# 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

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

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.

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
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Nando
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I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
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