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## SectionInverse Functions

Recall that in Brief Intro to Composite and Inverse Functions we gave the following definition of an inverse function:

###### Inverse Functions

Suppose the inverse of $f$ is a function, denoted by $f^{-1}\text{.}$ Then

\begin{equation*} f^{-1}(y) = x \text{ if and only if }f(x) = y. \end{equation*}

We also stated the following property about inverse functions:

###### Functions and Inverse Functions

Suppose $f^{-1}$ is the inverse function for $f\text{.}$ Then

\begin{equation*} f^{-1}\left(f(x)\right) = x\text{ and }f\left(f^{-1}( y)\right) = y \end{equation*}

as long as $x$ is in the domain of $f\text{,}$ and $y$ is in the domain of $f^{-1}\text{.}$

In this section we will explore the invertability of a function. In other words, by the end of this section we will be able to test if a function is invertable.

### SubsectionGraph of the Inverse Function

In Example161, we used a graph of $h$ to read values of $h^{-1}\text{.}$ But we can also plot the graph of $h^{-1}$ itself. Because $C$ is the input variable for $h^{-1}\text{,}$ we plot $C$ on the horizontal axis and $F$ on the vertical axis. To find some points on the graph of $h^{-1}\text{,}$ we interchange the coordinates of points on the graph of $h\text{.}$ The graph of $h^{-1}$ is shown in Figure192.

 $C=h(F)$ $F$ $C$ $14$ $-10$ $32$ $0$ $50$ $10$ $68$ $20$
 $F=h^{-1}(C)$ $C$ $F$ $-10$ $14$ $0$ $32$ $10$ $50$ $20$ $68$ ###### Example193

The Park Service introduced a flock of $12$ endangered pheasant into a wildlife preserve. After $t$ years, the population of the flock was given by

\begin{equation*} P = f (t) = 12 + 2t^3. \end{equation*}
1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $t = f^{-1}(P)\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.
Solution
1. The graph of $f$ is shown in Figure194, with $t$ on the horizontal axis and $P$ on the vertical axis. 2. We solve $P = 12 + 2t^3$ for $t$ in terms of $P\text{.}$ \begin{align*} 2t^3 \amp = P - 12\amp\amp\text{Substract 12 from both sides.}\\ t^3 \amp = \frac{P - 12}{2}\amp\amp\text{Divide both sides by 2.}\\ t \amp = \sqrt{\frac{P - 12}{2}}\amp\amp\text{Take cube roots.} \end{align*} The inverse function is $t = f^{-1}(P) =\sqrt{\dfrac{P - 12}{2}}\text{.}$ It tells us the number of years it takes for the pheasant population to grow to size $P\text{.}$
3. The graph of $f^{-1}$ is shown in Figure195, with $P$ on the horizontal axis and $t$ on the vertical axis.

The formula $T = f(L) = 2\pi \sqrt{\dfrac{L}{32}}$ gives the period in seconds, $T\text{,}$ of a pendulum as a function of its length in feet, $L\text{.}$

1. Graph the function on the domain $[0, 5]\text{.}$
2. Find a formula for the inverse function, $L = f^{-1}(T )\text{.}$ What is the meaning of the inverse function in this context?
3. Sketch a graph of the inverse function.

### SubsectionWhen Is the Inverse a Function?

We can always find the inverse of a function $y=f(x)$ simply by solving for $x$ thus interchanging the role of the input and output variables. In the preceding examples, this process created a new function. However, this process does not always lead to be a function.

For example, to find the inverse of $y = f (x) = x^2\text{,}$ we solve for $x$ to get $x = \pm\sqrt{y}\text{.}$ When we regard $y$ as the input and $x$ as the output, the relationship does not describe a function. This can be seen as plugging in $y=4 \text{,}$ for example, gives two outputs $x=2$ and $x=-2\text{.}$ The graphs of $f$ and its inverse are shown in Figure197. (Note that for the graph of the inverse, we plot $y$ on the horizontal axis and $x$ on the vertical axis.) Because the graph of the inverse does not pass the vertical line test, it is not a function.

For many applications, it is important to know whether or not the inverse of $f$ is a function. This can be determined from the graph of $f\text{.}$ When we interchange the roles of the input and output variables, horizontal lines of the form $y = k$ become vertical lines.

Thus, if the graph of the inverse is going to pass the vertical line test, the graph of the original function must pass the horizontal line test, namely, that no horizontal line should intersect the graph in more than one point. Notice that the graph of $f(x) = x^2$ does not pass the horizontal line test, so we would not expect its inverse to be a function.

###### Horizontal Line Test

If no horizontal line intersects the graph of a function more than once, then its inverse is also a function. Figure198Notice that as the line moves up the $y-$ axis, it only ever intersects the graph in a single place. This means this function is invertible. Figure199Notice that as the line moves up the $y-$ axis, it sometimes intersects the graph in more than one place. This means this function is not invertible.

We have been talking about how to tell if the inverse of a function is also a function, but in practice this is not the language typicaly used. Usually we ask this same question in the form "Is the function invertible?" The following definition explains this relationship:

###### Invertible Function

If $y=f(x)$ is a function such that its inverse, $x=f^{-1}(y)\text{,}$ is also a function then we say that $f(x)$ is an invertible function.

###### Example200

Which of the functions in Figure201 are invertible?

Solution

In each case, apply the horizontal line test to determine whether the function is invertible. Because no horizontal line intersects their graphs more than once, the functions pictured in Figures201(a) and (c) are invertible. The functions in Figures201(b) and (d) are not invertibe.

### SubsectionMathematical Properties of the Inverse Function

The inverse function $f^{-1}$ undoes the effect of the function $f\text{.}$ In Example152, the function $f(t) = 6 + 2t$ multiplies the input by $2$ and then adds $6$ to the result. The inverse function $f^{-1}(H) = \dfrac{H -6}{2}$ undoes those operations in reverse order: It subtracts $6$ from the input and then divides the result by $2\text{.}$

If we apply the function $f$ to a given input value and then apply the function $f^{-1}$ to the output from $f\text{,}$ the end result will be the original input value. For example, if we choose $t = 5$ as an input value, we find that \begin{align*} f(\alert{5})\amp= 6 + 2(\alert{5}) = \blert{16}\amp\amp\text{ Multiply by 2, then add 6.}\\ \text{and } f^{-1}(\blert{16}) \amp = \frac{\blert{16} - 6}{2} = \alert{5}.\amp\amp\text{Subtract 6, then divide by 2.} \end{align*}

We return to the original input value, $5\text{,}$ as illustrated in Figure202.

Example203 illustrates the fact that if $f^{-1}$ is the inverse function for $f\text{,}$ then $f$ is also the inverse function for $f^{-1}\text{.}$

###### Example203

Consider the function $f(x) = x^3 + 2$ and its inverse, $f^{-1}(y) = \sqrt{y - 2}\text{.}$

1. Show that the inverse function undoes the effect of $f$ on $x = 2\text{.}$
2. Show that $f$ undoes the effect of the inverse function on $y = -25\text{.}$
Solution
1. First evaluate the function $f$ for $x = 2\text{:}$
Then evaluate the inverse function $f^{-1}$ at $y = 10\text{:}$
\begin{equation*} f^{-1}(\blert{10}) = \sqrt{\blert{10} - 2} = \sqrt{8}= \alert{2}. \end{equation*}
We started and ended with $2\text{.}$
2. First evaluate the function $f^{-1}$ for $y = -25\text{:}$
\begin{equation*} f^{-1}(\alert{-25}) = \sqrt{-25 - 2} = \blert{-3}. \end{equation*}
Then evaluate the function $f$ for $x = -3\text{:}$
\begin{equation*} f (\blert{-3}) = (\blert{-3})^3 + 2 = \alert{-25}. \end{equation*}
We started and ended with $-25\text{.}$
###### Example204
1. Find a formula for the inverse of the function $f(x)=\dfrac{2}{x - 1}\text{.}$
2. Show that $f^{-1}$ undoes the effect of $f$ on $x = 3\text{.}$
3. Show that $f$ undoes the effect of $f^{-1}$ on $y = -2\text{.}$
Solution
1. To find the inverse, we solve for $x\text{:}$

\begin{equation*} \begin{aligned} y \amp= \frac{2}{x-1} \\ y(x-1) \amp= 2 \\ x-1 \amp= \frac{2}{y} \\ x \amp =\frac{2}{y}+1 \end{aligned} \end{equation*}

Therefore $f^{-1}(y)=\frac{2}{y}+1\text{.}$

2. First evaluate the function $f$ for $x=3\text{:}$

Then evaluate the inverse function $f^{-1}$ at $y=1\text{:}$

We started and ended with $3\text{.}$

3. First evaluate the inverse function $f^{-1}$ for $y=-2\text{:}$

Then evaluate the original function $f$ at $x=0\text{:}$

We started and ended with $-2\text{.}$

### SubsectionSymmetry

So far we have been careful to keep track of the input and output variables when we work with inverse functions. This is important when we are dealing with applications; the names of the variables are usually chosen because they have a meaning in the context of the application, and it would be confusing to change them.

However, we can also study inverse functions purely as mathematical objects. There is a relationship between the graph of a function and the graph of its inverse that is easier to see if we plot them both on the same set of axes.

A graph does not change if we change the names of the variables, so we can let $x$ represent the input for both functions, and let $y$ represent the output. Consider the function $C = h(F)$ from Example161, and its inverse function, $F = h^{-1}(C)\text{.}$ The formulas for these functions are \begin{align*} C \amp = h(F) = \frac{5}{9}(F - 32)\\ F \amp = h^{-1}(C) = 32 + \frac{9}{5}C. \end{align*} But their graphs are the same if we write them as \begin{align*} y \amp = h(x) =\frac{5}{9}(x - 32)\\ y \amp= h^{-1}(x) = 32 + \frac{9}{5}x. \end{align*} The graphs are shown in Figure205.

Now, for every point $(a, b)$ on the graph of $f\text{,}$ the point $(b, a)$ is on the graph of the inverse function. Observe in Figure205 that the points $(a, b)$ and $(b, a)$ are always located symmetrically across the line $y=x\text{.}$ The graphs are symmetric about the line $y=x$, which means that if we were to place a mirror along the line $y=x\text{,}$ each graph would be the reflection of the other.

###### Example206

Graph the function $f (x) = 2\sqrt{x + 4}$ on the domain $[-4, 12]\text{.}$ Graph its inverse function $f^{-1}$ on the same grid.

Solution

The graph of $f$ has the same shape as the graph of $y = \sqrt{x}\text{,}$ shifted $4$ units to the left and stretched vertically by a factor of $2\text{.}$ Figure207a shows the graph of $f\text{,}$ along with a table of values. By interchanging the rows of the table, we obtain points on the graph of the inverse function, shown in Figure207b.

If we use $x$ as the input variable for both functions, and $y$ as the output, we can graph $f$ and $f^{-1}$ on the same grid, as shown in Figure208. The two graphs are symmetric about the line $y = x\text{.}$ Graph the function $f (x) = x^3 + 2$ and its inverse $f^{-1}(x) = \sqrt{x - 2}$ on the same set of axes, along with the line $y = x\text{.}$