Finite Math Examples

$f (x) = \frac{x}{x - 8}$ , $g (x) = \frac{- 4}{x + 1}$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\frac{- 4}{x + 1})$ by substituting in the value of $g$ into $f$ .

$f (\frac{- 4}{x + 1}) = \frac{\frac{- 4}{x + 1}}{(\frac{- 4}{x + 1}) - 8}$

Step 3

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{- 4}{x + 1}) = \frac{- 4}{x + 1} \cdot \frac{1}{\frac{- 4}{x + 1} - 8}$

Step 4

Move the negative in front of the fraction.

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4}{x + 1} - 8}$

Step 5

Simplify the denominator.

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Step 5.1

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4}{x + 1} - 8 \cdot \frac{x + 1}{x + 1}}$

Step 5.2

Combine $- 8$ and $\frac{x + 1}{x + 1}$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4}{x + 1} + \frac{- 8 (x + 1)}{x + 1}}$

Step 5.3

Combine the numerators over the common denominator.

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 - 8 (x + 1)}{x + 1}}$

Step 5.4

Rewrite $\frac{- 4 - 8 (x + 1)}{x + 1}$ in a factored form.

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Step 5.4.1

Factor $- 4$ out of $- 4 - 8 (x + 1)$ .

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Step 5.4.1.1

Reorder $- 4$ and $- 8 (x + 1)$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 8 (x + 1) - 4}{x + 1}}$

Step 5.4.1.2

Factor $- 4$ out of $- 8 (x + 1)$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 (2 (x + 1)) - 4}{x + 1}}$

Step 5.4.1.3

Factor $- 4$ out of $- 4$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 (2 (x + 1)) - 4 \cdot 1}{x + 1}}$

Step 5.4.1.4

Factor $- 4$ out of $- 4 (2 (x + 1)) - 4 (1)$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 (2 (x + 1) + 1)}{x + 1}}$

Step 5.4.2

Apply the distributive property.

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 (2 x + 2 \cdot 1 + 1)}{x + 1}}$

Step 5.4.3

Multiply $2$ by $1$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 (2 x + 2 + 1)}{x + 1}}$

Step 5.4.4

Add $2$ and $1$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{\frac{- 4 (2 x + 3)}{x + 1}}$

Step 5.5

Move the negative in front of the fraction.

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{1}{- \frac{4 (2 x + 3)}{x + 1}}$

Step 6

Cancel the common factor of $1$ and $- 1$ .

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Step 6.1

Rewrite $1$ as $- 1 (- 1)$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot \frac{- 1 \cdot - 1}{- \frac{4 (2 x + 3)}{x + 1}}$

Step 6.2

Move the negative in front of the fraction.

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot (- \frac{1}{\frac{4 (2 x + 3)}{x + 1}})$

Step 7

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot (- (1 (\frac{x + 1}{4 (2 x + 3)})))$

Step 8

Multiply $\frac{x + 1}{4 (2 x + 3)}$ by $1$ .

$f (\frac{- 4}{x + 1}) = - \frac{4}{x + 1} \cdot (- \frac{x + 1}{4 (2 x + 3)})$

Step 9

Cancel the common factor of $4$ .

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Step 9.1

Move the leading negative in $- \frac{4}{x + 1}$ into the numerator.

$f (\frac{- 4}{x + 1}) = \frac{- 4}{x + 1} \cdot (- \frac{x + 1}{4 (2 x + 3)})$

Step 9.2

Move the leading negative in $- \frac{x + 1}{4 (2 x + 3)}$ into the numerator.

$f (\frac{- 4}{x + 1}) = \frac{- 4}{x + 1} \cdot \frac{- (x + 1)}{4 (2 x + 3)}$

Step 9.3

Factor $4$ out of $- 4$ .

$f (\frac{- 4}{x + 1}) = \frac{4 (- 1)}{x + 1} \cdot \frac{- (x + 1)}{4 (2 x + 3)}$

Step 9.4

Cancel the common factor.

$f (\frac{- 4}{x + 1}) = \frac{4 \cdot - 1}{x + 1} \cdot \frac{- (x + 1)}{4 (2 x + 3)}$

Step 9.5

Rewrite the expression.

$f (\frac{- 4}{x + 1}) = \frac{- 1}{x + 1} \cdot \frac{- (x + 1)}{2 x + 3}$

Step 10

Cancel the common factor of $x + 1$ .

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Step 10.1

Factor $x + 1$ out of $- (x + 1)$ .

$f (\frac{- 4}{x + 1}) = \frac{- 1}{x + 1} \cdot \frac{(x + 1) \cdot - 1}{2 x + 3}$

Step 10.2

Cancel the common factor.

$f (\frac{- 4}{x + 1}) = \frac{- 1}{x + 1} \cdot \frac{(x + 1) \cdot - 1}{2 x + 3}$

Step 10.3

Rewrite the expression.

$f (\frac{- 4}{x + 1}) = - \frac{- 1}{2 x + 3}$

Step 11

Move the negative in front of the fraction.

$f (\frac{- 4}{x + 1}) = \frac{1}{2 x + 3}$