Calculus Examples

$f (x) = x^{3} - 12 x$ , $(0, 4)$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate.

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Step 1.1.1.1.1

By the Sum Rule, the derivative of $x^{3} - 12 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 12 x]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 12 x]$

Step 1.1.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2} + \frac{d}{d x} [- 12 x]$

Step 1.1.1.2

Evaluate $\frac{d}{d x} [- 12 x]$ .

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Step 1.1.1.2.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x$ with respect to $x$ is $- 12 \frac{d}{d x} [x]$ .

$3 x^{2} - 12 \frac{d}{d x} [x]$

Step 1.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$3 x^{2} - 12 \cdot 1$

Step 1.1.1.2.3

Multiply $- 12$ by $1$ .

$f' (x) = 3 x^{2} - 12$

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 12$ .

$3 x^{2} - 12$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 12 = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$3 x^{2} - 12 = 0$

Step 1.2.2

Add $12$ to both sides of the equation.

$3 x^{2} = 12$

Step 1.2.3

Divide each term in $3 x^{2} = 12$ by $3$ and simplify.

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Step 1.2.3.1

Divide each term in $3 x^{2} = 12$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $3$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{12}{3}$

Step 1.2.3.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{12}{3}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $12$ by $3$ .

$x^{2} = 4$

Step 1.2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Step 1.2.5

Simplify $\pm \sqrt{4}$ .

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Step 1.2.5.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Step 1.2.5.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

Step 1.2.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Step 1.2.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

Step 1.2.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 1.4

Evaluate $x^{3} - 12 x$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = 2$ .

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Step 1.4.1.1

Substitute $2$ for $x$ .

${(2)}^{3} - 12 \cdot 2$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

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

$8 - 12 \cdot 2$

Step 1.4.1.2.1.2

Multiply $- 12$ by $2$ .

$8 - 24$

Step 1.4.1.2.2

Subtract $24$ from $8$ .

$- 16$

Step 1.4.2

Evaluate at $x = - 2$ .

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Step 1.4.2.1

Substitute $- 2$ for $x$ .

${(- 2)}^{3} - 12 \cdot - 2$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

Simplify each term.

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Step 1.4.2.2.1.1

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

$- 8 - 12 \cdot - 2$

Step 1.4.2.2.1.2

Multiply $- 12$ by $- 2$ .

$- 8 + 24$

Step 1.4.2.2.2

Add $- 8$ and $24$ .

$16$

Step 1.4.3

List all of the points.

$(2, - 16), (- 2, 16)$

Step 2

Exclude the points that are not on the interval.

$(2, - 16)$

Step 3

Use the first derivative test to determine which points can be maxima or minima.

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Step 3.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Step 3.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - 2)$ in the first derivative $3 x^{2} - 12$ to check if the result is negative or positive.

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Step 3.2.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = 3 {(- 4)}^{2} - 12$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

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

$f' (- 4) = 3 \cdot 16 - 12$

Step 3.2.2.1.2

Multiply $3$ by $16$ .

$f' (- 4) = 48 - 12$

Step 3.2.2.2

Subtract $12$ from $48$ .

$f' (- 4) = 36$

Step 3.2.2.3

The final answer is $36$ .

$36$

Step 3.3

Substitute any number, such as $0$ , from the interval $(- 2, 2)$ in the first derivative $3 x^{2} - 12$ to check if the result is negative or positive.

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Step 3.3.1

Replace the variable $x$ with $0$ in the expression.

$f' (0) = 3 {(0)}^{2} - 12$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Raising $0$ to any positive power yields $0$ .

$f' (0) = 3 \cdot 0 - 12$

Step 3.3.2.1.2

Multiply $3$ by $0$ .

$f' (0) = 0 - 12$

Step 3.3.2.2

Subtract $12$ from $0$ .

$f' (0) = - 12$

Step 3.3.2.3

The final answer is $- 12$ .

$- 12$

Step 3.4

Substitute any number, such as $4$ , from the interval $(2, \infty)$ in the first derivative $3 x^{2} - 12$ to check if the result is negative or positive.

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Step 3.4.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = 3 {(4)}^{2} - 12$

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

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

$f' (4) = 3 \cdot 16 - 12$

Step 3.4.2.1.2

Multiply $3$ by $16$ .

$f' (4) = 48 - 12$

Step 3.4.2.2

Subtract $12$ from $48$ .

$f' (4) = 36$

Step 3.4.2.3

The final answer is $36$ .

$36$

Step 3.5

Since the first derivative changed signs from positive to negative around $x = - 2$ , then $x = - 2$ is a local maximum.

$x = - 2$ is a local maximum

Step 3.6

Since the first derivative changed signs from negative to positive around $x = 2$ , then $x = 2$ is a local minimum.

$x = 2$ is a local minimum

Step 3.7

These are the local extrema for $f (x) = x^{3} - 12 x$ .

$x = - 2$ is a local maximum

$x = 2$ is a local minimum

$x = - 2$ is a local maximum

$x = 2$ is a local minimum

Step 4

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

No absolute maximum

Absolute Minimum: $(2, - 16)$

Step 5