Precalculus Examples

$g (x) = \frac{1}{2} {(2)}^{x - 1}$

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.

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

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

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

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

Step 2.1.2.2

Cancel the common factor of $2^{x + h - 1}$ and $2$ .

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Step 2.1.2.2.1

Factor $2$ out of $2^{x + h - 1}$ .

$g (x + h) = \frac{2 \cdot 2^{x + h - 2}}{2}$

Step 2.1.2.2.2

Cancel the common factors.

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Step 2.1.2.2.2.1

Factor $2$ out of $2$ .

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

Step 2.1.2.2.2.2

Cancel the common factor.

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

Step 2.1.2.2.2.3

Rewrite the expression.

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

Step 2.1.2.2.2.4

Divide $2^{x + h - 2}$ by $1$ .

$g (x + h) = 2^{x + h - 2}$

Step 2.1.2.3

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

$2^{x + h - 2}$

Step 2.2

Find the components of the definition.

$g (x + h) = 2^{x + h - 2}$

$g (x) = 2^{x - 2}$

$g (x + h) = 2^{x + h - 2}$

$g (x) = 2^{x - 2}$

Step 3

Plug in the components.

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

Step 4

Multiply $- 1$ by $2^{x - 2}$ .

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

Step 5