The inverse function takes an output of [latex]f[/latex] and returns an input for [latex]f[/latex]. So in the expression [latex]{f}^{-1}\left(70\right)[/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[/latex], 90 minutes, so [latex]{f}^{-1}\left(70\right)=90[/latex]. 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 [latex]f\left(a\right)=b[/latex], then [latex]{f}^{-1}\left(b\right)=a[/latex]. By this definition, if we are given [latex]{f}^{-1}\left(70\right)=a[/latex], then we are looking for a value [latex]a[/latex] so that [latex]f\left(a\right)=70[/latex]. In this case, we are looking for a [latex]t[/latex] so that [latex]f\left(t\right)=70[/latex], which is when [latex]t=90[/latex].

To evaluate [latex]g\left(3\right)[/latex], we find 3 on the x-axis and find the corresponding output value on the [latex]y[/latex]-axis. The point [latex]\left(3,1\right)[/latex] tells us that [latex]g\left(3\right)=1[/latex].

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

[latex]{ C }=\frac{5}{9}\left(F - 32\right)[/latex]
[latex]C\cdot \frac{9}{5}=F - 32[/latex]
[latex]F=\frac{9}{5}C+32[/latex]

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

[latex]C=h\left(F\right)=\frac{5}{9}\left(F - 32\right)[/latex],

then

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

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

[latex]\begin{align}&y=\frac{2}{x - 3}+4 && \text{Change }f(x)\text{ to }y. \\[1.5mm]&x=\frac{2}{y - 3}+4 && \text{Switch }x\text{ and }y. \\[1.5mm] &y - 4=\frac{2}{x - 3} && \text{Subtract 4 from both sides}. \\[1.5mm] &y - 3=\frac{2}{x - 4} && \text{Multiply both sides by }y - 3\text{ and divide by }x - 4. \\[1.5mm] &y=\frac{2}{x - 4}+3 && \text{Add 3 to both sides}.\\[-3mm]&\end{align}[/latex]

So [latex]{f}^{-1}\left(x\right)=\dfrac{2}{x - 4}+3[/latex].

Analysis of the Solution

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

[latex]\begin{align}&y=2+\sqrt{x - 4}\\[1.5mm]&x=2+\sqrt{y - 4}\\[1.5mm] &{\left(x - 2\right)}^{2}=y - 4 \\[1.5mm] &y={\left(x- 2\right)}^{2}+4 \end{align}[/latex]

So [latex]{f}^{-1}\left(x\right)={\left(x - 2\right)}^{2}+4[/latex].

The domain of [latex]f[/latex] is [latex]\left[4,\infty \right)[/latex]. Notice that the range of [latex]f[/latex] is [latex]\left[2,\infty \right)[/latex], so this means that the domain of the inverse function [latex]{f}^{-1}[/latex] is also [latex]\left[2,\infty \right)[/latex].

Analysis of the Solution

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

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 [latex]\left(0,\infty \right)[/latex] and range of [latex]\left(-\infty ,\infty \right)[/latex], so the inverse will have a domain of [latex]\left(-\infty ,\infty \right)[/latex] and range of [latex]\left(0,\infty \right)[/latex].

If we reflect this graph over the line [latex]y=x[/latex], the point [latex]\left(1,0\right)[/latex] reflects to [latex]\left(0,1\right)[/latex] and the point [latex]\left(4,2\right)[/latex] reflects to [latex]\left(2,4\right)[/latex]. Sketching the inverse on the same axes as the original graph gives us the result in the graph below.

The function and its inverse, showing reflection about the identity line