# Find 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.

For a function to have an **inverse function** the function to create a new function that is **one-to-one** and would have an inverse function.

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown below. 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 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 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 [latex]y\left(x\right)=a{x}^{2}[/latex]. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor *a*.

Our parabolic cross section has the equation

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:

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.

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 2*x*. The trough is 3 feet (36 inches) long, so the surface area will then be:

This example illustrates two important points:

- When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one.
- 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 [latex]{f}^{-1}\left(x\right)[/latex].

Warning: [latex]{f}^{-1}\left(x\right)[/latex] is not the same as the reciprocal of the function [latex]f\left(x\right)[/latex]. This use of –1 is reserved to denote inverse functions. To denote the reciprocal of a function [latex]f\left(x\right)[/latex], we would need to write [latex]{\left(f\left(x\right)\right)}^{-1}=\frac{1}{f\left(x\right)}[/latex].

An important relationship between inverse functions is that they "undo" each other. If [latex]{f}^{-1}[/latex] is the inverse of a function *f*, then *f* is the inverse of the function [latex]{f}^{-1}[/latex]. In other words, whatever the function *f* does to *x*, [latex]{f}^{-1}[/latex] undoes it—and vice-versa. More formally, we write

and

### A General Note: 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*.

### How To: 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.

- Replace [latex]f\left(x\right)[/latex] with
*y*. - Interchange
*x*and*y*. - Solve for
*y*, and rename the function [latex]{f}^{-1}\left(x\right)[/latex].

### Example 1: Verifying Inverse Functions

Show that [latex]f\left(x\right)=\frac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\frac{1}{x}-1[/latex] are inverses, for [latex]x\ne 0,-1[/latex] .

### Solution

We must show that [latex]{f}^{-1}\left(f\left(x\right)\right)=x[/latex] and [latex]f\left({f}^{-1}\left(x\right)\right)=x[/latex].

Therefore, [latex]f\left(x\right)=\frac{1}{x+1}[/latex] and [latex]{f}^{-1}\left(x\right)=\frac{1}{x}-1[/latex] are inverses.

### Try It 1

Show that [latex]f\left(x\right)=\frac{x+5}{3}[/latex] and [latex]{f}^{-1}\left(x\right)=3x - 5[/latex] are inverses.

Solution### Example 2: Finding the Inverse of a Cubic Function

Find the inverse of the function [latex]f\left(x\right)=5{x}^{3}+1[/latex].

### Solution

This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for *x*.

### Try It 2

Find the inverse function of [latex]f\left(x\right)=\sqrt[3]{x+4}[/latex].

Solution## Licenses & Attributions

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**Provided by:**OpenStax**Authored by:**Jay Abramson, et al..**Located at:**https://openstax.org/books/precalculus/pages/1-introduction-to-functions.**License:**CC BY: Attribution.**License terms:**Download For Free at : http://cnx.org/contents/[email protected]..

## Analysis of the Solution

Look at the graph of

fand [latex]{f}^{-1}[/latex]. Notice that the two graphs are symmetrical about the line [latex]y=x[/latex]. This is always the case when graphing a function and its inverse function.Also, since the method involved interchanging

xandy, notice corresponding points. If [latex]\left(a,b\right)[/latex] is on the graph off, then [latex]\left(b,a\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Since [latex]\left(0,1\right)[/latex] is on the graph off, then [latex]\left(1,0\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex]. Similarly, since [latex]\left(1,6\right)[/latex] is on the graph off, then [latex]\left(6,1\right)[/latex] is on the graph of [latex]{f}^{-1}[/latex].Figure 4