Precalculus Examples

$f (x) = 4 e^{2 x} - 1$

Step 1

Consider the difference quotient formula.

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

Step 2

Find the components of the definition.

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Step 2.1

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

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Step 2.1.1

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

$f (x + h) = 4 e^{2 (x + h)} - 1$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Apply the distributive property.

$f (x + h) = 4 e^{2 x + 2 h} - 1$

Step 2.1.2.2

The final answer is $4 e^{2 x + 2 h} - 1$ .

$4 e^{2 x + 2 h} - 1$

Step 2.2

Find the components of the definition.

$f (x + h) = 4 e^{2 x + 2 h} - 1$

$f (x) = 4 e^{2 x} - 1$

$f (x + h) = 4 e^{2 x + 2 h} - 1$

$f (x) = 4 e^{2 x} - 1$

Step 3

Plug in the components.

$\frac{f (x + h) - f (x)}{h} = \frac{4 e^{2 x + 2 h} - 1 - (4 e^{2 x} - 1)}{h}$

Step 4

Simplify the numerator.

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Step 4.1

Apply the distributive property.

$\frac{4 e^{2 x + 2 h} - 1 - (4 e^{2 x}) - - 1}{h}$

Step 4.2

Multiply $4$ by $- 1$ .

$\frac{4 e^{2 x + 2 h} - 1 - 4 e^{2 x} - - 1}{h}$

Step 4.3

Multiply $- 1$ by $- 1$ .

$\frac{4 e^{2 x + 2 h} - 1 - 4 e^{2 x} + 1}{h}$

Step 4.4

Add $- 1$ and $1$ .

$\frac{4 e^{2 x + 2 h} + 0 - 4 e^{2 x}}{h}$

Step 4.5

Add $4 e^{2 x + 2 h}$ and $0$ .

$\frac{4 e^{2 x + 2 h} - 4 e^{2 x}}{h}$

Step 4.6

Rewrite $4 e^{2 x + 2 h} - 4 e^{2 x}$ in a factored form.

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Step 4.6.1

Factor $4$ out of $4 e^{2 x + 2 h} - 4 e^{2 x}$ .

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Step 4.6.1.1

Factor $4$ out of $4 e^{2 x + 2 h}$ .

$\frac{4 e^{2 x + 2 h} - 4 e^{2 x}}{h}$

Step 4.6.1.2

Factor $4$ out of $- 4 e^{2 x}$ .

$\frac{4 e^{2 x + 2 h} + 4 (- e^{2 x})}{h}$

Step 4.6.1.3

Factor $4$ out of $4 e^{2 x + 2 h} + 4 (- e^{2 x})$ .

$\frac{4 (e^{2 x + 2 h} - e^{2 x})}{h}$

Step 4.6.2

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

$\frac{4 ({(e^{x + h})}^{2} - e^{2 x})}{h}$

Step 4.6.3

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

$\frac{4 ({(e^{x + h})}^{2} - {(e^{x})}^{2})}{h}$

Step 4.6.4

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 $b = e^{x}$ .

$\frac{4 (e^{x + h} + e^{x}) (e^{x + h} - e^{x})}{h}$

Step 5