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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

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 ${f}^{-1}$ (and read as "$f$ inverse"), will reverse this mapping. Function ${f}^{-1}$ takes $x$ to $1$, $y$ to $3$, and $z$ to $2$.
Reflection question
Which of the following is a true statement?

## Defining inverse functions

In general, if a function $f$ takes $a$ to $b$, then the inverse function, ${f}^{-1}$, takes $b$ to $a$.
From this, we have the formal definition of inverse functions:

## $f\left(a\right)=b\phantom{\rule{0.278em}{0ex}}⟺\phantom{\rule{0.278em}{0ex}}{f}^{-1}\left(b\right)=a$‍

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

### Example 1: Mapping diagram

Suppose function $h$ is defined by the mapping diagram above. What is ${h}^{-1}\left(9\right)$?

### Solution

We are given information about function $h$ and are asked a question about function ${h}^{-1}$. Since inverse functions reverse each other, we need to reverse our thinking.
Specifically, to find ${h}^{-1}\left(9\right)$, we can find the input of $h$ whose output is $9$. This is because if ${h}^{-1}\left(9\right)=x$, then by definition of inverses, $h\left(x\right)=9$.
From the mapping diagram, we see that $h\left(6\right)=9$, and so ${h}^{-1}\left(9\right)=6$.

Problem 1
${g}^{-1}\left(3\right)=$

### Example 2: Graph

This is the graph of function $g$. Let's find ${g}^{-1}\left(-7\right)$.

### Solution

To find ${g}^{-1}\left(-7\right)$, we can find the input of $g$ that corresponds to an output of $-7$. This is because if ${g}^{-1}\left(-7\right)=x$, then by definition of inverses, $g\left(x\right)=-7$.
From the graph, we see that $g\left(-3\right)=-7$.
Therefore, ${g}^{-1}\left(-7\right)=-3$.

Problem 2
What is ${h}^{-1}\left(4\right)$?

Challenge problem
Given that $f\left(x\right)=3x-2$, what is ${f}^{-1}\left(7\right)$?

## 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$$f\left(x\right)$
$-2$$\frac{1}{4}$
$-1$$\frac{1}{2}$
$0$$1$
$1$$2$
$2$$4$
We can reverse the inputs and outputs of function $f$ to find the inputs and outputs of function ${f}^{-1}$. So if $\left(a,b\right)$ is on the graph of $y=f\left(x\right)$, then $\left(b,a\right)$ will be on the graph of $y={f}^{-1}\left(x\right)$.
This gives us these graph and table of values of ${f}^{-1}$.
$x$${f}^{-1}\left(x\right)$
$\frac{1}{4}$$-2$
$\frac{1}{2}$$-1$
$1$$0$
$2$$1$
$4$$2$
Looking at the graphs together, we see that the graph of $y=f\left(x\right)$ and the graph of $y={f}^{-1}\left(x\right)$ are reflections across the line $y=x$.
This will be true in general; the graph of a function and its inverse are reflections over the line $y=x$.

Problem 3
This is the graph of $y=h\left(x\right)$.
Which is the best choice for the graph of $y={h}^{-1}\left(x\right)$?

Problem 4
The graph of $y=h\left(x\right)$ is a line segment joining the points $\left(5,1\right)$ and $\left(2,7\right)$.
Drag the endpoints of the solid segment below to graph $y={h}^{-1}\left(x\right)$.

## 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 $C=\frac{5}{9}\left(F-32\right)$ 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 $F=\frac{9}{5}C+32$, 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?
• 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:
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.
• Is it true that when you solve for an inverse of a function, you do PEMDAS backwards?
• 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.
• 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.
• 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.
• i dont understand
• A point on a line f(x), lets say (2,1), when flipped perpendicularly, makes (1,2). In the same way, when you extend the two lines(f(x) and f(x)inverse) to touch y = x, they're perpendicular
• i have trouble understanding inverses. Can someone help me?
i have trouble solving problems for the inverses.
• 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.
• What about 3D graphs...or complex planes? Do inverse functions math work or is it just vectors?
• Yes, inverse functions work in 3D graphs and complex planes, not just in vectors.
In mathematics, a 3D graph is a graph that shows a three-dimensional representation of a function or a set of data points. It is represented by three axes: x, y, and z. The x and y-axes represent the horizontal and vertical dimensions, respectively, while the z-axis represents the depth or height dimension.
A complex plane is a two-dimensional plane that represents complex numbers. It is represented by two axes: the real axis and the imaginary axis. The real axis represents the real part of the complex number, while the imaginary axis represents the imaginary part of the complex number.
Inverse functions can be graphed in 3D graphs and complex planes, just like in two-dimensional graphs. The graph of the inverse function is obtained by reflecting the original graph across the line y = x. The inverse function is defined only if the original function is one-to-one, which means that each input has a unique output.
Vectors are also used in 3D graphs, but they are not the only mathematical concept used. Vector functions are used to describe curves and surfaces in three-dimensional space.
In summary, inverse functions work in 3D graphs and complex planes, and they are graphed by reflecting the original graph across the line y = x. Vectors are also used in 3D graphs, but they are not the only mathematical concept used.
• What about when you have multiple outputs for the function how do you solve the inverse?