3.2 The Algebra of Functions

Learning Outcomes

Evaluate polynomial functions.
Evaluate a function with a sum or difference as the input such as $f(x+1)$ .
Evaluate the sum, difference, product, or quotient of two functions at given inputs.
Find the sum, difference, product, or quotient of two functions and state the domain.

Just as we have performed algebraic operations on polynomials, we can do the same with polynomial functions.

Evaluate Polynomial Functions

You can evaluate polynomial functions similar to how you have been evaluating expressions all along. To evaluate an expression for a value of the variable, you substitute the value for the variable every time it appears. Then use the order of operations to find the resulting value for the expression.

Example

Given the function $f(x)=3x^{2}-2x+1$ :

1) Find $f(-1)$

2) Find $f(x+1)$

Show Solution

$\begin{align} \require{color}f({\color{Green}{-1}})&=3({\color{Green}{-1}})^{2}-2({\color{Green}{-1}})+1 &&\color{blue}{\textsf{substitute -1 for each x in the polynomial}}\\ &= 3(1)-2(-1)+1 &&\color{blue}{\textsf{follow order of operations, evaluate exponents first, then multiply}}\\ &= 3+2+1 &&\color{blue}{\textsf{add}}\\ &= 6 \end{align}$

$\begin{align} \require{color}f({\color{Green}{x+1}})&=3({\color{Green}{x+1}})^{2}-2({\color{Green}{x+1}})+1 &&\color{blue}{\textsf{substitute x+1 for each x in the polynomial}}\\ &= 3(x+1)(x+1)-2(x+1)+1 &&\color{blue}{\textsf{follow order of operations}}\\ &= 3(x^2+2x+1)-2x-2+1\\ &= 3x^2+6x+3-2x-2+1 &&\color{blue}{\textsf{combine like terms}}\\ &= 3x^2+4x+2 \end{align}$

In the following video, we show more examples of evaluating polynomials for given values of the variable.

In this section, we will focus on how to perform algebraic operations on polynomial functions and introduce related notation.

The four basic operations on functions are adding, subtracting, multiplying, and dividing. If $f$ and $g$ are functions and $x$ is in the domain of both functions, then:

Addition: $(f + g)(x) = f(x)+ g(x)$

Subtraction: $(f − g)(x)= f(x) − g(x)$

Multiplication: $(f · g)(x)= f(x)\cdot g(x)$

Division: $\left(\dfrac{f}{ g}\right) (x) = \dfrac{f(x)}{g(x)}$ , provided $g(x) \not= 0$

We will focus on applying these operations to polynomial functions in this section.

Operations such as addition, subtraction, multiplication, and division can be used to produce a new function from two or more functions. The domain of the new function will be the intersection of the domains of the initial functions.

Add and Subtract Polynomial Functions

Adding and subtracting polynomial functions is the same as adding and subtracting polynomials. When you evaluate a sum or difference of functions, you can either evaluate first or perform the operation on the functions first as we will see. Our next examples describe the notation used to add and subtract polynomial functions.

ExAMPLE

Let $f(x)=3x^2-6x-2$ and $g(x)=4x-1$

Compute the following:

$1)\, f(-1)+g(4)\\$

$2)\, f(-1)-g(4)\\$

$3)\, g(-2)-g(3)\\$

Show Solution

$f(-1)+g(4)$

To find $f(-1),$ substitute $\require{color}\color{Green}{-1}$ in place of $x$ in the given function. To find $f(4)$ substitute $\require{color}\color{Green}{4}$ in place of $x$ in the given function.

$\begin{align}f(\color{Green}{-1}\color{black}{)} &=3(\color{Green}{-1}\color{black}{)^2-6(}\color{Green}{-1}\color{black}{)-2}\\ &= (3)(1)-6(-1)-2\\ &= 3+6-2\\ &= 7 \end{align}$

and

$\begin{align}g(\color{Green}{4}\color{black}{)} &=4(\color{Green}{4}\color{black}{)-1(}\\ &= 16-1\\ &=15 \end{align}$

We now have, $f(-1)=7$ and $g(4)=15$ .

Let’s substitute these in place of $f(-1)$ and $g(4)$ .

$\require{color} f(\color{Green}{-1}\color{black}{)+g(}\color{Green}{4}\color{black}{)}$

$\begin{align}&= 7+15\\ &= 22\end{align}\\\\$

$f(-1)-g(4)\\$

In part 1 we found $f(-1)=7$ and $g(4)=15$ .

Let’s substitute these in place of $f(-1)$ and $g(4)$ .

$f(-1)-g(4)$

$\begin{align}&= 7-15\\ &= -8\end{align}\\\\$

$g(-2)-g(3)\\$

To find $g(-2),$ substitute $\require{color}\color{Green}{-2}$ in place of $x$ in the given function. To find $g(3)$ substitute $\require{color}\color{Green}{3}$ in place of $x$ in the given function.

$\begin{align}g(\color{Green}{-2}\color{black}{)} &=4(\color{Green}{-2}\color{black}{)-1}\\ &= -8-1\\ &= -9 \end{align}$

and

$\begin{align}g(\color{Green}{3}\color{black}{)} &=4(\color{Green}{3}\color{black}{)-1}\\ &= 12-1\\ &=11 \end{align}$

We now have, $g(-2)=-9$ and $g(3)=11$ .

Let’s substitute these in place of $g(-2)$ and $g(3)$ .

$\require{color} g(\color{Green}{-2}\color{black}{)-g(}\color{Green}{3}\color{black}{)}$

$\begin{align}&= -9-11\\ &= -20\end{align}\\\\$

Here is a video with more examples like the one above.

Example

Let $f(x)=2x^3-5x+3$ and $h(x)=x-5$ ,

Find the following:

$1) (f+h)(x)$ and state the domain of the combined function.

$2) (h-f)(x)$ and state the domain of the combined function.

Show Solution

$\require{color}\begin{align}(f+h)(x)=f(x)+ h(x) & =(2x^3-5x+3)+(x-5) \\ & =2x^3-5x+x+3-5 && \color{blue}{\textsf{group like terms}} \\ & =2x^3-4x-2 && \color{blue}{\textsf{simplify}}\end{align}$

The domain of the combined function is the intersection of the domains of the original functions.
The domain of $f(x)=2x^3-5x+3$ is $(-\infty, \infty)$ because there are no restrictions on the domain.
The domain of $h(x)=x-5$ is $(-\infty, \infty)$ because there are also no restrictions on the domain.
The intersection of the two domains will also be $(-\infty, \infty)$ .

$\require{color}\begin{align}(h-f)(x)=h(x)-f(x) & =(x-5)-(2x^3-5x+3)\\ & =x-5-2x^2+5x-3 && \color{blue}{\textsf{distribute negative}} \\ & =-2x^2+6x-8 && \color{blue}{\textsf{simplify}}\end{align}$

In our next example, we will evaluate a sum and difference of functions and show that you can get to the same result in one of two ways.

Example

Let $f(x)=2x^3-5x+3$ and $h(x)=x-5$

Evaluate: $(f+h)(2)$

Show that you get the same result by

1) Evaluating the functions first, then performing the indicated operation on the result.

2) Performing the operation on the functions first then evaluating the result.

Show Solution

1) First, we will evaluate the functions separately:

$\require{color}\begin{align}f(\color{Green}{2}\color{black}{)} &= 2(\color{Green}{2}\color{black}{)}^3-5(\color{Green}{2}\color{black}{)+3}\\ &=16-10+3\\ &=9 \end{align}$

$\begin{align}h(\color{Green}{2}\color{black}{)} &= (\color{Green}{2}\color{black}{)-5}\\ &=-3 \end{align}$

Now we will perform the indicated operation using the results:

$\begin{align}(f+h)(2) &= f(2)+h(2)\\ &= 9+(-3)\\ &=6 \end{align}$

2) We can get the same result by adding the functions first and then evaluating the result at $x=2$ .

$\begin{align} (f+h)(x) &=f(x)+h(x)\\ &=(2{x}^{3}-5x+3)+(x-5)\\ &=2x^3-4x-2 \end{align}$ .

Now we can evaluate this result at $x=2$

$\require{color}\begin{align} (f+h)(\color{Green}{2}\color{black}{)} &=2(\color{Green}{2}\color{black}{)^3-4(}\color{Green}{2}\color{black}{)-2}\\ &=16-8-2\\ &=6 \end{align}$

Both methods give the same result, and both require about the same amount of work.

ExAMPLE

Given: $f(x)=x^2-3x\\$

Find:

$1)f(x+1)-f(2)\\$

$2)f(x+1)+f(x)\\$

$3)f(x+h)-f(x)\\$

Show Solution

$1)f(x+1)-f(2)\\$

To find $f(x+1),$ substitute $\require{color}\color{Green}{x+1}$ in place of $x$ in the given function. To find $f(2)$ substitute $\require{color}\color{Green}{2}$ in place of $x$ in the given function.

$\require{color}\begin{align}f(x) &= x^2-3x\\f(\color{Green}{x+1}\color{black}{)}&= (\color{Green}{x+1}\color{black}{)^{2}-3}(\color{Green}{x+1}\color{black}{)}\\ f(\color{Green}{2}\color{black}{)} &=(\color{Green}{2}\color{black}{)}^{2}\color{black}{-3(}\color{Green}{2}\color{black}{)}\end{align}$

We now have, $f(x+1)=\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}$ and $f(2)=\color{Plum}{(2)^2-3(2)}$ .

Let’s substitute these in place of $f(x+1)$ and $f(2)$ .

$\require{color}\color{BurntOrange}{f(x+1)}\color{black}{+}\color{Plum}{f(2)}$

$\require{color}\begin{align} &=\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}\color{black}{+}\color{Plum}{(2)^2-3(2)} &&\color{blue}{\textsf{substitute}}\\ &= \color{BurntOrange}{(x+1)(x+1)-3(x+1)}\color{black}{+}\color{Plum}{(2)(2)-3(2)}\\ &= \color{BurntOrange}{x^2+2x+1-3x-3}\,\color{black}{+} \color{Plum}{\,4-6} &&\color{blue}{\textsf{multiply}}\\ &= \color{BurntOrange}{x^2-x-2}\,\color{black}{+}\,\color{Plum}{(-2)} &&\color{blue}{\textsf{combine like terms}}\\ &= x^2-x-4 &&\color{blue}{\textsf{combine like terms}}\end{align}\\$

$2)f(x+1)+f(x)\\$

To find $f(x+1),$ substitute $\require{color}\color{Green}{x+1}$ in place of $x$ in the given function.

$\require{color}\begin{align}f(x) &= x^2-3x\\f(\color{Green}{x+1}\color{black}{)} &= (\color{Green}{x+1}\color{black}{)^{2}-3}(\color{Green}{x+1}\color{black}{)}\end{align}$

We now have, $f(x+1)=\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}$ and $f(x)=\color{Plum}{x^2-3x}$ .

Let’s substitute these in place of $f(x+1)$ and $f(x)$ .

$\require{color}\color{BurntOrange}{f(x+1)}\color{black}{+}\color{Plum}{f(x)}$

$\require{color}\begin{align} &=\color{BurntOrange}{{(x+1)}^{2}-3(x+1)}\color{black}{+}\color{Plum}{x^2-3x} &&\color{blue}{\textsf{substitute}}\\ &= \color{BurntOrange}{(x+1)(x+1)-3(x+1)}\color{black}{+}\color{Plum}{x^2-3x}\\ &= \color{BurntOrange}{x^2+2x+1-3x-3}\color{black}{+}\color{Plum}{x^2-3x} &&\color{blue}{\textsf{multiply}}\\ &= 2x^2-4x-2 &&\color{blue}{\textsf{combine like terms}}\end{align}\\$

$3)f(x+h)-f(x)\\$

To find $f(x+h),$ substitute $\require{color}\color{Green}{x+h}$ in place of $x$ in the given function.

$\require{color}\begin{align}f(x) &= x^2-3x\\f(\color{Green}{x+h}\color{black}{)} &= (\color{Green}{x+h}\color{black}{)^{2}-3}(\color{Green}{x+h}\color{black}{)}\end{align}$

We now have, $f(x+h)=\color{BurntOrange}{{(x+h)}^{2}-3}(x+h)$ and $f(x)=\color{Plum}{x^2-3x}$ .

Let’s substitute these in place of $f(x+h)$ and $f(x)$ .

$\require{color}\color{BurntOrange}{f(x+h)}\color{black}{-}\color{Plum}{f(x)}$

$\require{color}\begin{align} &=\color{BurntOrange}{{(x+h)}^{2}-3(x+h)}\color{black}{-}\color{Plum}{(x^2-3x)} &&\color{blue}{\textsf{substitute}}\\ &= \color{BurntOrange}{(x+h)(x+h)-3(x+h)}\color{black}{-}\color{Plum}{(x^2-3x)}\\ &= \color{BurntOrange}{x^2+2xh+h^2-3x-3h}\color{Plum}{-x^2+3x} &&\color{blue}{\textsf{distributive property}}\\ &= 2xh+h^2-3h &&\color{blue}{\textsf{combine like terms}}\end{align}\\$

Multiply and Divide Polynomial Functions

To multiply and divide polynomial functions, we can either evaluate the function first and then perform the indicated operation or vice-versa.

Example

Let $g(t)=t+7$ and $f(t)=5t^2-3$

1) Find $(g · f)(t)$

2) Evaluate $(g · f)(-1)$

3) Evaluate $(f \cdot g)(-1)$

Should Part 2 and Part 3 have the same answer? Explain.

Show Solution

$\begin{align}\require{color}(g · f)(t) & =(t+7)(5t^2-3) \\ & =t\cdot(5t^2)+t\cdot(-3)+7\cdot(5t^2)+7\cdot(-3) &&\color{blue}{\textsf{apply the distributive property}} \\ & =5t^3-3t+35t^2-21 &&\color{blue}{\textsf{simplify}}\\ & =5t^3+35t^2-3t-21 &&\color{blue}{\textsf{write in descending order}}\end{align}\\$

Evaluate $(g · f)(-1)$

$\begin{align}\require{color}(g · f)(t) &=5t^3+35t^2-3t-21 && \color{blue}{\textsf{from above}}\\(g · f)(-1) &=5(-1)^3+35(-1)^2-3(-1)-21 && \color{blue}{\textsf{evaluate exponents and multiply}}\\ &=-5+35+3-21 && \color{blue}{\textsf{add and subtract}}\\ &=12\end{align}$

Evaluate $(f \cdot g)(-1)$

We could take the same steps as we did above by finding $(f \cdot g)(t)$ and then substituting $-1$ in place of $t$ . Here is another way you can evaluate $(f \cdot g)(-1)$ .

$(f \cdot g)(-1)= f(-1) \cdot g(-1)$

Let’s first find $f(-1)$ and $g(-1)$ . Then multiply the results together.

$\begin{align}f(\color{Green}{-1}\color{black}{)} &= 5(\color{Green}{-1}\color{black}{)^2-3}\\ &= 5(1)-3\\ &= 5-3\\ &= 2 \end{align}$

$\begin{align}g(\color{Green}{-1}\color{black}{)} &= \color{Green}{-1}\color{black}{+7}\\ &= 6\end{align}$

Multiply results together.

$(f \cdot g)(-1) = f(-1) \cdot g(-1) = (2)\cdot (6) = 12$

Part 2 and Part 3 have the same answer because multiplication is commutative.

In the next example, we will divide polynomial functions and then evaluate the new function.

Example

Given $p(x)=4x$ and $r(x)=x+3$

Find $\left(\dfrac{p}{r}\right)(x)$ and state its domain. Then evaluate $\left(\dfrac{p}{r}\right)(-2)$

Show Solution

$\left(\dfrac{p}{r}\right)(x)=\dfrac{p(x)}{r(x)}=\dfrac{4x}{x+3},\\$

Domain: All real numbers except $x\not=-3.$ The value $x=-3$ will make the denominator equal to zero and therefore the expression will be undefined. $\\$

Domain in Interval Notation: $(-\infty, -3)\bigcup(-3, \infty)\\$

Domain in Set-Builder Notation: $\{ x | x\not=-3 \}\\\\$

Evaluate this quotient for $x = -2$ .

$\begin{align}\left(\dfrac{p}{r}\right)(-2) &=\dfrac{p(-2)}{r(-2)}\\\\ &=\dfrac{4(-2)}{-2+3}\\\\ &=\dfrac{-8}{1}\\\\ &=-8\end{align}$

In our next example we will find function values from a graph.

EXAMPLE

For the functions $\color{DarkRed}{f(x)}$ , red graph, and $\color{blue}{g(x)}$ , blue graph, pictured, find the function values.

$1)f(-1)$

$2)(f-g)(2)$

$3)(f+g)(0)$

$4)\left(\dfrac{f}{g}\right)(2)\\$

Show Solution

1) $\begin{align}\require{color}f(-1)=-2 &&\color{blue}{\textsf{when the input is -1, the output is -2}} \end{align}\\$

2) $\begin{align}\require{color}\left(f-g \right)(2) &=f(2)-g(2)\\ &=7-1 &&\color{blue}{\textsf{subtract}}\\ &=6 \end{align}\\$

3) $\begin{align}\require{color}\left(f+g \right)(0) &=f(0)+g(0)\\ &=-1+(-5) &&\color{blue}{\textsf{add}}\\ &=-6 \end{align}\\$

4) $\left(\dfrac{f}{g}\right)(2)=\dfrac{f(2)}{g(2)}=\dfrac{7}{1}=7$

Now try this next example:

Module 3: Polynomials and Polynomial Functions

Learning Outcomes

Evaluate Polynomial Functions

Example

Add and Subtract Polynomial Functions

ExAMPLE

Example

Example

ExAMPLE

Multiply and Divide Polynomial Functions

Example

Example

EXAMPLE