Algebra Examples

$f (x) = \frac{4}{x + 6}$ , $g (x) = \frac{24}{x}$

Step 1

Find the product of the functions.

Tap for more steps...

Step 1.1

Replace the function designators with the actual functions in $f (x) \cdot (g (x))$ .

$(\frac{4}{x + 6}) \cdot (\frac{24}{x})$

Step 1.2

Simplify.

Tap for more steps...

Step 1.2.1

Multiply $(\frac{4}{x + 6}) (\frac{24}{x})$ .

Tap for more steps...

Step 1.2.1.1

Multiply $\frac{4}{x + 6}$ by $\frac{24}{x}$ .

$\frac{4 \cdot 24}{(x + 6) x}$

Step 1.2.1.2

Multiply $4$ by $24$ .

$\frac{96}{(x + 6) x}$

Step 1.2.2

Reorder factors in $\frac{96}{(x + 6) x}$ .

$\frac{96}{x (x + 6)}$

Step 2

Set the denominator in $\frac{96}{x (x + 6)}$ equal to $0$ to find where the expression is undefined.

$x (x + 6) = 0$

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 6 = 0$

Step 3.2

Set $x$ equal to $0$ .

$x = 0$

Step 3.3

Set $x + 6$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.3.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 3.3.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 3.4

The final solution is all the values that make $x (x + 6) = 0$ true.

$x = 0, - 6$

Step 4

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 6) \cup (- 6, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 6}$

Step 5