Precalculus Examples

$A (x) = P {(0.82)}^{x}$

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.

$A (x + h) = P {(0.82)}^{x + h}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Remove parentheses.

$A (x + h) = P \cdot {0.82}^{x + h}$

Step 2.1.2.2

The final answer is $P \cdot {0.82}^{x + h}$ .

$P \cdot {0.82}^{x + h}$

Step 2.2

Find the components of the definition.

$A (x + h) = P \cdot {0.82}^{x + h}$

$A (x) = P \cdot {0.82}^{x}$

$A (x + h) = P \cdot {0.82}^{x + h}$

$A (x) = P \cdot {0.82}^{x}$

Step 3

Plug in the components.

$\frac{A (x + h) - A x}{h} = \frac{P \cdot {0.82}^{x + h} - (P \cdot {0.82}^{x})}{h}$

Step 4

Simplify the numerator.

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Step 4.1

Factor $P$ out of $P \cdot {0.82}^{x + h} - P \cdot {0.82}^{x}$ .

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Step 4.1.1

Factor $P$ out of $P \cdot {0.82}^{x + h}$ .

$\frac{P \cdot {0.82}^{x + h} - P \cdot {0.82}^{x}}{h}$

Step 4.1.2

Factor $P$ out of $- P \cdot {0.82}^{x}$ .

$\frac{P \cdot {0.82}^{x + h} + P (- 1 \cdot {0.82}^{x})}{h}$

Step 4.1.3

Factor $P$ out of $P \cdot {0.82}^{x + h} + P (- 1 \cdot {0.82}^{x})$ .

$\frac{P ({0.82}^{x + h} - 1 \cdot {0.82}^{x})}{h}$

Step 4.2

Rewrite $- 1 \cdot {0.82}^{x}$ as $- {0.82}^{x}$ .

$\frac{P ({0.82}^{x + h} - {0.82}^{x})}{h}$

Step 5