Multiplication and Division of Functions

Multiplication of Two Functions

The product function multiplies $f(x)$ and $g(x)$ at the same input $x$ . This produces the new function $(f\cdot g)$ .

Function Multiplication Visualization

Observe how the lines

f(x)=x

and

g(x)=2

are multiplied to become

(f \cdot g)(x)=2x

(f \cdot g)(x) = f(x) \cdot g(x)

As with a sum or difference, the product is defined only where both original functions are defined. Its domain is the intersection of their domains.

D_{f \cdot g} = D_f \cap D_g

Example of Multiplication

Use slightly different functions this time:

$f(x) = x^2$ , with domain $D_f = \{x | x \in \mathbb{R}\}$ (all real numbers).
$g(x) = x - 1$ , with domain $D_g = \{x | x \in \mathbb{R}\}$ (all real numbers).

Step 1: Determine the resulting function from multiplication

(f \cdot g)(x) = f(x) \cdot g(x)

Step 2: Determine the domain of the resulting function

We find the intersection of $D_f$ and $D_g$ :

D_{f \cdot g} = D_f \cap D_g

So, the resulting function from multiplication is $(f \cdot g)(x) = x^3 - x^2$ with the domain of all real numbers.

Division of Two Functions

Dividing function $f$ by function $g$ is also similar: we divide the result of $f(x)$ by $g(x)$ . The result is a new function $(\frac{f}{g})$ .

\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}

Division has one additional condition. Besides being in the domains of $f$ and $g$ , the input $x$ must satisfy $g(x)\neq0$ so that the denominator is nonzero.

The domain of the division function $(\frac{f}{g})$ is the intersection of domains $D_f$ and $D_g$ , but we must exclude all values of $x$ that cause $g(x) = 0$ .

D_{\frac{f}{g}}=(D_f\cap D_g)\setminus\{x\mid g(x)=0\}

The symbol $\setminus$ denotes set difference: remove the listed inputs from the domain intersection.

Example of Division

We use the same functions as in the multiplication example:

$f(x) = x^2$ , $D_f = \{x | x \in \mathbb{R}\}$
$g(x) = x - 1$ , $D_g = \{x | x \in \mathbb{R}\}$

Step 1: Determine the resulting function from division

\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{x-1}

Step 2: Determine the domain of the resulting function

First, find the intersection of $D_f$ and $D_g$ :

D_f \cap D_g = \{x | x \in \mathbb{R}\}

Second, find the value of $x$ that makes $g(x) = 0$ :

g(x) = 0

Third, exclude the value $x=1$ from the intersection of the domains:

D_{\frac{f}{g}}=\{x\mid x\in\mathbb{R}\}\setminus\{1\}

Or it can also be written as:

D_{\frac{f}{g}} = \{x | x \in \mathbb{R}, x \neq 1\}

So, the resulting function from division is $(\frac{f}{g})(x) = \frac{x^2}{x-1}$ with the domain of all real numbers except $x=1$ .

Practice Problems

Given the function $f(x) = \sqrt{x+4}$ with $D_f = \{x | x \ge -4, x \in \mathbb{R}\}$ and function $g(x) = x^2 - 9$ with $D_g = \{x | x \in \mathbb{R}\}$ .

Determine $(f \cdot g)(x)$ and its domain $D_{f \cdot g}$ .
Determine $(\frac{f}{g})(x)$ and its domain $D_{\frac{f}{g}}$ .
Calculate the value of $(f \cdot g)(5)$ .
Is $(\frac{f}{g})(3)$ defined? Explain.

Answer Key

Finding $(f \cdot g)(x)$ :
$(f \cdot g)(x) = f(x) \cdot g(x)$
Finding Domain $D_{f \cdot g}$ :
$D_{f \cdot g} = D_f \cap D_g$
So, $(f \cdot g)(x) = (x^2 - 9)\sqrt{x+4}$ with domain $\{x | x \ge -4, x \in \mathbb{R}\}$ .
Finding $(\frac{f}{g})(x)$ :

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+4}}{x^2 - 9}$

Finding Domain $D_{\frac{f}{g}}$ :

Domain intersection: $D_f \cap D_g = \{x | x \ge -4, x \in \mathbb{R}\}$ .

Find $x$ that makes $g(x)=0$ :
$g(x) = 0$
Exclude $x=3$ and $x=-3$ from the domain intersection:

$D_{\frac fg}=\{x\in\mathbb R\mid x\geq-4\}\setminus\{-3,3\}$

Or it can be written as:

$D_{\frac fg}=[-4,\infty)\setminus\{-3,3\}$

So, $(\frac{f}{g})(x) = \frac{\sqrt{x+4}}{x^2 - 9}$ with domain $\{x | x \ge -4, x \neq -3, x \neq 3\}$ .
Calculating $(f \cdot g)(5)$ :

We use the result from number 1: $(f \cdot g)(x) = (x^2 - 9)\sqrt{x+4}$ .

Since $5 \ge -4$ , $x=5$ is in the domain $D_{f \cdot g}$ .
$(f \cdot g)(5) = (5^2 - 9)\sqrt{5+4}$
Is $(\frac{f}{g})(3)$ defined?

Undefined. We look at the domain of $(\frac{f}{g})(x)$ from number 2, which is $\{x | x \ge -4, x \neq -3, x \neq 3\}$ . The value $x=3$ is explicitly excluded from the domain because it would cause the denominator $g(x) = x^2 - 9$ to become zero ( $3^2 - 9 = 0$ ). Division by zero is not allowed in mathematics.

Multiplication of Two Functions

The product function multiplies $f(x)$ and $g(x)$ at the same input $x$ . This produces the new function $(f\cdot g)$ .

Function Multiplication Visualization

Observe how the lines

f(x)=x

and

g(x)=2

are multiplied to become

(f \cdot g)(x)=2x

(f \cdot g)(x) = f(x) \cdot g(x)

As with a sum or difference, the product is defined only where both original functions are defined. Its domain is the intersection of their domains.

D_{f \cdot g} = D_f \cap D_g

Example of Multiplication

Use slightly different functions this time:

$f(x) = x^2$ , with domain $D_f = \{x | x \in \mathbb{R}\}$ (all real numbers).
$g(x) = x - 1$ , with domain $D_g = \{x | x \in \mathbb{R}\}$ (all real numbers).

Step 1: Determine the resulting function from multiplication

(f \cdot g)(x) = f(x) \cdot g(x)

Step 2: Determine the domain of the resulting function

We find the intersection of $D_f$ and $D_g$ :

D_{f \cdot g} = D_f \cap D_g

So, the resulting function from multiplication is $(f \cdot g)(x) = x^3 - x^2$ with the domain of all real numbers.

Division of Two Functions

Dividing function $f$ by function $g$ is also similar: we divide the result of $f(x)$ by $g(x)$ . The result is a new function $(\frac{f}{g})$ .

\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}

Division has one additional condition. Besides being in the domains of $f$ and $g$ , the input $x$ must satisfy $g(x)\neq0$ so that the denominator is nonzero.

The domain of the division function $(\frac{f}{g})$ is the intersection of domains $D_f$ and $D_g$ , but we must exclude all values of $x$ that cause $g(x) = 0$ .

D_{\frac{f}{g}}=(D_f\cap D_g)\setminus\{x\mid g(x)=0\}

The symbol $\setminus$ denotes set difference: remove the listed inputs from the domain intersection.

Example of Division

We use the same functions as in the multiplication example:

$f(x) = x^2$ , $D_f = \{x | x \in \mathbb{R}\}$
$g(x) = x - 1$ , $D_g = \{x | x \in \mathbb{R}\}$

Step 1: Determine the resulting function from division

\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{x-1}

Step 2: Determine the domain of the resulting function

First, find the intersection of $D_f$ and $D_g$ :

D_f \cap D_g = \{x | x \in \mathbb{R}\}

Second, find the value of $x$ that makes $g(x) = 0$ :

g(x) = 0

Third, exclude the value $x=1$ from the intersection of the domains:

D_{\frac{f}{g}}=\{x\mid x\in\mathbb{R}\}\setminus\{1\}

Or it can also be written as:

D_{\frac{f}{g}} = \{x | x \in \mathbb{R}, x \neq 1\}

So, the resulting function from division is $(\frac{f}{g})(x) = \frac{x^2}{x-1}$ with the domain of all real numbers except $x=1$ .

Practice Problems

Given the function $f(x) = \sqrt{x+4}$ with $D_f = \{x | x \ge -4, x \in \mathbb{R}\}$ and function $g(x) = x^2 - 9$ with $D_g = \{x | x \in \mathbb{R}\}$ .

Determine $(f \cdot g)(x)$ and its domain $D_{f \cdot g}$ .
Determine $(\frac{f}{g})(x)$ and its domain $D_{\frac{f}{g}}$ .
Calculate the value of $(f \cdot g)(5)$ .
Is $(\frac{f}{g})(3)$ defined? Explain.

Answer Key

Finding $(f \cdot g)(x)$ :
$(f \cdot g)(x) = f(x) \cdot g(x)$
Finding Domain $D_{f \cdot g}$ :
$D_{f \cdot g} = D_f \cap D_g$
So, $(f \cdot g)(x) = (x^2 - 9)\sqrt{x+4}$ with domain $\{x | x \ge -4, x \in \mathbb{R}\}$ .
Finding $(\frac{f}{g})(x)$ :

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x+4}}{x^2 - 9}$

Finding Domain $D_{\frac{f}{g}}$ :

Domain intersection: $D_f \cap D_g = \{x | x \ge -4, x \in \mathbb{R}\}$ .

Find $x$ that makes $g(x)=0$ :
$g(x) = 0$
Exclude $x=3$ and $x=-3$ from the domain intersection:

$D_{\frac fg}=\{x\in\mathbb R\mid x\geq-4\}\setminus\{-3,3\}$

Or it can be written as:

$D_{\frac fg}=[-4,\infty)\setminus\{-3,3\}$

So, $(\frac{f}{g})(x) = \frac{\sqrt{x+4}}{x^2 - 9}$ with domain $\{x | x \ge -4, x \neq -3, x \neq 3\}$ .
Calculating $(f \cdot g)(5)$ :

We use the result from number 1: $(f \cdot g)(x) = (x^2 - 9)\sqrt{x+4}$ .

Since $5 \ge -4$ , $x=5$ is in the domain $D_{f \cdot g}$ .
$(f \cdot g)(5) = (5^2 - 9)\sqrt{5+4}$
Is $(\frac{f}{g})(3)$ defined?

Undefined. We look at the domain of $(\frac{f}{g})(x)$ from number 2, which is $\{x | x \ge -4, x \neq -3, x \neq 3\}$ . The value $x=3$ is explicitly excluded from the domain because it would cause the denominator $g(x) = x^2 - 9$ to become zero ( $3^2 - 9 = 0$ ). Division by zero is not allowed in mathematics.