Calculus Examples

$f (x) = 4 x^{3} - 32$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $4 x^{3} - 32$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 32]$ .

$\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 32]$

Step 1.1.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 1.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 32]$

Step 1.1.2.2

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

$4 (3 x^{2}) + \frac{d}{d x} [- 32]$

Step 1.1.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [- 32]$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

Since $- 32$ is constant with respect to $x$ , the derivative of $- 32$ with respect to $x$ is $0$ .

$12 x^{2} + 0$

Step 1.1.3.2

Add $12 x^{2}$ and $0$ .

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

Step 1.2

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

$12 x^{2}$

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

$12 x^{2} = 0$

Step 2.2

Divide each term in $12 x^{2} = 0$ by $12$ and simplify.

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Step 2.2.1

Divide each term in $12 x^{2} = 0$ by $12$ .

$\frac{12 x^{2}}{12} = \frac{0}{12}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $12$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{12 x^{2}}{12} = \frac{0}{12}$

Step 2.2.2.1.2

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

$x^{2} = \frac{0}{12}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $0$ by $12$ .

$x^{2} = 0$

Step 2.3

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

$x = \pm \sqrt{0}$

Step 2.4

Simplify $\pm \sqrt{0}$ .

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Step 2.4.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.4.2

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

$x = \pm 0$

Step 2.4.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 3

The values which make the derivative equal to $0$ are $0$ .

$0$

Step 4

After finding the point that makes the derivative $f' (x) = 12 x^{2}$ equal to $0$ or undefined, the interval to check where $f (x) = 4 x^{3} - 32$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

$(- \infty, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 1$ in the expression.

$f' (- 1) = 12 {(- 1)}^{2}$

Step 5.2

Simplify the result.

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Step 5.2.1

Raise $- 1$ to the power of $2$ .

$f' (- 1) = 12 \cdot 1$

Step 5.2.2

Multiply $12$ by $1$ .

$f' (- 1) = 12$

Step 5.2.3

The final answer is $12$ .

$12$

Step 5.3

At $x = - 1$ the derivative is $12$ . Since this is positive, the function is increasing on $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = 12 {(1)}^{2}$

Step 6.2

Simplify the result.

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Step 6.2.1

One to any power is one.

$f' (1) = 12 \cdot 1$

Step 6.2.2

Multiply $12$ by $1$ .

$f' (1) = 12$

Step 6.2.3

The final answer is $12$ .

$12$

Step 6.3

At $x = 1$ the derivative is $12$ . Since this is positive, the function is increasing on $(0, \infty)$ .

Increasing on $(0, \infty)$ since $f' (x) > 0$

Step 7

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, 0), (0, \infty)$

Step 8