Precalculus Examples

$f (x) = - {(2)}^{x} - 1$ ; over the interval $2 \leq x \leq 5$

Step 1

Write $f (x) = - {(2)}^{x} - 1$ as an equation.

$y = - {(2)}^{x} - 1$

Step 2

Substitute using the average rate of change formula.

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Step 2.1

The average rate of change of a function can be found by calculating the change in $y$ values of the two points divided by the change in $x$ values of the two points.

$\frac{f (5) - f (2)}{(5) - (2)}$

Step 2.2

Substitute the equation $y = - {(2)}^{x} - 1$ for $f (5)$ and $f (2)$ , replacing $x$ in the function with the corresponding $x$ value.

$\frac{(- {(2)}^{5} - 1) - (- {(2)}^{2} - 1)}{(5) - (2)}$

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $2$ to the power of $5$ .

$\frac{- 1 \cdot 32 - 1 - (- 2^{2} - 1)}{5 - (2)}$

Step 3.1.2

Multiply $- 1$ by $32$ .

$\frac{- 32 - 1 - (- 2^{2} - 1)}{5 - (2)}$

Step 3.1.3

Simplify each term.

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Step 3.1.3.1

Raise $2$ to the power of $2$ .

$\frac{- 32 - 1 - (- 1 \cdot 4 - 1)}{5 - (2)}$

Step 3.1.3.2

Multiply $- 1$ by $4$ .

$\frac{- 32 - 1 - (- 4 - 1)}{5 - (2)}$

Step 3.1.4

Subtract $1$ from $- 4$ .

$\frac{- 32 - 1 - - 5}{5 - (2)}$

Step 3.1.5

Multiply $- 1$ by $- 5$ .

$\frac{- 32 - 1 + 5}{5 - (2)}$

Step 3.1.6

Subtract $1$ from $- 32$ .

$\frac{- 33 + 5}{5 - (2)}$

Step 3.1.7

Add $- 33$ and $5$ .

$\frac{- 28}{5 - (2)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $2$ .

$\frac{- 28}{5 - 2}$

Step 3.2.2

Subtract $2$ from $5$ .

$\frac{- 28}{3}$

Step 3.3

Move the negative in front of the fraction.

$- \frac{28}{3}$