# Inverse Functions

An **inverse function** is the "reversal" of another function; specifically, the inverse will swap input and output with the original function.

Given a function $f(x)$, the inverse is written $f^{-1}(x)$, but this should not be read as a negative exponent. Generally speaking, the inverse of a function is not the same as its reciprocal.

Understanding (and keeping straight) inverse functions and reciprocal functions comes down to understanding operations, identities, and inverses more broadly. You are already very familiar with these things from basic arithmetic. Given an object $x$ (say, a real number or a function), what we deem its inverse $x^{-1}$ will depend on the operation. Two objects are inverses if we can apply the operation to them and get back the identity element for that operation. Here are some examples:

If the operation is *addition*, the identity element is 0 because *adding* 0 to any object $x$ yields $x$ again: $0 + x = x$ and $x+0=x$. Thus, two objects are *additive inverses* if they can be *added* together to yield 0. Given a number or function $x$, the additive inverse is always $-x$.

If the operation is *multiplication*, the identity element is 1, because *multiplying* any other object $x$ with 1 yields $x$ again: $1 \cdot x = x$ and $x \cdot 1 = x$. Therefore, two objects are *multiplicative inverses* if they can be multiplied together to yield 1. Given a nonzero number or function $x,$ the multiplicative inverse is always $1/x$, otherwise known as the *reciprocal*.

In the case of functional inverses, the operation is *function composition*. It is less obvious now, but the identity element for composition is $x$. To see this, observe that if $g(x)$ is composed with the *function* $x$, we get $g(x)\circ (x) = g((x)) =g(x)$ and $x \circ g(x) = g(x)$. Thus, given functions $f(x)$ and $g(x)$, $g(x)$ is the inverse of $f(x)$ if and only if $g(f(x))=x$ and $f(g(x))=x$. In this case, we write $g(x)=f^{-1}(x)$.

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## Basic Example

Consider the following function:

$x$ | $f(x)$ | $x$ | $f^{-1}(x)$ | |

1 | 4 | 4 | 1 | |

2 | 2 | 2 | 2 | |

3 | 2 | 2 | 3 | |

4 | 1 | 1 | 4 | |

5 | 7 | 7 | 5 |

To determine $f^{-1} (1)$, we need to find all values of $x$ such that $f(x) = 1$. From the table, this only happens when $x = 4 ,$ so $f^{-1} (1) = 4$.

To determine $f^{-1} (2)$, we need to find all values of $x$ such that $f(x) = 2$. From the table, this occurs with $f(2) = 2, f(3) = 2$. Thus, $f^{-1} (2) = \{ 2, 3 \}$.

Hence, the inverse of a function need not be a function itself, since it could take on multiple values.

## Finding the Inverse of a Function Graphically

Given the point $(x,y)$ on the graph of $f$, we know that the point $(y,x)$ is on the graph of $f^{-1}$. These two points are related by being symmetric in the line $y =x$. Hence, the graph of $y = f^{-1}(x)$ is the reflection of the graph of $y = f(x)$ about the line $y = x$.

This offers us a way to graph the inverse function (if the axes are drawn to the same scale).

1) First, we fold the piece of paper along the line $y = x$.

2) Next, we trace out the graph of $f(x)$ to leave an imprint on the other side.

3) Finally, we draw out the imprint, which will give us the graph of $f ^ {-1} (x)$.

Let's look at an example.

## In the figure below, the blue curve is the graph of the function $y = x^2 - 4x + 3$, and the black line is $y = x$. Draw the graph of the inverse of the blue curve.

1) Since both the axes are drawn to the same scale, so we can simply fold the graph about $y = x$ to get the red part.

2) Now we trace the rest of the graph to get the complete graph of the inverse.

3) The red curve is the inverse of the blue curve. Note that the curves are reflections of each other about the line $y = x$. We have found the inverse of the function graphically!

## When is the inverse a function?

As mentioned, the inverse of a function need not be a function. In order for it to be a function, given any $y$ value, there is at most one $x$ value such that $f(x) = y$. This means that when we look at a horizontal line $y = c$, it cuts the graph in at most 1 point. This is also known as the *horizontal line test*.

More formally, we say that

The inverse of the function $f$ is a function, if and only if $f$ is a bijective function. $_\square$

If $f$ weren't injective, then there would exist an $f(x)$ for two values of $x$, which we call $x_1$ and $x_2$. Then $f^{-1}(f(x))$ will have two values $x_1$ and $x_2$. However, a function cannot have two outputs for one input. Then it is no longer a function. Therefore, $f$ has to be injective.

If $f$ is not surjective, then there would exist a value $k \in X$ such that $f(x) = k$ has no real solution for $x$. This is because there exists some value which belongs to the codomain of the function, but not in the range of the function. Then $f^{-1}(k)$ would not be defined as $k$ is not in the range of $f^{-1}$. There must be no such value in the domain of $f^{-1}$ for which $f^{-1}$ is undefined. Therefore the range of $f$ must be equal to the codomain of $f$. Therfore $f$ must be a surjective function.

Since $f$ must be both an injective and a surjective function, it is a bijective function. $_\square$

## Finding the Inverse of a Function Algebraically

To find the inverse of $f$, we take the following steps:

1) Check that it satisfies the horizontal line test.

2) Substitute $x$ with $f^{-1} (x)$ and $f(x)$ with $x$. Make $f^{-1}(x)$ the subject of the formula.

3) Determine the domain of $f ^ {-1}$, which is the codomain of $f$. Note that this might not be the range of $f$.

4) Determine the range of $f^{-1}$, which is the domain of $f$.

Let's work through the following examples:

## Find the inverse of $f: \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = 2x + 8 .$

Step 1: This graph satisfies the horizontal line test.

Step 2: The substitution gives us $x = 2 f^{-1} (x) + 8$.

Make $f^{-1} (x)$ the subject of the formula by subtracting 8 and dividing by 2, to get $f^{-1} (x) = \frac{ x- 8 } { 2 }$.Step 3: The domain of $f^{-1} (x)$ is the codomain of $f$. From the graph, this is $\mathbb{R}$.

Step 4: The range of $f^{-1} (x)$ is the domain of $f$, which is $\mathbb{R}$.

Hence, the inverse of $f$ is

$f^{-1} : \mathbb{R} \rightarrow \mathbb{R}, \quad f^{-1} (x) = \frac{ x - 8 } { 2 } .\ _\square$

## Find the inverse of $g: [0, \infty) \rightarrow [0,\infty), \quad g(x) = \sqrt{ \frac{x}{2} }.$

Step 1: This graph satisfies the horizontal line test.

Step 2: The substitution gives us $x = \sqrt{ \frac { g^{-1} (x)}{2} }$.

Make $g^{-1} (x)$ the subject of the formula by squaring and multiplying by 2, to get $g^{-1} (x) = 2x^2$.Step 3: The domain of $g^{-1} (x)$ is the codomain of $g$. From the graph, this is $[0, \infty )$.

Step 4: The range of $g^{-1} (x)$ is the domain of $g$, which is $[0, \infty )$.

Hence, the inverse of $g$ is $g^{-1} : [0, \infty) \rightarrow [0, \infty ) , \quad g^{-1} (x) = 2x^2 .\ _\square$

In this example, we see why it is important to determine the codomain of the original function, instead of relying just on the given range.

## Find the inverse of

$h : [-1, 0] \rightarrow [0,1], \quad h(x) = - x^2.$

Step 1: This graph satisfies the horizontal line test.

Step 2: The substitution gives us $x = - [h^{-1} (x) ] ^2$.

For us to make $h^{-1} (x)$ the subject of the formula, we have to take square roots and obtain that $h ^ {-1} (x) = \pm \sqrt{ - x }$. It is important for us to determine which square root we want. In this case, since $h (-1) = 1$, it indicates that we want the negative square root.Step 3: The domain of $h^{-1} (x)$ is the codomain of $h$. From the graph, this is $[0, 1 ]$.

Step 4: The range of $h^{-1} (x)$ is the domain of $h$, which is $[-1,0]$.

Hence, the inverse of $h$ is $h^{-1} : [0, 1] \rightarrow [-1,0] , \quad h^{-1} (x) = - \sqrt{-x}.\ _\square$

## Find the inverse of $f: \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = 4e^{2x} + 3.$

Step 1: This graph satisfies the horizontal line test.

Step 2: The substitution gives us $x = 4 e^{ 2 f^{-1} (x) } + 3$.

Making $f^{-1} (x)$ the subject of the formula gives us $f^{-1} (x) = \frac {\ln \frac{ x-3} { 4} } { 2} = \ln \sqrt { \frac{ x-3} { 4} }$.Step 3: The domain of $f^{-1} (x)$ is the codomain of $f$. From the graph, this is $[3, \infty ]$.

Step 4: The range of $f^{-1} (x)$ is the domain of $f$, which is $\mathbb{R}$.

Hence, the inverse of $f$ is $f^{-1} : (3, \infty) \rightarrow \mathbb{R} , \quad f^{-1} (x) = \ln \sqrt{ \frac{ x-3} { 4} }.\ _\square$

## Find the inverse of $f: (-\infty,0] \rightarrow [0,\infty), \quad f(x) = 2\ln(x^2+1).$

Step 1: This graph satisfies the horizontal line test.

Step 2: The substitution gives us $x = 2\ln\left(\left(f^{-1} (x)\right)^2+1\right)$.

Making $f^{-1} (x)$ the subject of the formula gives us $f^{-1} (x) =\pm\sqrt{e^{\frac{1}{2}x}-1}$. Again, we have to determine which square root we want. Since $f(-1)=2\ln2\Rightarrow f^{-1}(2\ln2)=-1,$ the square root we are looking for is the negative one.Step 3: The domain of $f^{-1} (x)$ is the codomain of $f$. From the graph, this is $[0, \infty )$.

Step 4: The range of $f^{-1} (x)$ is the domain of $f$, which is $(-\infty,0]$.

Hence, the inverse of $f$ is $f^{-1} : [0, \infty) \rightarrow (-\infty,0], \quad f^{-1} (x) =-\sqrt{e^{\frac{1}{2}x}-1}.\ _\square$

## Three functions $f,g$, and $h$ satisfy $h(x)=g\left(f(x)\right)$. If we are given:

$\begin{aligned} h(x)&=x^2+2x-4\\ g(x)&=3e^x-2, \end{aligned}$ what is the equation for $f(x)$?

From the given relationship between the three functions, we have

$\begin{aligned} h(x)&=g\left(f(x)\right)\\ \Rightarrow f(x)&=g^{-1}\left(h(x)\right). \end{aligned}$

We can find $g^{-1}$ using the same process we have been using, as shown below:

Step 1: This graph satisfies the horizontal line test.

Step 2: The substitution gives us $x =3e^{g^{-1}(x)}-2$. Making $g^{-1}(x)$ the subject of the formula gives $g^{-1}(x)=\ln\frac{x+2}{3}$.

Step 3: The domain of $g^{-1} (x)$ is the codomain of $g$. From the graph, this is $[-2, \infty )$.

Step 4: The range of $g^{-1} (x)$ is the domain of $g$, which is $\mathbb{R}$.

Hence, the inverse of $g$ is $g^{-1} : [-2, \infty) \rightarrow \mathbb{R}, \quad g^{-1} (x) =\ln\frac{x+2}{3}.$

Now we can find $f$ as follows: $f(x)=g^{-1}\left(h(x)\right)=\ln\frac{h(x)+2}{3}=\ln\frac{x^2+2x-2}{3}.\ _\square$

## Advanced Problems

## The function $g : A \rightarrow B$ is defined as $g(x) = x^2 - 6x + 8$. The set $A$ contains the numbers in the interval $(-\infty, a]$. Find the maximum possible value of $a$ and the set $B$ such that $g^{-1}$ exists.

For $g^{-1}$ to exist, $g$ must be bijective. The maximum value of $a$ such that $g$ is injective is the $x$-coordinate of the vertex point $= -\frac{-6}{2} = 3$. Therefore $A = (-\infty, 3]$ and $a=3.$

For $g$ to be surjective, $B$ must be equal to its range, which is $[g(3), \infty)=[-1, \infty). \ _\square$

## Find the expression for $g^{-1}(x)$ from the previous example, and also its domain and range.

The domain of $g^{-1}$ is equal to the range of $g$, which is $[-1, \infty)$ as calculated in the previous example. The range of $g^{-1}$ is equal to the domain of $g$, which is equal to $(-\infty, 3]$.

To find $g^{-1}$ we make substitutions

$x = \left(g^{-1}(x)\right)^2 - 6 g^{-1}(x) + 8.$

We need to make $g^{-1}$ the subject of the formula. The easiest way to do this would be to make it a perfect square:

$\begin{aligned} x &= \left(g^{-1}(x)\right)^2 - 6 g^{-1}(x) + 9 -1\\ &= \left(g^{-1}(x) - 3\right)^2 - 1\\ \left(g^{-1}(x) - 3\right)^2 &= x + 1\\ g^{-1}(x) -3 &= \pm \sqrt{x + 1} \\ g^{-1}(x) &= 3 \pm \sqrt{x + 1}. \end{aligned}$

However, this is not the final answer. This is not even a function. If we consider the calculated range of $g^{-1}$, we realize that we have to take only $g^{-1}(x) = 3 - \sqrt{x+1}$ and not the other (positive) case. Its domain and range agree with the above calculated ones, and it is definitely the inverse of $g$. This can be verified by finding the expression of $g\left(g^{-1}(x)\right)$ which should be equal to $x$. $_\square$

**Cite as:**Inverse Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/functions-inverses/