3.8: Inverses and Radical Functions

Finding the Inverse of a Polynomial Function

Two functions \(f\) and \(g\) are inverse functions if for every coordinate pair in \(f\), \((a,b)\), there exists a corresponding coordinate pair in the inverse function, \(g\), \((b, a)\). In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test.

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure \(\PageIndex{2}\). We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Figure \(\PageIndex{2}\)

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with \(x\) measured horizontally and \(y\) measured vertically, with the origin at the vertex of the parabola (Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\)

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form \(y(x)=ax^2\). Our equation will need to pass through the point \((6, 18)\), from which we can solve for the stretch factor \(a\).

\[ \begin{align*} 18&=a6^2 \\[4pt] a &=\dfrac{18}{36} \\[4pt] &=\dfrac{1}{2} \end{align*}\]

Our parabolic cross section has the equation

\(y(x)=\dfrac{1}{2}x^2\)

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth \(y\), the width will be given by \(2x\), so we need to solve the equation above for \(x\) and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive \(x\) values. On this domain, we can find an inverse by solving for the input variable:

\[ \begin{align*} y&=\dfrac{1}{2}x^2 \\[4pt] 2y&=x^2 \\[4pt] x&=\pm \sqrt{2y} \end{align*}\]

This is not a function as written. We are limiting ourselves to positive \(x\) values, so we eliminate the negative solution, giving us the inverse function we’re looking for.

\(y=\dfrac{x^2}{2}\), \(x>0\)

Because \(x\) is the distance from the center of the parabola to either side, the entire width of the water at the top will be \(2x\). The trough is 3 feet (36 inches) long, so the surface area will then be:

\[ \begin{align*} \text{Area} &=l⋅w \\[4pt] &=36⋅2x \\[4pt] &=72x \\[4pt] &=72\sqrt{2y} \end{align*}\]

This example illustrates two important points:

1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions.

Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation \(f^{−1}(x)\).

Warning: \(f^{−1}(x)\) is not the same as the reciprocal of the function \(f(x)\). This use of “–1” is reserved to denote inverse functions. To denote the reciprocal of a function \(f(x)\), we would need to write:

\[ (f(x))^{−1}=\frac{1}{f(x)}.\]

An important relationship between inverse functions is that they “undo” each other. If \(f^{−1}\) is the inverse of a function \(f\), then \(f\) is the inverse of the function \(f^{−1}\). In other words, whatever the function \(f\) does to \(x\), \(f^{−1}\) undoes it—and vice-versa.

\(f^{−1}(f(x))=x\), for all \(x\) in the domain of \(f\)

and

\(f(f^{−1}(x))=x\), for all \(x\) in the domain of \(f^{−1}\)

Note that the inverse switches the domain and range of the original function.

VERIFYING TWO FUNCTIONS ARE INVERSES OF ONE ANOTHER

Two functions, \(f\) and \(g\), are inverses of one another if for all \(x\) in the domain of \(f\) and \(g\),

\(g(f(x))=f(g(x))=x\)

Howto: Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one

1. Replace \(f(x)\) with \(y\).
2. Interchange \(x\) and \(y\).
3. Solve for \(y\), and rename the function \(f^{−1}(x)\).

Solving Applications of Radical Functions

Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section.

Example \(\PageIndex{6}\): Solving an Application with a Cubic Function

A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by

\[V=\dfrac{2}{3}\pi r^3 \nonumber\]

Find the inverse of the function \(V=\frac{2}{3}\pi r^3\) that determines the volume \(V\) of a cone and is a function of the radius \(r\). Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use \(\pi=3.14\).

Solution

Start with the given function for \(V\). Notice that the meaningful domain for the function is \(r>0\) since negative radii would not make sense in this context nor would a radius of \(0\). Also note the range of the function (hence, the domain of the inverse function) is \(V>0\). Solve for \(r\) in terms of \(V\), using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable.

\[ V =\dfrac{2}{3}\pi r^3\nonumber \]

Solve for \(r^3\).

\[r^3 =\dfrac{3V}{2\pi} \nonumber\]

Solve for \(r\).

\[ r=\sqrt[3]{\dfrac{3V}{2\pi}} \nonumber \]

This is the result stated in the section opener. Now evaluate this for \(V=100\) and \(\pi=3.14\).

\[ \begin{align*} r&=\sqrt[3]{\dfrac{3V}{2\pi}} \\[4pt] &=\sqrt[3]{\dfrac{3⋅100}{2⋅3.14}} \\[4pt] &≈\sqrt[3]{47.7707} \\[4pt] &≈3.63 \end{align*}\]

Therefore, the radius is about 3.63 ft.