Precalculus Examples

$f (x) = 3 x \ln (x)$ ; $[1, e^{2}]$

Step 1

Write $f (x) = 3 x \ln (x)$ as an equation.

$y = 3 x \ln (x)$

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.

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

Step 2.2

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

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

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Use logarithm rules to move $2$ out of the exponent.

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

Step 3.1.2

The natural logarithm of $e$ is $1$ .

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

Step 3.1.3

Multiply $2$ by $1$ .

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

Step 3.1.4

Multiply $2$ by $3$ .

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

Step 3.1.5

Multiply $- 1 \cdot 3 \cdot 1$ .

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Step 3.1.5.1

Multiply $- 1$ by $3$ .

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

Step 3.1.5.2

Multiply $- 3$ by $1$ .

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

Step 3.1.6

The natural logarithm of $1$ is $0$ .

$\frac{6 e^{2} - 3 \cdot 0}{e^{2} - (1)}$

Step 3.1.7

Multiply $- 3$ by $0$ .

$\frac{6 e^{2} + 0}{e^{2} - (1)}$

Step 3.1.8

Add $6 e^{2}$ and $0$ .

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

Step 3.2

Simplify the denominator.

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Step 3.2.1

Rewrite $1$ as $1^{2}$ .

$\frac{6 e^{2}}{e^{2} - 1^{2}}$

Step 3.2.2

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

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