Precalculus Examples

$f (x) = e^{x - 3} + 2$ , $[3, 6]$

Step 1

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

$y = e^{x - 3} + 2$

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.

$\frac{f (6) - f (3)}{(6) - (3)}$

Step 2.2

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

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Subtract $3$ from $6$ .

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

Step 3.1.2

Simplify each term.

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Step 3.1.2.1

Subtract $3$ from $3$ .

$\frac{e^{3} + 2 - (e^{0} + 2)}{6 - (3)}$

Step 3.1.2.2

Anything raised to $0$ is $1$ .

$\frac{e^{3} + 2 - (1 + 2)}{6 - (3)}$

Step 3.1.3

Add $1$ and $2$ .

$\frac{e^{3} + 2 - 1 \cdot 3}{6 - (3)}$

Step 3.1.4

Multiply $- 1$ by $3$ .

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

Step 3.1.5

Subtract $3$ from $2$ .

$\frac{e^{3} - 1}{6 - (3)}$

Step 3.1.6

Rewrite $e^{3} - 1$ in a factored form.

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Step 3.1.6.1

Rewrite $1$ as $1^{3}$ .

$\frac{e^{3} - 1^{3}}{6 - (3)}$

Step 3.1.6.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = e$ and $b = 1$ .

$\frac{(e - 1) (e^{2} + e \cdot 1 + 1^{2})}{6 - (3)}$

Step 3.1.6.3

Simplify.

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Step 3.1.6.3.1

Multiply $e$ by $1$ .

$\frac{(e - 1) (e^{2} + e + 1^{2})}{6 - (3)}$

Step 3.1.6.3.2

One to any power is one.

$\frac{(e - 1) (e^{2} + e + 1)}{6 - (3)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $3$ .

$\frac{(e - 1) (e^{2} + e + 1)}{6 - 3}$

Step 3.2.2

Subtract $3$ from $6$ .

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