Example: Interpreting the Inverse of a Tabular Function

A function $f\left(t\right)$ is given below, showing distance in miles that a car has traveled in $t$ minutes. Find and interpret ${f}^{-1}\left(70\right)$.

$t\text{ (minutes)}$	30	50	70	90
$f\left(t\right)\text{ (miles)}$	20	40	60	70

Answer: The inverse function takes an output of $f$ and returns an input for $f$. So in the expression ${f}^{-1}\left(70\right)$, 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function $f$, 90 minutes, so ${f}^{-1}\left(70\right)=90$. The interpretation of this is that, to drive 70 miles, it took 90 minutes. Alternatively, recall that the definition of the inverse was that if $f\left(a\right)=b$, then ${f}^{-1}\left(b\right)=a$. By this definition, if we are given ${f}^{-1}\left(70\right)=a$, then we are looking for a value $a$ so that $f\left(a\right)=70$. In this case, we are looking for a $t$ so that $f\left(t\right)=70$, which is when $t=90$.

Example: Evaluating a Function and Its Inverse from a Graph at Specific Points

A function $g\left(x\right)$ is given below. Find $g\left(3\right)$ and ${g}^{-1}\left(3\right)$.

Answer: To evaluate $g\left(3\right)$, we find 3 on the x-axis and find the corresponding output value on the y-axis. The point $\left(3,1\right)$ tells us that $g\left(3\right)=1$. To evaluate ${g}^{-1}\left(3\right)$, recall that by definition ${g}^{-1}\left(3\right)$ means the value of x for which $g\left(x\right)=3$. By looking for the output value 3 on the vertical axis, we find the point $\left(5,3\right)$ on the graph, which means $g\left(5\right)=3$, so by definition, ${g}^{-1}\left(3\right)=5$.

Example: Inverting the Fahrenheit-to-Celsius Function

Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature.

$C=\frac{5}{9}\left(F - 32\right)$

Answer:

$\begin{array}{c}\hfill{ C }=\frac{5}{9}\left(F - 32\right)\hfill \\ C\cdot \frac{9}{5}=F - 32\hfill \\ F=\frac{9}{5}C+32\hfill \end{array}$

By solving in general, we have uncovered the inverse function. If

$C=h\left(F\right)=\frac{5}{9}\left(F - 32\right)$,

then

$F={h}^{-1}\left(C\right)=\frac{9}{5}C+32$.

In this case, we introduced a function $h$ to represent the conversion because the input and output variables are descriptive, and writing ${C}^{-1}$ could get confusing.

Example: Solving to Find an Inverse Function

Find the inverse of the function $f\left(x\right)=\frac{2}{x - 3}+4$.

Answer: $$\begin{array}{l}y=\frac{2}{x - 3}+4\hfill & \text{Set up an equation}.\hfill \\ y - 4=\frac{2}{x - 3}\hfill & \text{Subtract 4 from both sides}.\hfill \\ x - 3=\frac{2}{y - 4}\hfill & \text{Multiply both sides by }x - 3\text{ and divide by }y - 4.\hfill \\ x=\frac{2}{y - 4}+3\hfill & \text{Add 3 to both sides}.\hfill \end{array}$$ So ${f}^{-1}\left(y\right)=\frac{2}{y - 4}+3$ or ${f}^{-1}\left(x\right)=\frac{2}{x - 4}+3$.

Analysis of the Solution

The domain and range of $f$ exclude the values 3 and 4, respectively. $f$ and ${f}^{-1}$ are equal at two points but are not the same function, as we can see by creating the table below.

$x$	1	2	5	${f}^{-1}\left(y\right)$
$f\left(x\right)$	3	2	5	$y$

Example: Solving to Find an Inverse with Radicals

Find the inverse of the function $f\left(x\right)=2+\sqrt{x - 4}$.

Answer: $$\begin{array}{l}y=2+\sqrt{x - 4}\hfill \\ {\left(y - 2\right)}^{2}=x - 4\hfill \\ x={\left(y - 2\right)}^{2}+4\hfill \end{array}$$ So ${f}^{-1}\left(x\right)={\left(x - 2\right)}^{2}+4$. The domain of $f$ is $\left[4,\infty \right)$. Notice that the range of $f$ is $\left[2,\infty \right)$, so this means that the domain of the inverse function ${f}^{-1}$ is also $\left[2,\infty \right)$.

Analysis of the Solution

The formula we found for ${f}^{-1}\left(x\right)$ looks like it would be valid for all real $x$. However, ${f}^{-1}$ itself must have an inverse (namely, $f$ ) so we have to restrict the domain of ${f}^{-1}$ to $\left[2,\infty \right)$ in order to make ${f}^{-1}$ a one-to-one function. This domain of ${f}^{-1}$ is exactly the range of $f$.

Example: Finding the Inverse of a Function Using Reflection about the Identity Line

Given the graph of $f\left(x\right)$, sketch a graph of ${f}^{-1}\left(x\right)$.

Answer: This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of $\left(0,\infty \right)$ and range of $\left(-\infty ,\infty \right)$, so the inverse will have a domain of $\left(-\infty ,\infty \right)$ and range of $\left(0,\infty \right)$. If we reflect this graph over the line $y=x$, the point $\left(1,0\right)$ reflects to $\left(0,1\right)$ and the point $\left(4,2\right)$ reflects to $\left(2,4\right)$. Sketching the inverse on the same axes as the original graph gives us the result in the graph below.