Calculus Examples

f (x) = x^{4} - 12 x^{2} + 36

$f (x) = x^{4} - 12 x^{2} + 36$

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

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of

x^{4} - 12 x^{2} + 36

$x^{4} - 12 x^{2} + 36$ with respect to

x

$x$ is

\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]$ .

\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]$

Step 1.1.1.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 4

$n = 4$ .

4 x^{3} + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]

$4 x^{3} + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]$

4 x^{3} + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]

$4 x^{3} + \frac{d}{d x} [- 12 x^{2}] + \frac{d}{d x} [36]$

Step 1.1.2

Evaluate

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

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

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Step 1.1.2.1

Since

- 12

$- 12$ is constant with respect to

x

$x$ , the derivative of

- 12 x^{2}

$- 12 x^{2}$ with respect to

x

$x$ is

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

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

4 x^{3} - 12 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [36]

$4 x^{3} - 12 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [36]$

Step 1.1.2.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 2

$n = 2$ .

4 x^{3} - 12 (2 x) + \frac{d}{d x} [36]

$4 x^{3} - 12 (2 x) + \frac{d}{d x} [36]$

Step 1.1.2.3

Multiply

2

$2$ by

- 12

$- 12$ .

4 x^{3} - 24 x + \frac{d}{d x} [36]

$4 x^{3} - 24 x + \frac{d}{d x} [36]$

4 x^{3} - 24 x + \frac{d}{d x} [36]

$4 x^{3} - 24 x + \frac{d}{d x} [36]$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

Since

36

$36$ is constant with respect to

x

$x$ , the derivative of

36

$36$ with respect to

x

$x$ is

0

$0$ .

4 x^{3} - 24 x + 0

$4 x^{3} - 24 x + 0$

Step 1.1.3.2

Add

4 x^{3} - 24 x

$4 x^{3} - 24 x$ and

0

$0$ .

$f' (x) = 4 x^{3} - 24 x$

Step 1.2

The first derivative of

$f (x)$ with respect to

$x$ is

$4 x^{3} - 24 x$ .

$4 x^{3} - 24 x$

Step 2

Set the first derivative equal to

$0$ then solve the equation

$4 x^{3} - 24 x = 0$ .

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Step 2.1

Set the first derivative equal to

$0$ .

$4 x^{3} - 24 x = 0$

Step 2.2

Factor

$4 x$ out of

$4 x^{3} - 24 x$ .

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Step 2.2.1

Factor

$4 x$ out of

$4 x^{3}$ .

$4 x (x^{2}) - 24 x = 0$

Step 2.2.2

Factor

$4 x$ out of

$- 24 x$ .

$4 x (x^{2}) + 4 x (- 6) = 0$

Step 2.2.3

Factor

$4 x$ out of

$4 x (x^{2}) + 4 x (- 6)$ .

$4 x (x^{2} - 6) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$x = 0$

$x^{2} - 6 = 0$

Step 2.4

Set

$x$ equal to

$0$ .

$x = 0$

Step 2.5

Set

$x^{2} - 6$ equal to

$0$ and solve for

$x$ .

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Step 2.5.1

Set

$x^{2} - 6$ equal to

$0$ .

$x^{2} - 6 = 0$

Step 2.5.2

Solve

$x^{2} - 6 = 0$ for

$x$ .

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Step 2.5.2.1

Add

$6$ to both sides of the equation.

$x^{2} = 6$

Step 2.5.2.2

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

$x = \pm \sqrt{6}$

Step 2.5.2.3

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

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Step 2.5.2.3.1

First, use the positive value of the

$\pm$ to find the first solution.

$x = \sqrt{6}$

Step 2.5.2.3.2

Next, use the negative value of the

$\pm$ to find the second solution.

$x = - \sqrt{6}$

Step 2.5.2.3.3

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

$x = \sqrt{6}, - \sqrt{6}$

Step 2.6

The final solution is all the values that make

$4 x (x^{2} - 6) = 0$ true.

$x = 0, \sqrt{6}, - \sqrt{6}$

Step 3

The values which make the derivative equal to

$0$ are

$0, \sqrt{6}, - \sqrt{6}$ .

$0, \sqrt{6}, - \sqrt{6}$

Step 4

Split

$(- \infty, \infty)$ into separate intervals around the

$x$ values that make the derivative

$0$ or undefined.

$(- \infty, - \sqrt{6}) \cup (- \sqrt{6}, 0) \cup (0, \sqrt{6}) \cup (\sqrt{6}, \infty)$

Step 5

Substitute a value from the interval

$(- \infty, - \sqrt{6})$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable

$x$ with

$- 3.4494898$ in the expression.

$f' (- 3.4494898) = 4 {(- 3.4494898)}^{3} - 24 \cdot - 3.4494898$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Raise

$- 3.4494898$ to the power of

$3$ .

$f' (- 3.4494898) = 4 \cdot - 41.04540972 - 24 \cdot - 3.4494898$

Step 5.2.1.2

Multiply

$4$ by

$- 41.04540972$ .

$f' (- 3.4494898) = - 164.18163891 - 24 \cdot - 3.4494898$

Step 5.2.1.3

Multiply

$- 24$ by

$- 3.4494898$ .

$f' (- 3.4494898) = - 164.18163891 + 82.7877552$

Step 5.2.2

Add

$- 164.18163891$ and

$82.7877552$ .

$f' (- 3.4494898) = - 81.39388371$

Step 5.2.3

The final answer is

$- 81.39388371$ .

$- 81.39388371$

Step 5.3

$x = - 3.4494898$ the derivative is

$- 81.39388371$ . Since this is negative, the function is decreasing on

$(- \infty, - \sqrt{6})$ .

Decreasing on

$(- \infty, - \sqrt{6})$ since

$f' (x) < 0$

Decreasing on

$(- \infty, - \sqrt{6})$ since

$f' (x) < 0$

Step 6

Substitute a value from the interval

$(- 2.4494898, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable

$x$ with

$- 1.2247449$ in the expression.

$f' (- 1.2247449) = 4 {(- 1.2247449)}^{3} - 24 \cdot - 1.2247449$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Raise

$- 1.2247449$ to the power of

$3$ .

$f' (- 1.2247449) = 4 \cdot - 1.83711743 - 24 \cdot - 1.2247449$

Step 6.2.1.2

Multiply

$4$ by

$- 1.83711743$ .

$f' (- 1.2247449) = - 7.34846974 - 24 \cdot - 1.2247449$

Step 6.2.1.3

Multiply

$- 24$ by

$- 1.2247449$ .

$f' (- 1.2247449) = - 7.34846974 + 29.3938776$

Step 6.2.2

Add

$- 7.34846974$ and

$29.3938776$ .

$f' (- 1.2247449) = 22.04540785$

Step 6.2.3

The final answer is

$22.04540785$ .

$22.04540785$

Step 6.3

$x = - 1.2247449$ the derivative is

$22.04540785$ . Since this is positive, the function is increasing on

$(- 2.4494898, 0)$ .

Increasing on

$(- \sqrt{6}, 0)$ since

$f' (x) > 0$

Increasing on

$(- \sqrt{6}, 0)$ since

$f' (x) > 0$

Step 7

Substitute a value from the interval

$(0, \sqrt{6})$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable

$x$ with

$1.2247449$ in the expression.

$f' (1.2247449) = 4 {(1.2247449)}^{3} - 24 \cdot 1.2247449$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise

$1.2247449$ to the power of

$3$ .

$f' (1.2247449) = 4 \cdot 1.83711743 - 24 \cdot 1.2247449$

Step 7.2.1.2

Multiply

$4$ by

$1.83711743$ .

$f' (1.2247449) = 7.34846974 - 24 \cdot 1.2247449$

Step 7.2.1.3

Multiply

$- 24$ by

$1.2247449$ .

$f' (1.2247449) = 7.34846974 - 29.3938776$

Step 7.2.2

Subtract

$29.3938776$ from

$7.34846974$ .

$f' (1.2247449) = - 22.04540785$

Step 7.2.3

The final answer is

$- 22.04540785$ .

$- 22.04540785$

Step 7.3

$x = 1.2247449$ the derivative is

$- 22.04540785$ . Since this is negative, the function is decreasing on

$(0, \sqrt{6})$ .

Decreasing on

$(0, \sqrt{6})$ since

$f' (x) < 0$

Decreasing on

$(0, \sqrt{6})$ since

$f' (x) < 0$

Step 8

Substitute a value from the interval

$(\sqrt{6}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable

$x$ with

$3.4494898$ in the expression.

$f' (3.4494898) = 4 {(3.4494898)}^{3} - 24 \cdot 3.4494898$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise

$3.4494898$ to the power of

$3$ .

$f' (3.4494898) = 4 \cdot 41.04540972 - 24 \cdot 3.4494898$

Step 8.2.1.2

Multiply

$4$ by

$41.04540972$ .

$f' (3.4494898) = 164.18163891 - 24 \cdot 3.4494898$

Step 8.2.1.3

Multiply

$- 24$ by

$3.4494898$ .

$f' (3.4494898) = 164.18163891 - 82.7877552$

Step 8.2.2

Subtract

$82.7877552$ from

$164.18163891$ .

$f' (3.4494898) = 81.39388371$

Step 8.2.3

The final answer is

$81.39388371$ .

$81.39388371$

Step 8.3

$x = 3.4494898$ the derivative is

$81.39388371$ . Since this is positive, the function is increasing on

$(\sqrt{6}, \infty)$ .

Increasing on

$(\sqrt{6}, \infty)$ since

$f' (x) > 0$

Increasing on

$(\sqrt{6}, \infty)$ since

$f' (x) > 0$

Step 9

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

Increasing on:

$(- \sqrt{6}, 0), (\sqrt{6}, \infty)$

Decreasing on:

$(- \infty, - \sqrt{6}), (0, \sqrt{6})$

Step 10

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