# 5.7 Inverses and radical functions

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In this section, you will:
• Find the inverse of an invertible polynomial function.
• Restrict the domain to find the inverse of a polynomial function.

A mound of gravel is in the shape of a cone with the height equal to twice the radius.

The volume is found using a formula from elementary geometry.

$\begin{array}{ccc}V& \hfill =& \frac{1}{3}\pi {r}^{2}h\hfill \\ & =& \frac{1}{3}\pi {r}^{2}\left(2r\right)\hfill \\ & =& \frac{2}{3}\pi {r}^{3}\hfill \end{array}$

We have written the volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of the radius $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula

$r=\sqrt{\frac{3V}{2\pi }}$

This function is the inverse of the formula for $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}r.$

In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.

## Finding the inverse of a polynomial function

Two functions $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are inverse functions if for every coordinate pair in $\text{\hspace{0.17em}}f,\left(a,b\right),\text{\hspace{0.17em}}$ there exists a corresponding coordinate pair in the inverse function, $\text{\hspace{0.17em}}g,\left(b,\text{\hspace{0.17em}}a\right).\text{\hspace{0.17em}}$ 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 [link] . 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.

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 $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ measured horizontally and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ measured vertically, with the origin at the vertex of the parabola. See [link] .

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 $\text{\hspace{0.17em}}y\left(x\right)=a{x}^{2}.\text{\hspace{0.17em}}$ Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor $\text{\hspace{0.17em}}a.$

$\begin{array}{ccc}\hfill 18& =& a{6}^{2}\hfill \\ \hfill a& =& \frac{18}{36}\hfill \\ & =& \frac{1}{2}\hfill \end{array}$

Our parabolic cross section has the equation

$y\left(x\right)=\frac{1}{2}{x}^{2}$

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 $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ the width will be given by $\text{\hspace{0.17em}}2x,\text{\hspace{0.17em}}$ so we need to solve the equation above for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values. On this domain, we can find an inverse by solving for the input variable:

$\begin{array}{ccc}\hfill y& =& \frac{1}{2}{x}^{2}\hfill \\ \hfill 2y& =& {x}^{2}\hfill \\ \hfill x& =& ±\sqrt{2y}\hfill \end{array}$

This is not a function as written. We are limiting ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values, so we eliminate the negative solution, giving us the inverse function we’re looking for.

#### Questions & Answers

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Khay By By Danielrosenberger By    By Eric Crawford By Mistry Bhavesh By  By Anindyo Mukhopadhyay