# 7.10: Function Operations

**At Grade**Created by: CK-12

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**Practice**Operations on Functions

The area of a rectangle is

### Guidance

As you saw in the Review Queue, we have already dealt with adding, subtracting, and multiplying functions. To add and subtract, you combine like terms (see the *Adding and Subtracting Polynomials* concept). When multiplying, you either FOIL or use the “box” method (see the *Multiplying Polynomials* concept). When you add, subtract, or multiply functions, it is exactly the same as what you would do with polynomials, except for the notation. Notice, in the Review Queue, we didn’t write out the entire function, just

Distribute the negative sign to the second function and combine like terms. Be careful!

#### Example A

If

**Solution:** First, even though the

To divide the two functions, we will place

To find the **restriction(s)** on this function, we need to determine what value(s) of

Now we will introduce a new way to manipulate functions; composing them. When you **compose** two functions, we put one function into the other, where ever there is an

#### Example B

Using

**Solution:** For

Now, substitute in the actual function for

To find the domain of

Now, to find \begin{align*}g(f(x))\end{align*}, we would put \begin{align*}f(x)\end{align*} into \begin{align*}g(x)\end{align*} everywhere there is an \begin{align*}x\end{align*}-value.

\begin{align*}g(f(x))&= \frac{1}{2} \left[f(x) \right]^2 \\ &= \frac{1}{2} \Big [ \sqrt{x-8} \Big ]^2 \\ &= \frac{1}{2}(x-8) \\ &= \frac{1}{2}x-4\end{align*}

Notice that \begin{align*}f(g(x)) \ne g(f(x))\end{align*}. It is possible that \begin{align*}f \circ g=g \circ f\end{align*} and is a special case, addressed in the next concept. To find the domain of \begin{align*}g(f(x))\end{align*}, we will determine where \begin{align*}x\end{align*} is defined. \begin{align*}g(f(x))\end{align*} is a line, so we would think that the domain is all real numbers. However, while simplifying the composition, the square and square root canceled out. Therefore, any restriction on \begin{align*}f(x)\end{align*} or \begin{align*}g(x)\end{align*} would still exist. The domain would be all real numbers such that \begin{align*}x \ge 8\end{align*} from the domain of \begin{align*}f(x)\end{align*}. *Whenever operations cancel, the original restrictions from the inner function still exist*. As with the case of \begin{align*}f(g(x))\end{align*}, no simplifying occurred, so the domain was unique to that function.

#### Example C

If \begin{align*}f(x)=x^4-1\end{align*} and \begin{align*}g(x)=2 \sqrt[4]{x+1}\end{align*}, find \begin{align*}g \circ f\end{align*} and the restrictions on the domain.

**Solution:** Recall that \begin{align*}g \circ f\end{align*} is another way of writing \begin{align*}g(f(x))\end{align*}. Let’s plug \begin{align*}f\end{align*} into \begin{align*}g\end{align*}.

\begin{align*}g \circ f&=2 \sqrt[4]{f \left(x \right)+1} \\ &=2 \sqrt[4]{\left(x^4-1 \right)+1} \\ &=2 \sqrt[4]{x^4} \\ &=2 \left | x \right |\end{align*}

The final function, \begin{align*}g \circ f \ne 2x\end{align*} because \begin{align*}x\end{align*} is being raised to the \begin{align*}4^{th}\end{align*} power, which will always yield a positive answer. Therefore, even when \begin{align*}x\end{align*} is negative, the answer will be positive. For example, if \begin{align*}x=-2\end{align*}, then \begin{align*}g \circ f=2 \sqrt[4]{\left(-2 \right)^4}=2 \cdot 2=4.\end{align*}. An absolute value function has no restrictions on the domain. *This will always happen when even roots and powers cancel.* The range of this function is going to be all positive real numbers because the absolute value is never negative.

Recall, the previous example, however. The restrictions, if there are any, from the inner function, \begin{align*}f(x)\end{align*}, still exist. Because there are no restrictions on \begin{align*}f(x)\end{align*}, the domain of \begin{align*}g \circ f\end{align*} remains all real numbers.

**Intro Problem Revisit** If we set *g* equal to \begin{align*}2x^2\end{align*} and *f* equal to \begin{align*}\sqrt{x + 3}\end{align*}, to find the width, we need to find \begin{align*}\frac{g}{f}\end{align*}.

\begin{align*}\frac{2x^2}{\sqrt{x + 3}}\\ \frac{2x^2}{\sqrt{x + 3}}\cdot \frac{\sqrt{x + 3}}{\sqrt{x + 3}}\\ \frac{2x^2\sqrt{x + 3}}{x + 3}\end{align*}

Therefore the width of the rectangle is \begin{align*}\frac{2x^2\sqrt{x + 3}}{x + 3}\end{align*}, and \begin{align*}x = -3\end{align*} is a restriction on the answer.

### Guided Practice

\begin{align*}f(x)=5x^{-1}\end{align*} and \begin{align*}g(x)=4x+7\end{align*}. Find:

1. \begin{align*}fg\end{align*}

2. \begin{align*}g-f\end{align*}

3. \begin{align*}\frac{f}{g}\end{align*}

4. \begin{align*}g(f(x))\end{align*} and the domain

5. \begin{align*}f \circ f\end{align*}

#### Answers

1. \begin{align*}fg\end{align*} is the product of \begin{align*}f(x)\end{align*} and \begin{align*}g(x)\end{align*}.

\begin{align*}fg&=5x^{-1}(4x+7) \\ &=20x^0+35x^{-1} \\ &=20+35x^{-1} \ or \ \frac{20x+35}{x}\end{align*}

Both representations are correct. Discuss with your teacher how s/he would like you to leave your answer.

2. Subtract \begin{align*}f(x)\end{align*} from \begin{align*}g(x)\end{align*} and simplify, if possible.

\begin{align*}g-f&=(4x+7)-5x^{-1} \\ &=4x+7-5x^{-1} \ or \ \frac{4x^2+7x-5}{x}\end{align*}

3. Divide \begin{align*}f(x)\end{align*} by \begin{align*}g(x)\end{align*}. Don’t forget to include the restriction(s).

\begin{align*}\frac{f}{g}&= \frac{5x^{-1}}{4x+7} \\ &= \frac{5}{x(4x+7)}; \ x \ne 0, - \frac{7}{4}\end{align*}

Recall the properties of exponents. Anytime there is a negative exponent, it should be moved into the denominator. We set each factor in the denominator equal to zero to find the restrictions.

4. \begin{align*}g(f(x))\end{align*} is a composition function. Let’s plug \begin{align*}f(x)\end{align*} into \begin{align*}g(x)\end{align*} everywhere there is an \begin{align*}x\end{align*}.

\begin{align*}g(f(x))&=4f(x)+7 \\ &=4(5x^{-1})+7 \\ &=20x^{-1}+7 \ or \ \frac{20+7x}{x}\end{align*}

The domain of \begin{align*}f(x)\end{align*} is all real numbers except \begin{align*}x \ne 0\end{align*}, because we cannot divide by zero. Therefore, the domain of \begin{align*}g(f(x))\end{align*} is all real numbers except \begin{align*}x \ne 0\end{align*}.

5. \begin{align*}f \circ f\end{align*} is a composite function on itself. We will plug \begin{align*}f(x)\end{align*} into \begin{align*}f(x)\end{align*} everywhere there is an \begin{align*}x\end{align*}.

\begin{align*}f(f(x))&=5(f(x))^{-1} \\ &=5(5x^{-1})^{-1} \\ &=5 \cdot 5^{-1}x^1 \\ &=x\end{align*}

### Explore More

For problems 1-8, use the following functions to form the indicated compositions and clearly indicate any restrictions to the domain of the composite function.

\begin{align*}f(x)=x^2+5 \qquad g(x)=3 \sqrt{x-5} \qquad h(x)=5x+1\end{align*}

- \begin{align*}f+h\end{align*}
- \begin{align*}h-g\end{align*}
- \begin{align*}\frac{f}{g}\end{align*}
- \begin{align*}fh\end{align*}
- \begin{align*}f \circ g\end{align*}
- \begin{align*}h(f(x))\end{align*}
- \begin{align*}g \circ f\end{align*}
- \begin{align*}f \circ g \circ h\end{align*}

For problems 9-16, use the following functions to form the indicated compositions and clearly indicate any restrictions to the domain of the composite function.

\begin{align*}p(x)= \frac{5}{x} \qquad q(x)=5 \sqrt{x} \qquad r(x)= \frac{\sqrt{x}}{5} \qquad s(x)= \frac{1}{5}x^2\end{align*}

- \begin{align*}ps\end{align*}
- \begin{align*}\frac{q}{r}\end{align*}
- \begin{align*}q+r\end{align*}
- \begin{align*}p(q(x))\end{align*}
- \begin{align*}s(q(x))\end{align*}
- \begin{align*}q \circ s\end{align*}
- \begin{align*}q \circ p \circ s\end{align*}
- \begin{align*}p \circ r\end{align*}

composite function

A composite function is a function formed by using the output of one function as the input of another function . Composite functions are written in the form or .Difference

The result of a subtraction operation is called a difference.Function

A function is a relation where there is only one output for every input. In other words, for every value of , there is only one value for .Horizontal Asymptote

A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote.Restriction

A restriction is a value of the domain where cannot be defined.Sum

The sum is the result after two or more amounts have been added together.Vertical Asymptote

A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach.### Image Attributions

## Description

## Learning Objectives

Here you'll learn to add, subtract, multiply, divide and compose two or more functions.

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## Date Created:

Mar 12, 2013## Last Modified:

Jun 04, 2015## Vocabulary

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