If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x))?

A function can be conceptualized as a 'black box'. The input, or *x* value is placed into the box, and the box performs a specific set of operations on it. Once the operations are complete, the output (the "*f(x)*" or "*y*" value) is retrieved. Once the output is retrieved, the box is ready to work on the next input.

Using this idea, **function composition** can be seen as a box inside of a box. The input *x* value goes into the inner box, and then the output of the *inner* box is used as the input of the *outer* box.

### Composition of Functions

Functions are often described in terms of “input” and “output.” For example, consider the function *f*(*x*) = 2*x* + 3. When we input an *x* value, we output a *y* value, or a function value. We find the output by taking the input *x*, multiplying by 2, and adding 3. We can do this for any value of *x*. Now consider a second function *g*(*x*) = 5*x*. For this function too, we can take an *x* value, input the *x* into *g*(*x*), and obtain an output. What happens if we take the output of *g* and use it as the input of *f*?

### Examples

#### Example 1

Earlier, you were given a problem about finding a **composite function**.

If *f(x) = x + 2*, and *g(x) = 2x + 4*, what is *f(g(x))*?

*f(g(x))* = *f(2x + 4)* = *(2x + 4) + 2* = *2x + 6*

#### Example 2

Given the function definition above, *g*(*x*) = 5*x*. Therefore if *x* = 4, then we have *g*(4) = 5(4) = 20. What happens if we then take the output of 20 and use it as the input of *f*?

Substituting 20 in for *x* in *f*(*x*) = 2*x* + 3 gives: *f*(20) = 2(20) + 3 = 43.

The table below shows several examples of this same process:

x |
Output from g |
Output from f |
---|---|---|

2 | 10 | 23 |

3 | 15 | 33 |

4 | 20 | 43 |

5 | 25 | 53 |

Examining the values in the table, we can see a pattern: all of the final output values from *f* are 3 more than 10 times the initial input. We have created a new function called *h*(*x*) out of *f*(*x*) = 2*x* + 3 in which *g*(*x*) = 5*x* is the input:

*h*(*x*) = *f*(5*x*) = 2(5*x*) + 3 = 10*x* + 3

When we input one function into another, we call this the **composition** of the two functions. Formally, we write the composed function as *f*(*g*(*x*)) = 10*x* + 3 or write it as (*f* o *g*)*x* = 10*x* + 3

#### Example 3

Find *f*(*g*(*x*)) and *g*(*f*(*x*)):

*f*(x) = 3*x*+ 1 and*g*(*x*) =*x*^{2}

*f*(*g*(*x*)) = *f*(*x*^{2}) = 3(*x*^{2}) + 1 = 3*x*^{2} + 1

*g*(*f*(*x*)) = *g*(3*x* + 1) = (3*x* + 1)^{2} = 9*x*^{2} + 6*x* + 1

In both cases, the resulting function is quadratic.

*f*(x) = 2*x*+ 4 and*g*(*x*) = (1/2)*x*- 2

*f*(*g*(*x*)) = 2((1/2)*x* - 2) + 4 = (2/2)*x* - 4 + 4 = (2/2)*x* = *x*

*g*(*f*(*x*)) = *g*(2*x* + 4) = (1/2)(2*x* + 4) - 2 = *x*+ 2 - 2 = *x*.

In this case, the composites were equal to each other, and they both equal *x*, the original input into the function. This means that there is a special relationship between these two functions. We will examine this relationship in Chapter 3. It is important to note, however, that *f*(*g*(*x*) is not necessarily equal to *g*(*f*(*x*)).

#### Example 4

Decompose the function *f*(*x*) = (3*x* - 1)^{2} - 5 into a quadratic function *g*(*x*) and a linear function *h*(*x*).

When we compose functions, we are combining two (or more) functions by inputting the output of one function into another. We can also *decompose* a function. Consider the function *f*(*x*) = (2*x* + 1)^{2}. We can decompose this function into an “inside” and an “outside” function. For example, we can construct *f*(*x*) = (2*x*+ 1)^{2} with a linear function and a quadratic function. If *g*(*x*) = *x*^{2} and *h*(*x*) = (2*x* + 1), then *f*(*x*) = *g*(*h*(*x*)). The linear function *h*(*x*) = (2*x* + 1) is the inside function, and the quadratic function *g*(*x*) = *x*^{2} is the outside function.

Let *h*(*x*) = 3*x* - 1 and *g*(*x*) = *x*^{2} - 5. Then *f*(*x*) = *g*(*h*(*x*)) because *g*(*h*(*x*)) = *g*(3*x* - 1) = (3*x* - 1)^{2} - 5.

The decomposition of a function is not necessarily unique. For example, there are many ways that we could express a linear function as the composition of other linear functions.

#### Example 5

Given:

\begin{align*}f(x) = 5x + 3\end{align*}

\begin{align*}g(x) = 3x^2\end{align*}

Find: \begin{align*}f(g(4))\end{align*}

To find *f*(*g*(4)), we need to know what \begin{align*}g(4)\end{align*} is, so we know what to substitute into \begin{align*}f(x)\end{align*}:

Substitute 4 for *x* for the function *g*(*x*), giving: \begin{align*}3 \cdot 4^2\end{align*}

Simplify: \begin{align*}3 \cdot 16 = 48\end{align*}

\begin{align*}\therefore g(4) = 48\end{align*}

Substitute 48 for the *x* in the function \begin{align*}f(x)\end{align*} giving: \begin{align*}5(48) +3\end{align*}

Simplify: \begin{align*}240 + 3 = 243\end{align*}

\begin{align*}\therefore f(g(4)) = 243\end{align*}

#### Example 6

Given:

\begin{align*}h(n) = 7n +1 + 4(g(n))\end{align*}

\begin{align*}g(t) = -t\end{align*}

\begin{align*}f(x) = -2x + g(x)\end{align*}

Find: \begin{align*}f(h(-5))\end{align*}

First, let's solve for the value of the inner function, \begin{align*}h(-5)\end{align*}. Then we'll know what to plug into the outer function.

\begin{align*}h(-5) = (7)(-5)+1+4(g(-5))\end{align*}

To solve for the value of *h*, we need to solve \begin{align*}g(-5)\end{align*}

\begin{align*}g(-5) = - (-5)\end{align*}

\begin{align*}\therefore g(-5) = 5\end{align*}

Now we have: \begin{align*}h(-5)=(7)(-5)+1+(4)(5)\end{align*}

Simplify to get: \begin{align*}h(-5)=-14\end{align*}

Now we know that \begin{align*}h(-5) = -14\end{align*}. That tells us that \begin{align*}f(h(-5))\end{align*} is \begin{align*}f(-14)\end{align*}

Find \begin{align*}f(-14) = (-2)(-14)+g(-14)\end{align*}

So to solve for the value of \begin{align*}f(-14)\end{align*}, we need to solve for the value of \begin{align*}g(-14)\end{align*}

\begin{align*}g(-14) = - (-14)\end{align*}

\begin{align*}\therefore g(-14) = 14\end{align*}

Now we can finish up!

\begin{align*}f(-14) = (-2)(-14) + 14\end{align*}

\begin{align*}\therefore f(-14) = 42\end{align*}

### Review

For problems 1-4:

\begin{align*}f(x) = 2x - 1\end{align*} \begin{align*}g(x) = 3x\end{align*} \begin{align*}h(x) = x^2 + 1\end{align*}

- Find: \begin{align*}f(g(-3))\end{align*}
- Find: \begin{align*}f(h(7))\end{align*}
- Find: \begin{align*}h(g(-4))\end{align*}
- Find: \begin{align*} f(g(h(2)))\end{align*}

Evaluate each composition below:

- Given: \begin{align*}f(x) = -5x + 2\end{align*} and \begin{align*}g(x) = \frac{1}{2}x + 4\end{align*}. Find \begin{align*}f(g(12))\end{align*}.
- Given: \begin{align*}g(x) = -3x + 6\end{align*} and \begin{align*}h(x) = 9x + 3\end{align*}. Find \begin{align*}g(h(\frac{1}{3}))\end{align*}.
- Given: \begin{align*}f(x) = -\frac{1}{5}x + 4\end{align*} and \begin{align*}g(x) = 4x^2\end{align*}. Find \begin{align*}f(g(10))\end{align*}.
- Given \begin{align*}g(x) = 3|x -4|+ 6\end{align*} and \begin{align*}h(x) = -x^3\end{align*}. Find \begin{align*}h(g(4))\end{align*}.
- Given \begin{align*}f(x) = \sqrt{x + 2}\end{align*} and \begin{align*}g(x) = |2x|\end{align*}. Find \begin{align*}g(f(-7))\end{align*}.
- Given \begin{align*}f(x) = -3x + 2\end{align*} and given \begin{align*}g(x) = 2x^2\end{align*} and given \begin{align*}h(x) = 4|7 - x| + 6\end{align*}. Find \begin{align*}f(g(h(1)))\end{align*}.
- Given \begin{align*}f(x) = (-3)\end{align*} and given \begin{align*}g(x) = \sqrt{2x}\end{align*} and given \begin{align*}h(x) = |4x| - 12\end{align*}. Find \begin{align*}f(h(g(18)))\end{align*}.
- Are compositions commutative? In other words, does \begin{align*}f(g(x)) = g(f(x))\end{align*} ?
- Given: \begin{align*}f(x) = -2^2 - 5x\end{align*} and \begin{align*}h(x) = 3x + 2\end{align*}. Find \begin{align*}f(h(x))\end{align*}.
- Two functions are inverses of each other if \begin{align*} f(g(x)) = x\end{align*} and \begin{align*}g(f(x)) = x\end{align*} If \begin{align*}f(x) = x + 3\end{align*}, find its inverse: \begin{align*}g(x)\end{align*}
- A toy manufacturer has a new product to sell. The number of units to be sold,
*n*, is a function of the price*p*such that: \begin{align*}n(p) = 30 - 25p\end{align*}. The revenue*r*earned from the sales is a function of the number of units sold*n*such that: \begin{align*}r(n) = 1000 - \frac{1}{4}x^2\end{align*} . Find the function for revenue in terms of price,*p*.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 1.16.