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## Algebra (all content)

### Course: Algebra (all content) > Unit 7

Lesson 19: Introduction to inverses of functions (Algebra 2 level)# Intro to inverse functions

CCSS.Math: ,

Learn what the inverse of a function is, and how to evaluate inverses of functions that are given in tables or graphs.

**Inverse functions**, in the most general sense, are functions that "reverse" each other.

For example, here we see that function f takes 1 to x, 2 to z, and 3 to y.

The inverse of f, denoted f, start superscript, minus, 1, end superscript (and read as "f inverse"), will

*reverse*this mapping. Function f, start superscript, minus, 1, end superscript takes x to 1, y to 3, and z to 2.## Defining inverse functions

In general, if a function f takes a to b, then the inverse function, f, start superscript, minus, 1, end superscript, takes b to a.

From this, we have the formal definition of inverse functions:

## f, left parenthesis, a, right parenthesis, equals, b, \Longleftrightarrow, f, start superscript, minus, 1, end superscript, left parenthesis, b, right parenthesis, equals, a

Let's dig further into this definition by working through a couple of examples.

### Example 1: Mapping diagram

Suppose function h is defined by mapping diagram above. What is h, start superscript, minus, 1, end superscript, left parenthesis, 9, right parenthesis?

### Solution

We are given information about function h and are asked a question about function h, start superscript, minus, 1, end superscript. Since inverse functions reverse each other, we need to

*reverse*our thinking.Specifically, to find h, start superscript, minus, 1, end superscript, left parenthesis, 9, right parenthesis, we can find the input of h whose output is 9. This is because if h, start superscript, minus, 1, end superscript, left parenthesis, 9, right parenthesis, equals, x, then by definition of inverses, h, left parenthesis, x, right parenthesis, equals, 9.

From the mapping diagram, we see that h, left parenthesis, 6, right parenthesis, equals, 9, and so h, start superscript, minus, 1, end superscript, left parenthesis, 9, right parenthesis, equals, 6.

### Check your understanding

### Example 2: Graph

This is the graph of function g. Let's find g, start superscript, minus, 1, end superscript, left parenthesis, minus, 7, right parenthesis.

### Solution

To find g, start superscript, minus, 1, end superscript, left parenthesis, minus, 7, right parenthesis, we can find the input of g that corresponds to an output of minus, 7. This is because if g, start superscript, minus, 1, end superscript, left parenthesis, minus, 7, right parenthesis, equals, x, then by definition of inverses, g, left parenthesis, x, right parenthesis, equals, minus, 7.

From the graph, we see that g, left parenthesis, minus, 3, right parenthesis, equals, minus, 7.

Therefore, g, start superscript, minus, 1, end superscript, left parenthesis, minus, 7, right parenthesis, equals, minus, 3.

### Check your understanding

## A graphical connection

The examples above have shown us the algebraic connection between a function and its inverse, but there is also a graphical connection!

Consider function f, given in the graph and in a table of values.

x | f, left parenthesis, x, right parenthesis |
---|---|

minus, 2 | start fraction, 1, divided by, 4, end fraction |

minus, 1 | start fraction, 1, divided by, 2, end fraction |

0 | 1 |

1 | 2 |

2 | 4 |

We can reverse the inputs and outputs of function f to find the inputs and outputs of function f, start superscript, minus, 1, end superscript. So if left parenthesis, a, comma, b, right parenthesis is on the graph of y, equals, f, left parenthesis, x, right parenthesis, then left parenthesis, b, comma, a, right parenthesis will be on the graph of y, equals, f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis.

This gives us these graph and table of values of f, start superscript, minus, 1, end superscript.

x | f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis |
---|---|

start fraction, 1, divided by, 4, end fraction | minus, 2 |

start fraction, 1, divided by, 2, end fraction | minus, 1 |

1 | 0 |

2 | 1 |

4 | 2 |

Looking at the graphs together, we see that the graph of y, equals, f, left parenthesis, x, right parenthesis and the graph of y, equals, f, start superscript, minus, 1, end superscript, left parenthesis, x, right parenthesis are reflections across the line y, equals, x.

This will be true in general; the graph of a function and its inverse are reflections over the line y, equals, x.

## Check your understanding

## Why study inverses?

It may seem arbitrary to be interested in inverse functions but in fact we use them all the time!

Consider that the equation C, equals, start fraction, 5, divided by, 9, end fraction, left parenthesis, F, minus, 32, right parenthesis can be used to convert the temperature in degrees Fahrenheit, F, to a temperature in degrees Celsius, C.

But suppose we wanted an equation that did the reverse – that converted a temperature in degrees Celsius to a temperature in degrees Fahrenheit. This describes the function F, equals, start fraction, 9, divided by, 5, end fraction, C, plus, 32, or the inverse function.

On a more basic level, we solve many equations in mathematics, by "isolating the variable". When we isolate the variable, we "undo" what is around it. In this way, we are using the idea of inverse functions to solve equations.

## Want to join the conversation?

- how would I find the inverse function of a quadratic, such as 2x^2+2x-1?(39 votes)
- You can find the inverse of any function y=f(x) by reflecting it across the line y=x. The quadratic you list is not one-to-one, so you will have to restrict the domain to make it invertible.

Algebraically reflecting a graph across the line y=x is the same as switching the x and y variables and then resolving for y in terms of x.

As you progress in your ability to find inverse functions you can see Sal solve for an inverse of a quadratic function here:

https://www.khanacademy.org/math/algebra2/manipulating-functions/finding-inverse-functions/v/function-inverses-example-2

But i highly recommend you make sure you can find the inverse of a linear function first before tackling quadratics and the associated domain restriction complications that they bring. If after working through that video and the subsequent examples, you would be better served posting your question there if you still aren't sure.(75 votes)

- Is it true that when you solve for an inverse of a function, you do PEMDAS backwards?(14 votes)
- Nice question!

Yes you could think of it that way. If a function can be constructed by starting with x and performing a sequence of (reversible) operations, then its inverse can be constructed by starting with x and**both**reversing each operation and reversing the order of operations.

Example: Suppose f(x) = 7(x - 5)^3. Note that f(x) is constructed by starting with x, subtracting 5, cubing, and then multiplying by 7.

Then f^-1(x) is constructed by starting with x, dividing by 7, taking the cube root, and then adding 5.

So f^-1(x) = cuberoot(x/7) + 5.(40 votes)

- Why is the inverse always a reflection? Is it simply two lines that have the same set of reversed relationships, because plugging in the answer does not make a full restitution, instead it gives the same original value of x in a different line? Is there another reason for this? I am fascinated.(7 votes)
- We can think of a function as a collection of points in the plane. Each point has the form (x, y). If we consider the inverse function, it will contain each of these points, but with the coordinates switched.

So if (a, b) is on our original function, then (b, a) is on the inverse. Let's look at how we get from (a, b) to (b, a). Draw a line segment between them.

The slope of this line segment is then (b-a)/(a-b)=(-1)(b-a)/(b-a)= -1. That's interesting; if we have a point on a function and want to find the corresponding point on the inverse function, we slide along a line of slope -1. But how far do we slide?

Let's find the midpoint of our line segment. In the x-direction, we go from a to b. So the midpoint has the x-coordinate (a+b)/2. In the y-direction, we go from b to a. So the midpoint has y-coordinate (b+a)/2. Same as the x-coordinate!

So the midpoint of the segment must lie on the line y=x. Notice that y=x has a slope of 1, and our segment has a slope of -1. So the two are perpendicular.

So what we've done to move from (a, b) to (b, a) is reflect over the line y=x.(14 votes)

- i have trouble understanding inverses. Can someone help me?

i have trouble solving problems for the inverses.(4 votes)- An inverse function essentially undoes the effects of the original function. If f(x) says to multiply by 2 and then add 1, then the inverse f(x) will say to subtract 1 and then divide by 2. If you want to think about this graphically, f(x) and its inverse function will be reflections across the line y = x.

To find the inverse of a function you just have to switch the x and the y and then solve for y.

For example, what is the inverse of y = 2x + 1?

y = 2x + 1

x = 2y + 1. (Switch the x and y)

2y = x - 1

y = (x-1)/2. And we're done.(7 votes)

- How did you get -3 for the second example? I see no correlation between the -7 and -3...(5 votes)
- you should put -7 on the left side of the equation; f(x)=-7=3x-2 and solve the equation for x, you get 3. Now, knowing that x is the reverse function of y or f(x) which is 3, so f-1(x)=x=3(1 vote)

- domain of f(x) is the range of inverse function and domain of inverse function is the range of f(x). but it is not true in some cases like f(x) = √2x-3. if we see domain of this function is x>=3/2 and inverse of this function is x^2/2+3/2 domain of this function is all real numbers . so, acc. to above line it says range of f(x) is all real numbers but in actual and it is x>=0.(3 votes)
- If f(x) = √(2x-3)

Domain = x>= 3/2; Range = y >= 0

Then for the inverse of f(x) = x^2/2+3/2

Domain = x >= 0; Range = y >= 3/2

If you widen the domain for the inverse function to x = any real number, then you will have input values for the inverse that can not be used in the original function. If you truly want the 2 functions to be inverses, you need to maintain the restrictions on domain/range for the 2 function.

Hope this makes sense.(5 votes)

- Is there f^-2 instead of f^-1?(3 votes)
- could you in any way use these inverse functions to make a circle? I wouldnt think so, cause i cant find one, but can anyone else(2 votes)
- The basic equation for a circle is not a function. Since it is not a function, it can't have an inverse function.(2 votes)

- is ono-to-one the only function that is invertible?(0 votes)
- of course, yes only one-to- one functions are invertible because if a function is not one-to-one it's inverse won't be a function meaning it doesn't exist.(6 votes)

- what would be a possible use of the imaginary number /i/ in this format of a function?(2 votes)