt{x-3}$$

##### Problem 5
Step 1

Replace f(x) with y.

$$y = 5 \sqrt{x}$$

Step 2

Switch x and y.

$$x = 5 \sqrt{y}$$

Step 3

Solve for new 'y'.

$$x{\color{Red}{^2}} = \big(5 \sqrt{y} \big) {\color{Red}{^2}} \\ x^2 = 5{\color{Red}{^2}} \cdot (\sqrt{y}){\color{Red}{^2}} \\ x^2 = 25y \\ \frac{1}{25} \cdot x^2 = \frac{1}{25} \cdot 25y \\ \frac{1}{25} \cdot x^2 = y$$

Step 4

Replace 'y' with f-1(x).

$$f^{-1} (x) = \frac{1}{25} \cdot x^2$$