Precalculus Examples

Given the function $h (x) = x^{2} - 7 x + 6$ , determine the average rate of change of the function over the interval $3 \leq x \leq 7$

Step 1

Write $h (x) = x^{2} - 7 x + 6$ as an equation.

$y = x^{2} - 7 x + 6$

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.

$\frac{f (7) - f (3)}{(7) - (3)}$

Step 2.2

Substitute the equation $y = x^{2} - 7 x + 6$ for $f (7)$ and $f (3)$ , replacing $x$ in the function with the corresponding $x$ value.

$\frac{({(7)}^{2} - 7 \cdot 7 + 6) - ({(3)}^{2} - 7 \cdot 3 + 6)}{(7) - (3)}$

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $7$ to the power of $2$ .

$\frac{49 - 7 \cdot 7 + 6 - (3^{2} - 7 \cdot 3 + 6)}{7 - (3)}$

Step 3.1.2

Multiply $- 7$ by $7$ .

$\frac{49 - 49 + 6 - (3^{2} - 7 \cdot 3 + 6)}{7 - (3)}$

Step 3.1.3

Simplify each term.

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Step 3.1.3.1

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

$\frac{49 - 49 + 6 - (9 - 7 \cdot 3 + 6)}{7 - (3)}$

Step 3.1.3.2

Multiply $- 7$ by $3$ .

$\frac{49 - 49 + 6 - (9 - 21 + 6)}{7 - (3)}$

Step 3.1.4

Subtract $21$ from $9$ .

$\frac{49 - 49 + 6 - (- 12 + 6)}{7 - (3)}$

Step 3.1.5

Add $- 12$ and $6$ .

$\frac{49 - 49 + 6 - - 6}{7 - (3)}$

Step 3.1.6

Multiply $- 1$ by $- 6$ .

$\frac{49 - 49 + 6 + 6}{7 - (3)}$

Step 3.1.7

Subtract $49$ from $49$ .

$\frac{0 + 6 + 6}{7 - (3)}$

Step 3.1.8

Add $0$ and $6$ .

$\frac{6 + 6}{7 - (3)}$

Step 3.1.9

Add $6$ and $6$ .

$\frac{12}{7 - (3)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $3$ .

$\frac{12}{7 - 3}$

Step 3.2.2

Subtract $3$ from $7$ .

$\frac{12}{4}$

Step 3.3

Divide $12$ by $4$ .

$3$