Precalculus Examples

$12 + 2 e^{- 2 x}$

Step 1

Write $12 + 2 e^{- 2 x}$ as a function.

$f (x) = 12 + 2 e^{- 2 x}$

Step 2

Consider the difference quotient formula.

$\frac{f (x + h) - f (x)}{h}$

Step 3

Find the components of the definition.

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Step 3.1

Evaluate the function at $x = x + h$ .

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Step 3.1.1

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = 12 + 2 e^{- 2 (x + h)}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Apply the distributive property.

$f (x + h) = 12 + 2 e^{- 2 x - 2 h}$

Step 3.1.2.2

The final answer is $12 + 2 e^{- 2 x - 2 h}$ .

$12 + 2 e^{- 2 x - 2 h}$

Step 3.2

Find the components of the definition.

$f (x + h) = 12 + 2 e^{- 2 x - 2 h}$

$f (x) = 12 + 2 e^{- 2 x}$

$f (x + h) = 12 + 2 e^{- 2 x - 2 h}$

$f (x) = 12 + 2 e^{- 2 x}$

Step 4

Plug in the components.

$\frac{f (x + h) - f (x)}{h} = \frac{12 + 2 e^{- 2 x - 2 h} - (12 + 2 e^{- 2 x})}{h}$

Step 5

Simplify the numerator.

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Step 5.1

Apply the distributive property.

$\frac{12 + 2 e^{- 2 x - 2 h} - 1 \cdot 12 - (2 e^{- 2 x})}{h}$

Step 5.2

Multiply $- 1$ by $12$ .

$\frac{12 + 2 e^{- 2 x - 2 h} - 12 - (2 e^{- 2 x})}{h}$

Step 5.3

Multiply $2$ by $- 1$ .

$\frac{12 + 2 e^{- 2 x - 2 h} - 12 - 2 e^{- 2 x}}{h}$

Step 5.4

Subtract $12$ from $12$ .

$\frac{2 e^{- 2 x - 2 h} + 0 - 2 e^{- 2 x}}{h}$

Step 5.5

Add $2 e^{- 2 x - 2 h}$ and $0$ .

$\frac{2 e^{- 2 x - 2 h} - 2 e^{- 2 x}}{h}$

Step 5.6

Rewrite $2 e^{- 2 x - 2 h} - 2 e^{- 2 x}$ in a factored form.

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Step 5.6.1

Factor $2$ out of $2 e^{- 2 x - 2 h} - 2 e^{- 2 x}$ .

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Step 5.6.1.1

Factor $2$ out of $2 e^{- 2 x - 2 h}$ .

$\frac{2 e^{- 2 x - 2 h} - 2 e^{- 2 x}}{h}$

Step 5.6.1.2

Factor $2$ out of $- 2 e^{- 2 x}$ .

$\frac{2 e^{- 2 x - 2 h} + 2 (- e^{- 2 x})}{h}$

Step 5.6.1.3

Factor $2$ out of $2 e^{- 2 x - 2 h} + 2 (- e^{- 2 x})$ .

$\frac{2 (e^{- 2 x - 2 h} - e^{- 2 x})}{h}$

Step 5.6.2

Rewrite $e^{- 2 x - 2 h}$ as ${(e^{- x - h})}^{2}$ .

$\frac{2 ({(e^{- x - h})}^{2} - e^{- 2 x})}{h}$

Step 5.6.3

Rewrite $e^{- 2 x}$ as ${(e^{- x})}^{2}$ .

$\frac{2 ({(e^{- x - h})}^{2} - {(e^{- x})}^{2})}{h}$

Step 5.6.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = e^{- x - h}$ and $b = e^{- x}$ .

$\frac{2 (e^{- x - h} + e^{- x}) (e^{- x - h} - e^{- x})}{h}$

Step 6