The inverse function reverses which quantity is input and which quantity is output, so if $f(2)=4$, then $f^{-1}(4)=2$.

Alternatively, if you want to re-name the inverse function $g(x)$, then $g(4)=2$.

We would expect $f^{-1}$ to swap the domain and range of $f$, so $f^{-1}$ would have domain $(0,\infty)$ and range $(−\infty,\infty)$.

The inverse function takes an output of $f$ and returns an input for $f$. So in the expression $f^{-1}(70)$, the 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 $f^{-1}(70)=90$. Interpreting this, it means that to drive 70 miles, it took 90 minutes.

Alternatively, recall the definition of the inverse was that if $f(a)=b$ then $f^{-1}(b)=a$. By this definition, if you are given $f^{-1}(70)=a$ then you are looking for a value a so that $f(a)=70$. In this case, we are looking for a $t$ so that $f(t)=70$, which is when $t=90$.

To evaluate $g(3)$, we find 3 on the horizontal axis and find the corresponding output value on the vertical axis. The point $(3, 1)$ tells us that $g(3)=1$.

To evaluate $g^{-1}(3)$, recall that by definition $g^{-1}(3)$ means $g(x)=3$. By looking for the output value 3 on the vertical axis we find the point $(5, 3)$ on the graph, which means $g(5)=3$, so by definition

$$g^{-1}(3)=5.$$

A quick Google search would find the inverse function, but alternatively, Betty might look back at how she solved for the Fahrenheit temperature for a specific Celsius value, and repeat the process in general

$$C=\frac{5}{9}(f - 32)$$ $$C \cdot \frac{9}{5}=f - 32$$ $$f=\frac{9}{5}C + 32$$

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

$$C=h(f)=\frac{5}{9}(f - 32)$$

Then

$$f=h^{- 1}(C)=\frac{9}{5}C + 32$$

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

The graph clearly fails the horizontal line test. We want to choose a portion of the function that still gives us all possible output values (from $[0, \infty)$), but leaves us with a one-to-one portion of the graph.

We can limit the domain to $[0, \infty)$ as pictured here, and find an inverse on this limited domain.

You may have already guessed that since we undo a square with a square root, the inverse of $h(x)=x^2$ on this domain is $h^{- 1}(x) = \sqrt{x}$.

You can also solve for the inverse function algebraically. If $h(x)=x^2$, we can introduce the variable $y$ to represent the output values, allowing us to write $y=x^2$. To find the inverse we solve for the input variable

To solve for x we take the square root of each side.

$$\sqrt{y}=\sqrt{x^{2}}$$

and get

$$\sqrt{y}=|x|,$$

so

$$x=\pm \sqrt{y}.$$

We have restricted $x$ to being non-negative, so we’ll use the positive square root, $x=\sqrt{y}$ or $h^{- 1}(y)=\sqrt{y}$. In cases like this where the variables are not descriptive, it is common to see the inverse function rewritten with the variable x: $h^{- 1}(x)=\sqrt{x}$. Rewriting the inverse using the variable x is often required for graphing inverse functions using calculators or computers.

Note that the domain and range of the square root function do correspond with the range and domain of the quadratic function on the limited domain. In fact, if we graph h(x) on the restricted domain and $h^{-1}(x)$ on the same axes, we can notice symmetry: the graph of $h^{-1}(x)$ is the graph of  $h(x)$ reflected over the line $y=x$.