Precalculus Examples

Given the function $g (x) = - x^{2} + x + 5$ , determine the average rate of change of the function over the interval $- 4 \leq x \leq 3$

Step 1

Write $g (x) = - x^{2} + x + 5$ as an equation.

$y = - x^{2} + x + 5$

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.

$\frac{f (3) - f (- 4)}{(3) - (- 4)}$

Step 2.2

Substitute the equation $y = - x^{2} + x + 5$ for $f (3)$ and $f (- 4)$ , replacing $x$ in the function with the corresponding $x$ value.

$\frac{- {(3)}^{2} + 3 + 5 - (- {(- 4)}^{2} - 4 + 5)}{(3) - (- 4)}$

Step 3

Simplify the expression.

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Step 3.1

Remove parentheses.

$\frac{- {(3)}^{2} + 3 + 5 - (- {(- 4)}^{2} - 4 + 5)}{(3) - (- 4)}$

Step 3.2

Simplify the numerator.

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Step 3.2.1

Raise $3$ to the power of $2$ .

$\frac{- 1 \cdot 9 + 3 + 5 - (- {(- 4)}^{2} - 4 + 5)}{3 - (- 4)}$

Step 3.2.2

Multiply $- 1$ by $9$ .

$\frac{- 9 + 3 + 5 - (- {(- 4)}^{2} - 4 + 5)}{3 - (- 4)}$

Step 3.2.3

Simplify each term.

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Step 3.2.3.1

Raise $- 4$ to the power of $2$ .

$\frac{- 9 + 3 + 5 - (- 1 \cdot 16 - 4 + 5)}{3 - (- 4)}$

Step 3.2.3.2

Multiply $- 1$ by $16$ .

$\frac{- 9 + 3 + 5 - (- 16 - 4 + 5)}{3 - (- 4)}$

Step 3.2.4

Subtract $4$ from $- 16$ .

$\frac{- 9 + 3 + 5 - (- 20 + 5)}{3 - (- 4)}$

Step 3.2.5

Add $- 20$ and $5$ .

$\frac{- 9 + 3 + 5 - - 15}{3 - (- 4)}$

Step 3.2.6

Multiply $- 1$ by $- 15$ .

$\frac{- 9 + 3 + 5 + 15}{3 - (- 4)}$

Step 3.2.7

Add $- 9$ and $3$ .

$\frac{- 6 + 5 + 15}{3 - (- 4)}$

Step 3.2.8

Add $- 6$ and $5$ .

$\frac{- 1 + 15}{3 - (- 4)}$

Step 3.2.9

Add $- 1$ and $15$ .

$\frac{14}{3 - (- 4)}$

Step 3.3

Simplify the denominator.

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Step 3.3.1

Multiply $- 1$ by $- 4$ .

$\frac{14}{3 + 4}$

Step 3.3.2

Add $3$ and $4$ .

$\frac{14}{7}$

Step 3.4

Divide $14$ by $7$ .

$2$