Calculus Examples

$f (x) = \frac{1}{2 x - 1}$ , $g (x) = \frac{1}{x + 2}$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (\frac{1}{x + 2})$ by substituting in the value of $g$ into $f$ .

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

Step 3

Simplify the denominator.

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Step 3.1

Combine $2$ and $\frac{1}{x + 2}$ .

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

Step 3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

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

Step 3.3

Combine $- 1$ and $\frac{x + 2}{x + 2}$ .

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

Step 3.4

Combine the numerators over the common denominator.

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

Step 3.5

Rewrite $\frac{2 - (x + 2)}{x + 2}$ in a factored form.

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Step 3.5.1

Apply the distributive property.

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

Step 3.5.2

Multiply $- 1$ by $2$ .

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

Step 3.5.3

Subtract $2$ from $2$ .

$f (\frac{1}{x + 2}) = \frac{1}{\frac{- x + 0}{x + 2}}$

Step 3.5.4

Add $- x$ and $0$ .

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

Step 3.6

Move the negative in front of the fraction.

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

Step 4

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

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Step 4.1

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

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

Step 4.2

Move the negative in front of the fraction.

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

Step 5

Multiply the numerator by the reciprocal of the denominator.

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

Step 6

Multiply $\frac{x + 2}{x}$ by $1$ .

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