Precalculus Examples

$f (x) = \frac{x}{x - 6}$ , $g (x) = \frac{- 8}{x + 2}$

Step 1

Find the product of the functions.

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Step 1.1

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

$(\frac{x}{x - 6}) \cdot (\frac{- 8}{x + 2})$

Step 1.2

Simplify.

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Step 1.2.1

Simplify the expression.

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Step 1.2.1.1

Move the negative in front of the fraction.

$\frac{x}{x - 6} \cdot (- \frac{8}{x + 2})$

Step 1.2.1.2

Rewrite using the commutative property of multiplication.

$- \frac{x}{x - 6} \cdot \frac{8}{x + 2}$

Step 1.2.2

Multiply $\frac{8}{x + 2}$ by $\frac{x}{x - 6}$ .

$- \frac{8 x}{(x + 2) (x - 6)}$

Step 2

Set the denominator in $\frac{8 x}{(x + 2) (x - 6)}$ equal to $0$ to find where the expression is undefined.

$(x + 2) (x - 6) = 0$

Step 3

Solve for $x$ .

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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 $0$ .

$x + 2 = 0$

$x - 6 = 0$

Step 3.2

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 3.2.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 3.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.3

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

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Step 3.3.1

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

$x - 6 = 0$

Step 3.3.2

Add $6$ to both sides of the equation.

$x = 6$

Step 3.4

The final solution is all the values that make $(x + 2) (x - 6) = 0$ true.

$x = - 2, 6$

Step 4

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 6) \cup (6, \infty)$

Set-Builder Notation:

${x | x \neq - 2, 6}$

Step 5