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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.

Gravel in the shape of a cone.

The volume is found using a formula from elementary geometry.

V = 1 3 π r 2 h = 1 3 π r 2 ( 2 r ) = 2 3 π r 3

We have written the volume V in terms of the radius r . 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 = 3 V 2 π 3

This function is the inverse of the formula for V in terms of 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 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 [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.

Diagram of a parabolic trough that is 18” in height, 3’ in length, and 12” in width.

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. See [link] .

Graph of a parabola.

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 ) = a x 2 . Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a .

18 = a 6 2 a = 18 36 = 1 2

Our parabolic cross section has the equation

y ( x ) = 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 2 x , 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:

y = 1 2 x 2 2 y = x 2 x = ± 2 y

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.

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Practice Key Terms 1

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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