Multiplication and Division of Functions

Multiplication of Two Functions

Multiplying two functions, $f$ and $g$, is as easy as multiplying two numbers. We just multiply the result of $f(x)$ by $g(x)$ for the same value of $x$. The result is a new function $(f \cdot g)$.

Just like addition and subtraction, this multiplication machine $(f \cdot g)$ can only process raw materials (values of $x$) that can be processed by both original machines, $f$ and $g$. So, its domain is the intersection of the domain of $f$ and the domain of $g$.

Example of Multiplication

Let's use slightly different functions this time:

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

Step 1: Determine the resulting function from multiplication

Step 2: Determine the domain of the resulting function

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

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 $(\frac{f}{g})$.

Now, there's a very important additional rule! We know that division by zero is not allowed. So, besides the value of $x$ needing to be in the domain of both $f$ and $g$, the value of $g(x)$ (the divisor function) cannot be equal to zero.

Therefore, 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 $g(x) = 0$.

The $-$ sign here means "minus" or "excluded".

Example of Division

We use the same functions as in the multiplication example:

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

Step 1: Determine the resulting function from division

Step 2: Determine the domain of the resulting function

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

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

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

Or it can also be written as:

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 $D_g = \{x | x \in \mathbb{R}\}$.

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

Answer Key

1. Finding $(f \cdot g)(x)$:

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

   $$= (\sqrt{x+4}) \cdot (x^2 - 9)$$

   $$= (x^2 - 9)\sqrt{x+4}$$

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

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

   $$= \{x | x \ge -4, x \in \mathbb{R}\} \cap \{x | x \in \mathbb{R}\}$$

   $$= \{x | x \ge -4, x \in \mathbb{R}\}$$

   So, $(f \cdot g)(x) = (x^2 - 9)\sqrt{x+4}$ with domain $\{x | x \ge -4, x \in \mathbb{R}\}$.

2. 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 $g(x)=0$:

   $$g(x) = 0$$

   $$x^2 - 9 = 0$$

   $$(x-3)(x+3) = 0$$

   $$x = 3 \text{ or } x = -3$$

   Exclude $x=3$ and $x=-3$ from the domain intersection:

   $$D_{\frac{f}{g}} = \{x | x \ge -4, x \in \mathbb{R}\} - \{-3, 3\}$$

   Or it can be written as:

   $$D_{\frac{f}{g}} = \{x | x \ge -4, x \in \mathbb{R}, x \neq -3, x \neq 3\}$$

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

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 $D_{f \cdot g}$.

   $$(f \cdot g)(5) = (5^2 - 9)\sqrt{5+4}$$

   $$= (25 - 9)\sqrt{9}$$

   $$= (16)(3)$$

   $$= 48$$

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.