Precalculus Examples

$f (x) = {(2)}^{x + 2} + 1$

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.

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

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Remove parentheses.

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

Step 2.1.2.2

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

$2^{x + h + 2} + 1$

Step 2.2

Find the components of the definition.

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

$f (x) = 2^{x + 2} + 1$

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

$f (x) = 2^{x + 2} + 1$

Step 3

Plug in the components.

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

Step 4

Simplify the numerator.

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Step 4.1

Apply the distributive property.

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

Step 4.2

Multiply $- 1$ by $1$ .

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

Step 4.3

Subtract $1$ from $1$ .

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

Step 4.4

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

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

Step 5