Precalculus Examples

$f (x) = 4 x - x^{2} + 5$ $2 \leq x \leq 4$

Step 1

Write $f (x) = 4 x - x^{2} + 5$ as an equation.

$y = 4 x - x^{2} + 5$

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.

$\frac{f (4) - f (2)}{(4) - (2)}$

Step 2.2

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

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Multiply $4$ by $4$ .

$\frac{16 - 4^{2} + 5 - (4 (2) - 2^{2} + 5)}{4 - (2)}$

Step 3.1.2

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

$\frac{16 - 1 \cdot 16 + 5 - (4 (2) - 2^{2} + 5)}{4 - (2)}$

Step 3.1.3

Multiply $- 1$ by $16$ .

$\frac{16 - 16 + 5 - (4 (2) - 2^{2} + 5)}{4 - (2)}$

Step 3.1.4

Simplify each term.

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Step 3.1.4.1

Multiply $4$ by $2$ .

$\frac{16 - 16 + 5 - (8 - 2^{2} + 5)}{4 - (2)}$

Step 3.1.4.2

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

$\frac{16 - 16 + 5 - (8 - 1 \cdot 4 + 5)}{4 - (2)}$

Step 3.1.4.3

Multiply $- 1$ by $4$ .

$\frac{16 - 16 + 5 - (8 - 4 + 5)}{4 - (2)}$

Step 3.1.5

Subtract $4$ from $8$ .

$\frac{16 - 16 + 5 - (4 + 5)}{4 - (2)}$

Step 3.1.6

Add $4$ and $5$ .

$\frac{16 - 16 + 5 - 1 \cdot 9}{4 - (2)}$

Step 3.1.7

Multiply $- 1$ by $9$ .

$\frac{16 - 16 + 5 - 9}{4 - (2)}$

Step 3.1.8

Subtract $16$ from $16$ .

$\frac{0 + 5 - 9}{4 - (2)}$

Step 3.1.9

Add $0$ and $5$ .

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

Step 3.1.10

Subtract $9$ from $5$ .

$\frac{- 4}{4 - (2)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $2$ .

$\frac{- 4}{4 - 2}$

Step 3.2.2

Subtract $2$ from $4$ .

$\frac{- 4}{2}$

Step 3.3

Divide $- 4$ by $2$ .

$- 2$