10.1 Finding Composite and Inverse Functions - Intermediate Algebra 2e | OpenStax (2024)

Find and Evaluate Composite Functions

Before we introduce the functions, we need to look at another operation on functions called composition. In composition, the output of one function is the input of a second function. For functions $f$ and $g,$ the composition is written $f\circ g$ and is defined by $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right).$

We read $f\left(g\left(x\right)\right)$ as $\text{“}f$ of $g$ of $x\text{.”}$

To do a composition, the output of the first function, $g\left(x\right),$ becomes the input of the second function, f, and so we must be sure that it is part of the domain of f.

We have actually used composition without using the notation many times before. When we graphed quadratic functions using translations, we were composing functions. For example, if we first graphed $g\left(x\right)=x^2$ as a parabola and then shifted it down vertically four units, we were using the composition defined by $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ where $f\left(x\right)=x-4.$

The next example will demonstrate that $\left(f\circ g\right)\left(x\right),$ $\left(g\circ f\right)\left(x\right)$ and $\left(f·g\right)\left(x\right)$ usually result in different outputs.

Example 10.1

For functions $f\left(x\right)=4x-5$ and $g\left(x\right)=2x+3,$ find: ⓐ $\left(f\circ g\right)\left(x\right),$ ⓑ $\left(g\circ f\right)\left(x\right),$ and ⓒ $\left(f·g\right)\left(x\right).$

Solution

ⓐ

Use the definition of $\left(f\circ g\right)\left(x\right).$
Distribute.
Simplify.

ⓑ

Use the definition of $\left(f\circ g\right)\left(x\right).$
Distribute.
Simplify.

Notice the difference in the result in part ⓐ and part ⓑ.

ⓒ Notice that $\left(f·g\right)\left(x\right)$ is different than $\left(f\circ g\right)\left(x\right).$ In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them.

$\begin{array}{cccccc}\text{Use the definition of}\left(f·g\right)\left(x\right).\hfill & & & & & \left(f·g\right)\left(x\right)=f\left(x\right)·g\left(x\right)\hfill \\ \text{Substitute}f\left(x\right)=4x-5\text{and}g\left(x\right)=2x+3.\hfill & & & & & \left(f·g\right)\left(x\right)=\left(4x-5\right)·\left(2x+3\right)\hfill \\ \text{Multiply.}\hfill & & & & & \left(f·g\right)\left(x\right)=8x^2+2x-15\hfill \end{array}$

In the next example we will evaluate a composition for a specific value.

Example 10.2

For functions $f\left(x\right)=x^2-4,$ and $g\left(x\right)=3x+2,$ find: ⓐ $\left(f\circ g\right)\left(\mathrm{-3}\right),$ ⓑ $\left(g\circ f\right)\left(\mathrm{-1}\right),$ and ⓒ $\left(f\circ f\right)\left(2\right).$

Solution

ⓐ

Use the definition of $\left(f\circ g\right)\left(\mathrm{-3}\right).$
Simplify.
Simplify.

ⓑ

Use the definition of $\left(g\circ f\right)\left(\mathrm{-1}\right).$
Simplify.
Simplify.

ⓒ

Use the definition of $\left(f\circ f\right)\left(2\right).$
Simplify.
Simplify.

Determine Whether a Function is One-to-One

When we first introduced functions, we said a function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x-value is matched with only one y-value.

We used the birthday example to help us understand the definition. Every person has a birthday, but no one has two birthdays and it is okay for two people to share a birthday. Since each person has exactly one birthday, that relation is a function.

A function is one-to-one if each value in the range has exactly one element in the domain. For each ordered pair in the function, each y-value is matched with only one x-value.

Our example of the birthday relation is not a one-to-one function. Two people can share the same birthday. The range value August 2 is the birthday of Liz and June, and so one range value has two domain values. Therefore, the function is not one-to-one.

To help us determine whether a relation is a function, we use the vertical line test. A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. Also, if any vertical line intersects the graph in more than one point, the graph does not represent a function.

The vertical line is representing an x-value and we check that it intersects the graph in only one y-value. Then it is a function.

To check if a function is one-to-one, we use a similar process. We use a horizontal line and check that each horizontal line intersects the graph in only one point. The horizontal line is representing a y-value and we check that it intersects the graph in only one x-value. If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function. This is the horizontal line test.

We can test whether a graph of a relation is a function by using the vertical line test. We can then tell if the function is one-to-one by applying the horizontal line test.

Find the Inverse of a Function

Let’s look at a one-to one function, $f$, represented by the ordered pairs $\left\{\left(0,5\right),\left(1,6\right),\left(2,7\right),\left(3,8\right)\right\}.$ For each $x$-value, $f$ adds 5 to get the $y$-value. To ‘undo’ the addition of 5, we subtract 5 from each $y$-value and get back to the original $x$-value. We can call this “taking the inverse of $f$” and name the function $f^{\mathrm{-1}}.$

Notice that that the ordered pairs of $f$ and $f^{\mathrm{-1}}$ have their $x$-values and $y$-values reversed. The domain of $f$ is the range of $f^{\mathrm{-1}}$ and the domain of $f^{\mathrm{-1}}$ is the range of $f.$

In the next example we will find the inverse of a function defined by ordered pairs.

We just noted that if $f\left(x\right)$ is a one-to-one function whose ordered pairs are of the form $\left(x,y\right),$ then its inverse function $f^{\mathrm{-1}}\left(x\right)$ is the set of ordered pairs $\left(y,x\right).$

So if a point $\left(a,b\right)$ is on the graph of a function $f\left(x\right),$ then the ordered pair $\left(b,a\right)$ is on the graph of $f^{\mathrm{-1}}\left(x\right).$ See Figure 10.2.

Figure 10.2

The distance between any two pairs $\left(a,b\right)$ and $\left(b,a\right)$ is cut in half by the line $y=x.$ So we say the points are mirror images of each other through the line $y=x.$

Since every point on the graph of a function $f\left(x\right)$ is a mirror image of a point on the graph of $f^{\mathrm{-1}}\left(x\right),$ we say the graphs are mirror images of each other through the line $y=x.$ We will use this concept to graph the inverse of a function in the next example.

When we began our discussion of an inverse function, we talked about how the inverse function ‘undoes’ what the original function did to a value in its domain in order to get back to the original x-value.

We can use this property to verify that two functions are inverses of each other.

Example 10.7

Verify that $f\left(x\right)=5x-1$ and $g\left(x\right)=\frac{x+1}{5}$ are inverse functions.

Solution

The functions are inverses of each other if $g\left(f\left(x\right)\right)=x$ and $f\left(g\left(x\right)\right)=x.$

Substitute $5x-1$ for $f\left(x\right).$
Simplify.
Simplify.
Substitute $\frac{x+1}{5}$ for $g\left(x\right).$
Simplify.
Simplify.

Since both $g\left(f\left(x\right)\right)=x$ and $f\left(g\left(x\right)\right)=x$ are true, the functions $f\left(x\right)=5x-1$ and $g\left(x\right)=\frac{x+1}{5}$ are inverse functions. That is, they are inverses of each other.

We have found inverses of function defined by ordered pairs and from a graph. We will now look at how to find an inverse using an algebraic equation. The method uses the idea that if $f\left(x\right)$ is a one-to-one function with ordered pairs $\left(x,y\right),$ then its inverse function $f^{\mathrm{-1}}\left(x\right)$ is the set of ordered pairs $\left(y,x\right).$

If we reverse the x and y in the function and then solve for y, we get our inverse function.

We summarize the steps below.