Calculus Examples

$f (x) = x^{3} + 18 x^{2} + 2$

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 $x^{3} + 18 x^{2} + 2$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [18 x^{2}] + \frac{d}{d x} [2]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (18 x^{2}) + \frac{d}{d x} (2)$

Step 1.1.1.2

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

$f' (x) = 3 x^{2} + \frac{d}{d x} (18 x^{2}) + \frac{d}{d x} (2)$

Step 1.1.2

Evaluate $\frac{d}{d x} [18 x^{2}]$ .

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Step 1.1.2.1

Since $18$ is constant with respect to $x$ , the derivative of $18 x^{2}$ with respect to $x$ is $18 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 x^{2} + 18 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (2)$

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 $n = 2$ .

$f' (x) = 3 x^{2} + 18 (2 x) + \frac{d}{d x} (2)$

Step 1.1.2.3

Multiply $2$ by $18$ .

$f' (x) = 3 x^{2} + 36 x + \frac{d}{d x} (2)$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$f' (x) = 3 x^{2} + 36 x + 0$

Step 1.1.3.2

Add $3 x^{2} + 36 x$ and $0$ .

$f' (x) = 3 x^{2} + 36 x$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} + 36 x$ .

$3 x^{2} + 36 x$

Step 2

Set the first derivative equal to $0$ then solve the equation $3 x^{2} + 36 x = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$3 x^{2} + 36 x = 0$

Step 2.2

Factor $3 x$ out of $3 x^{2} + 36 x$ .

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Step 2.2.1

Factor $3 x$ out of $3 x^{2}$ .

$3 x (x) + 36 x = 0$

Step 2.2.2

Factor $3 x$ out of $36 x$ .

$3 x (x) + 3 x (12) = 0$

Step 2.2.3

Factor $3 x$ out of $3 x (x) + 3 x (12)$ .

$3 x (x + 12) = 0$

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 $0$ .

$x = 0$

$x + 12 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $x + 12$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 12$ equal to $0$ .

$x + 12 = 0$

Step 2.5.2

Subtract $12$ from both sides of the equation.

$x = - 12$

Step 2.6

The final solution is all the values that make $3 x (x + 12) = 0$ true.

$x = 0, - 12$

Step 3

The values which make the derivative equal to $0$ are $0, - 12$ .

$0, - 12$

Step 4

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

$(- \infty, - 12) \cup (- 12, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 12)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 13$ in the expression.

$f' (- 13) = 3 {(- 13)}^{2} + 36 (- 13)$

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 $- 13$ to the power of $2$ .

$f' (- 13) = 3 \cdot 169 + 36 (- 13)$

Step 5.2.1.2

Multiply $3$ by $169$ .

$f' (- 13) = 507 + 36 (- 13)$

Step 5.2.1.3

Multiply $36$ by $- 13$ .

$f' (- 13) = 507 - 468$

Step 5.2.2

Subtract $468$ from $507$ .

$f' (- 13) = 39$

Step 5.2.3

The final answer is $39$ .

$39$

Step 5.3

At $x = - 13$ the derivative is $39$ . Since this is positive, the function is increasing on $(- \infty, - 12)$ .

Increasing on $(- \infty, - 12)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 12, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 6$ in the expression.

$f' (- 6) = 3 {(- 6)}^{2} + 36 (- 6)$

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 $- 6$ to the power of $2$ .

$f' (- 6) = 3 \cdot 36 + 36 (- 6)$

Step 6.2.1.2

Multiply $3$ by $36$ .

$f' (- 6) = 108 + 36 (- 6)$

Step 6.2.1.3

Multiply $36$ by $- 6$ .

$f' (- 6) = 108 - 216$

Step 6.2.2

Subtract $216$ from $108$ .

$f' (- 6) = - 108$

Step 6.2.3

The final answer is $- 108$ .

$- 108$

Step 6.3

At $x = - 6$ the derivative is $- 108$ . Since this is negative, the function is decreasing on $(- 12, 0)$ .

Decreasing on $(- 12, 0)$ since $f' (x) < 0$

Step 7

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 7.1

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

$f' (1) = 3 {(1)}^{2} + 36 (1)$

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

One to any power is one.

$f' (1) = 3 \cdot 1 + 36 (1)$

Step 7.2.1.2

Multiply $3$ by $1$ .

$f' (1) = 3 + 36 (1)$

Step 7.2.1.3

Multiply $36$ by $1$ .

$f' (1) = 3 + 36$

Step 7.2.2

Add $3$ and $36$ .

$f' (1) = 39$

Step 7.2.3

The final answer is $39$ .

$39$

Step 7.3

At $x = 1$ the derivative is $39$ . Since this is positive, the function is increasing on $(0, \infty)$ .

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

Step 8

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

Increasing on: $(- \infty, - 12), (0, \infty)$

Decreasing on: $(- 12, 0)$

Step 9