Skip to main content

Section 4.1 Inverse Functions

Recall that in Brief Intro to Composite and Inverse FunctionsΒ 2.2 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 invertibility of a function. In other words, by the end of this section we will be able to test if a function is invertible.

Subsection 4.1.1 Graph of the Inverse Function

In ExampleΒ 2.25, 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 FigureΒ 4.1.
\(C=h(F)\)
\(F\) \(C\)
\(14\) \(-10\)
\(32\) \(0\)
\(50\) \(10\)
\(68\) \(20\)
\(F=h^{-1}(C)\)
\(C\) \(F\)
\(-10\) \(14\)
\(0\) \(32\)
\(10\) \(50\)
\(20\) \(68\)
described in detail following the image
graph of Fahrenheit vs Celsius
Figure 4.1.

Example 4.2.

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 FigureΒ 4.3, with \(t\) on the horizontal axis and \(P\) on the vertical axis.
    described in detail following the image
    graph of pheasant population
    Figure 4.3.
  2. We solve \(P = 12 + 2t^3\) for \(t\) in terms of \(P\text{.}\)
    \begin{align*} 2t^3 \amp = P - 12\amp\amp\text{Subtract 12 from both sides.}\\ t^3 \amp = \frac{P - 12}{2}\amp\amp\text{Divide both sides by 2.}\\ t \amp = \sqrt[3]{\frac{P - 12}{2}}\amp\amp\text{Take cube roots.} \end{align*}
    The inverse function is \(t = f^{-1}(P) =\sqrt[3]{\dfrac{P - 12}{2}}\text{.}\) 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 FigureΒ 4.4, with \(P\) on the horizontal axis and \(t\) on the vertical axis.
    described in detail following the image
    graph of time to achieve pheasant population
    Figure 4.4.

Checkpoint 4.5.

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, \(L\text{.}\)

(b)

Find a formula for the inverse function, \(L = f^{-1}(T )\text{.}\) What is the meaning of the inverse function in this context?

Subsection 4.1.2 When 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 FigureΒ 4.6. (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.
described in detail following the image
graphs of x-squared and its inverse
Figure 4.6.
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.
described in detail following the image
Horizontal line test
Figure 4.7. Notice 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.
described in detail following the image
Horizontal line test
Figure 4.8. Notice 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 typically 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.

Example 4.9.

Subsection 4.1.3 Mathematical Properties of the Inverse Function

The inverse function \(f^{-1}\) undoes the effect of the function \(f\text{.}\) In ExampleΒ 2.16, 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*}
described in detail following the image
function and inverse diagram
Figure 4.11.
We return to the original input value, \(5\text{,}\) as illustrated in FigureΒ 4.11.
ExampleΒ 4.12 illustrates the fact that if \(f^{-1}\) is the inverse function for \(f\text{,}\) then \(f\) is also the inverse function for \(f^{-1}\text{.}\)

Example 4.12.

Consider the function \(f(x) = x^3 + 2\) and its inverse, \(f^{-1}(y) = \sqrt[3]{y - 2}\text{.}\)
  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 \(x = 2\text{:}\)
    \begin{equation*} f(\alert{2}) = \alert{2}^3 + 2 = \blert{10}. \end{equation*}
    Then evaluate the inverse function \(f^{-1}\) at \(y = 10\text{:}\)
    \begin{equation*} f^{-1}(\blert{10}) = \sqrt[3]{\blert{10} - 2} = \sqrt[3]{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[3]{-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 \(-25\text{.}\)
ExampleΒ 4.13 illustrates finding an inverse of a function. The general process is to move operations (or undo them) so the input variable (\(x\)) is what is alone on one side of the equation.

Example 4.13.

  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 \(x=3\text{:}\)
    \begin{equation*} f(\alert{3})=\dfrac{2}{\alert{3}-1}=\blert{1} \end{equation*}
    Then evaluate the inverse function \(f^{-1}\) at \(y=1\text{:}\)
    \begin{equation*} f^{-1}(\blert{1})=\frac{2}{\blert{1}}+1=\alert{3} \end{equation*}
    We started and ended with \(3\text{.}\)
  3. First evaluate the inverse function \(f^{-1}\) for \(y=-2\text{:}\)
    \begin{equation*} f^{-1}(\alert{-2})=\frac{2}{\alert{-2}}+1=\blert{0} \end{equation*}
    Then evaluate the original function \(f\) at \(x=0\text{:}\)
    \begin{equation*} f(\blert{0})=\frac{2}{\blert{0}-1}=\alert{-2} \end{equation*}
    We started and ended with \(-2\text{.}\)

Subsection 4.1.4 Symmetry

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 ExampleΒ 2.25, 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 FigureΒ 4.14.
described in detail following the image
graphs of both function and inverse
Figure 4.14.
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 FigureΒ 4.14 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.

Example 4.15.

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 \(2\text{.}\) FigureΒ 4.16a 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 FigureΒ 4.16b.
described in detail following the image
graph and table of a function
Figure 4.16.
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 FigureΒ 4.17. The two graphs are symmetric about the line \(y = x\text{.}\)
described in detail following the image
function and inverse on the same grid
Figure 4.17.

Checkpoint 4.18.

Graph the function \(f (x) = x^3 + 2\) and its inverse \(f^{-1}(x) = \sqrt[3]{x - 2}\) on the same set of axes, along with the line \(y = x\text{.}\)

Exercises 4.1.5 Exercises

1. Invertible or Not.

a) Let \(f\) be the function that sends a person to their favorite color. Does \(f\) have an inverse?
b) Let \(g\) be the function that sends a student at UNL to their NUID. Does \(g\) have an inverse?

2. Invertible or Not.

Determine whether the functions below are invertible. If so, find a formula for the inverse function. If not, enter ’NONE’ in the answer blank.
a) \(y = a(x)=-4x+1\)
\(\hspace{3mm} a^{-1} (y) =\)
b) \(y = b(x)=x^4\)
\(\hspace{3mm} b^{-1} (y) =\)
c) \(y = c(x)=\sqrt{x^2+3}\)
\(\hspace{3mm} c^{-1} ( y) =\)
d) \(y = d(x)=\frac{2x-1}{x}\)
\(\hspace{3mm} d^{-1} (y) =\)

3. Interpretting Inverses.

Suppose a company’s profit is a linear function of the number of units of their product that are sold. Selling 4,000 units yields a profit of $16200, and selling 6,000 units yields a profit of $22200.
a) Find a formula for \(P=f(x)\text{,}\) the company’s profit if \(x\) units are sold.
\(f(x) =\)
b) As a real-world interpretation, the slope is:
As a real-world interpretation, the \(P\)-intercept is:
c) Find a formula for \(f^{-1}(P)\text{.}\)
\(f^{-1} (P) =\)
d) Evaluate \(f^{-1}(50,000)\text{.}\)
\(f^{-1}(50,000) =\)
e) What is a valid real-world interpretation of \(f^{-1}(50,000)\text{?}\)
  • the number of units the company must sell to have a profit of 50,000 dollars
  • the profit the company makes by selling 50,000 units
  • the inverse of the function does not have a logical real-world interpretation

4. Interpretting Inverses.

Let the population of the city Metropolis (in millions) be given by \(P=f(t)=0.005 t^3 +2.1\text{,}\) where \(t\) is the number of years since 1982.
a) Evaluate \(f(35)\text{.}\)
\(f(35) =\)
What is a valid interpretation of \(f(35)\text{?}\)
  • the population of Metropolis in millions 35 years after 1982
  • the number of years after 1982 when the population will be 35 million
  • the number of people that move to Metropolis every year in millions
b) Find a formula for \(f^{-1}(P)\text{.}\) [Be sure to use the variable P in your formula].
\(f^{-1}(P) =\)
c) Evaluate \(f^{-1}(35)\text{.}\)
\(f^{-1}(35) =\)
What is a valid interpretation of \(f^{-1}(35)\text{?}\)
  • the population of Metropolis 35 years after 1982
  • the number of years after 1982 at which point the population will be 35 million
  • the inverse function does not have a real-world interpretation

5. Matching Inverse Functionss.

For this question, consider only the positive square root of the function if applicable.
Consider the following functions. Match the function to its inverse (if the inverse is one of the other functions). If the function has no inverse or if the inverse is not one of the other functions, select N/A.
\begin{equation*} f(x)=-6+4 x^{3} \hspace{0.25 in} g(x)=\sqrt[3]{\frac{x-6}{4}} \hspace{0.25 in} h(x)=\sqrt[3]{\frac{x}{4}}-6 \end{equation*}
\begin{equation*} p(x)=6+4 x^{3} \hspace{0.25 in} q(x)=\sqrt[3]{\frac{x+6}{4}} \hspace{0.25 in} r(x)=\sqrt[3]{\frac{x}{4}}+6 \end{equation*}

6. Interpretting Inverses.

A rock is dropped into the ocean from a boat. The depth, in feet, of the rock at time \(t\) can be modeled by \(d = D(t)=- 8 t^2+224 t\) for \(0\leq t\leq 14\text{,}\) where \(t\) is time in minutes since the rock entered the water.
a) Evaluate \(D(14)\)
We can interpret \(D(14)\) in real-world terms as:
b) Evaluate \(D^{-1}(700)\)
We can interpret \(D^{-1} (700)\) in real-world terms as:
c) An interpretation of \(1568 -D(t)\) in real-world terms is:

7. Interpretting Inverses.

a) Suppose the temperature of the water, \(F=T(d)\text{,}\) in degrees Fahrenheit, is decreasing exponentially as a scuba diver descends on a dive to the bottom of the ocean, where \(d\) is her depth in feet below the surface of the water. In particular, \(T(0)=85\text{,}\) and \(T(360)=72\text{.}\)
a) Find an exponential function modeling \(T(d)\) (round your growth factor to at least four decimal places).
\(T(d) =\)
b) Which of the following is a formula for \(T^{-1}(F)\text{?}\)
c) Which function would you use if you wanted to know at what depth the temperature of the water is 60 degrees Fahrenheit?

8. Inverting Steps.

Let \(y = f(x) = 6\sqrt{x+1}-4\text{.}\) When we are computing what happens to \(x\text{,}\) we first add 1 to \(x\text{,}\) then take the square root, and so on. Represent this chain of computations visually by labeling the arrows in the diagram below. (In the possible answers below, let ’?’ be the result from the previous step.)
A.1
A.2
A.3
A.4
b) Now, reverse these steps. This is essentially solving for \(y\text{.}\)
A.4 in Reverse
A.3 in Reverse
A.2 in Reverse
A.1 in Reverse
c) Using the steps you identified in part b), write a formula for \(f^{-1}(y)\text{.}\)
\(f^{-1}(y) =\)