Precalculus Examples

$f (x) = 3 x^{2} + 2 x + 1$ for $2 \leq x \leq 4$

Step 1

Write $f (x) = 3 x^{2} + 2 x + 1$ as an equation.

$y = 3 x^{2} + 2 x + 1$

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.

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

Step 2.2

Substitute the equation $y = 3 x^{2} + 2 x + 1$ for $f (4)$ and $f (2)$ , replacing $x$ in the function with the corresponding $x$ value.

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

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

$\frac{3 \cdot 16 + 2 (4) + 1 - (3 \cdot 2^{2} + 2 (2) + 1)}{4 - (2)}$

Step 3.1.2

Multiply $3$ by $16$ .

$\frac{48 + 2 (4) + 1 - (3 \cdot 2^{2} + 2 (2) + 1)}{4 - (2)}$

Step 3.1.3

Multiply $2$ by $4$ .

$\frac{48 + 8 + 1 - (3 \cdot 2^{2} + 2 (2) + 1)}{4 - (2)}$

Step 3.1.4

Simplify each term.

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Step 3.1.4.1

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

$\frac{48 + 8 + 1 - (3 \cdot 4 + 2 (2) + 1)}{4 - (2)}$

Step 3.1.4.2

Multiply $3$ by $4$ .

$\frac{48 + 8 + 1 - (12 + 2 (2) + 1)}{4 - (2)}$

Step 3.1.4.3

Multiply $2$ by $2$ .

$\frac{48 + 8 + 1 - (12 + 4 + 1)}{4 - (2)}$

Step 3.1.5

Add $12$ and $4$ .

$\frac{48 + 8 + 1 - (16 + 1)}{4 - (2)}$

Step 3.1.6

Add $16$ and $1$ .

$\frac{48 + 8 + 1 - 1 \cdot 17}{4 - (2)}$

Step 3.1.7

Multiply $- 1$ by $17$ .

$\frac{48 + 8 + 1 - 17}{4 - (2)}$

Step 3.1.8

Add $48$ and $8$ .

$\frac{56 + 1 - 17}{4 - (2)}$

Step 3.1.9

Add $56$ and $1$ .

$\frac{57 - 17}{4 - (2)}$

Step 3.1.10

Subtract $17$ from $57$ .

$\frac{40}{4 - (2)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $2$ .

$\frac{40}{4 - 2}$

Step 3.2.2

Subtract $2$ from $4$ .

$\frac{40}{2}$

Step 3.3

Divide $40$ by $2$ .

$20$