# Finding inverse functions

CCSS Math: HSF.BF.B.4a
Learn how to find the formula of the inverse function of a given function. For example, find the inverse of f(x)=3x+2.
Inverse functions, in the most general sense, are functions that "reverse" each other. For example, if $f$ takes $a$ to $b$, then the inverse, $f^{-1}$, must take $b$ to $a$.
Or in other words, $f(a)=b \iff f^{-1}(b)=a$.
In this article we will learn how to find the formula of the inverse function when we have the formula of the original function.

## Before we start...

In this lesson, we will find the inverse function of $f(x)=3x+2$.
Before we do that, let's first think about how we would find $f^{-1}(8)$.
To find $f^{-1}(8)$, we need to find the input of $f$ that corresponds to an output of $8$. This is because if $f^{-1}(8)=x$, then by definition of inverses, $f(x)=8$.
\begin{aligned} f(x) &= 3 x+2\\\\ 8 &= 3 x+2 &&\small{\gray{\text{Let f(x)=8}}} \\\\6&=3x &&\small{\gray{\text{Subtract 2 from both sides}}}\\\\ 2&=x &&\small{\gray{\text{Divide both sides by 3}}} \end{aligned}
So $f(2)=8$ which means that $f^{-1}(8)=2$

## Finding inverse functions

We can generalize what we did above to find $f^{-1}(y)$ for any $y$.
Note that while the choice of variable is arbitrary, we must choose something different than $x$ as this is already involved in the function.
We can change it back to $x$ in the end if desired.
To find $f^{-1}(y)$, we can find the input of $f$ that corresponds to an output of $y$. This is because if $f^{-1}(y)=x$ then by definition of inverses, $f(x)=y$.
\begin{aligned} f(x) &= 3 x+2\\\\ y &= 3 x+2 &&\small{\gray{\text{Let f(x)=y}}} \\\\y-2&=3x &&\small{\gray{\text{Subtract 2 from both sides}}}\\\\ \dfrac{y-2}{3}&=x &&\small{\gray{\text{Divide both sides by 3}}} \end{aligned}
So $f^{-1}(y)=\dfrac{y-2}{3}$.
Since the choice of the variable is arbitrary, we can write this as $f^{-1}(x)=\dfrac{x-2}{3}$.
Because inverse functions reverse the inputs and outputs, another way to find $f^{-1}$ is by switching $x$ and $y$ initially, then solving for $y$ to write the inverse in function form.
\begin{aligned} f(x)&=3x+2\\\\ \goldD y &= 3\blueD x+2 &&\small{\gray{\text{Replace f(x) with y}}}\\\\ \blueD x &= 3\goldD y+2 &&\small{\gray{\text{Switch x and y}}} \\\\x-2&=3y &&\small{\gray{\text{Subtract 2 from both sides}}}\\\\ \dfrac{x-2}{3}&=y &&\small{\gray{\text{Divide both sides by 3}}} \\\\ \end{aligned}
So $f^{-1}(x)=\dfrac{x-2}{3}$.
Notice the similarities among the two methods. In each case, the equation is solved for the variable that is not isolated. The main difference involves when the variables are reversed. (The first method does this at the end while the second method does this initially.)

### 1) Linear function

Find the inverse of $g(x)=2x-5$.
$g^{-1}(x)=$

To start, replace $g(x)$ with $y$. Then solve for $x$.
\begin{aligned} g(x)&= 2x-5 \\\\ y &= 2x-5 &&\small{\gray{\text{Replace g(x) with y}}}\\\\ y+5 &= 2x &&\small{\gray{\text{Add 5 to both sides}}}\\\\ \dfrac{y+5}{2}&=x &&\small{\gray{\text{Divide both sides by 2}}} \\\\ \dfrac{y+5}{2}&=g^{-1}(y) &&\small{\gray{\text{Replace x with }g^{-1}(y)}} \end{aligned}
Since the choice of variable is arbitrary, we can now switch $y$ to $x$ to write the inverse in terms of $x$.
$g^{-1}(x)=\dfrac{x+5}{2}$

### 2) Cubic function

Find the inverse of $h(x)=x^3+2$.
$h^{-1}(x)=$

To start, replace $h(x)$ with $y$. Then solve for $x$.
Since the choice of variable is arbitrary, we can now switch $y$ to $x$ to write the inverse in terms of $x$.
$h^{-1}(x)=\sqrt[3]{x-2}$

### 3) Cube-root function

Find the inverse of $f(x)=4\cdot \sqrt[\Large3]{x}$.
$f^{-1}(x)=$

To start, replace $f(x)$ with $y$. Then solve for $x$.
\begin{aligned} f(x)&= 4\sqrt[3]{x} \\\\ y &= 4\sqrt[3]{x} &&\small{\gray{\text{Replace f(x) with y}}}\\\\ \dfrac{y}{4} &=\sqrt[3]{x}&&\small{\gray{\text{Divide both sides by 4}}}\\\\ \left(\dfrac{y}{4}\right)^3&=x&&\small{\gray{\text{Cube both sides}}} \\\\ \left(\dfrac{y}{4}\right)^3&=f^{-1}(y)&&\small{\gray{\text{Replace x with }f^{-1}(y)}} \\\\ \end{aligned}
Since the choice of variable is arbitrary, we can now switch $y$ to $x$ to write the inverse in terms of $x$.
So $f^{-1}(x)=\left(\dfrac{x}{4}\right)^3$ or $f^{-1}(x)=\dfrac{x^3}{64}$

### 4) Rational functions

Find the inverse of $g(x)=\dfrac{x-3}{x-2}$.
$g^{-1}(x)=$

To start, replace $g(x)$ with $y$. Then solve for $x$.
\begin{aligned} g(x)&= \dfrac{x-3}{x-2} \\\\\\ y&= \dfrac{x-3}{x-2} &&\small{\gray{\text{Replace g(x) with y}}}\\\\\\ y(x-2) &= x-3 &&\small{\gray{\text{Multiply both sides by x-2}}}\\\\ yx-2y&= x-3 &&\small{\gray{\text{Distribute}}} \\\\ yx-x&=2y-3&&\small{\gray{\text{Group terms with x on one side}}}\\\\ x(y-1)&=2y-3&&\small{\gray{\text{Factor out an x}}}\\\\ x&=\dfrac{2y-3}{y-1} &&\small{\gray{\text{Divide both sides by y-1}}}\\\\ g^{-1}(y)&=\dfrac{2y-3}{y-1} &&\small{\gray{\text{Replace x with }g^{-1}(y)}}\end{aligned}
Since the choice of variable is arbitrary, we can now switch $y$ to $x$ to write the inverse in terms of $x$.
$g^{-1}(x)=\dfrac{2x-3}{x-1}$

### 5) Challenge problem

Match each function with the type of its inverse.
Function
Inverse type
• $f(x)=3x-5$
• $g(x)=x^3-7$
• $h(x)=\dfrac{x+2}{x-3}$
• $j(x)=\sqrt{x+2}$
$f(x)=3x-5$Linear
$g(x)=x^3-7$Cube root
$h(x)=\dfrac{x+2}{x-3}$Rational
$j(x)=\sqrt{x+2}$Quadratic