Calculus Examples

$f (x) = x^{3} - 8 x^{2} - 20 x$

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 $x^{3} - 8 x^{2} - 20 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 8 x^{2}] + \frac{d}{d x} [- 20 x]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 8 x^{2}) + \frac{d}{d x} (- 20 x)$

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 $n = 3$ .

$f' (x) = 3 x^{2} + \frac{d}{d x} (- 8 x^{2}) + \frac{d}{d x} (- 20 x)$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 8 x^{2}]$ .

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Step 1.1.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x^{2}$ with respect to $x$ is $- 8 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 x^{2} - 8 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 20 x)$

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 $n = 2$ .

$f' (x) = 3 x^{2} - 8 (2 x) + \frac{d}{d x} (- 20 x)$

Step 1.1.2.3

Multiply $2$ by $- 8$ .

$f' (x) = 3 x^{2} - 16 x + \frac{d}{d x} (- 20 x)$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 20 x]$ .

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Step 1.1.3.1

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x$ with respect to $x$ is $- 20 \frac{d}{d x} [x]$ .

$f' (x) = 3 x^{2} - 16 x - 20 \frac{d}{d x} x$

Step 1.1.3.2

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

$f' (x) = 3 x^{2} - 16 x - 20 \cdot 1$

Step 1.1.3.3

Multiply $- 20$ by $1$ .

$f' (x) = 3 x^{2} - 16 x - 20$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} - 16 x - 20$ .

$3 x^{2} - 16 x - 20$

Step 2

Set the first derivative equal to $0$ then solve the equation $3 x^{2} - 16 x - 20 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$3 x^{2} - 16 x - 20 = 0$

Step 2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.3

Substitute the values $a = 3$ , $b = - 16$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (3 \cdot - 20)}}{2 \cdot 3}$

Step 2.4

Simplify.

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Step 2.4.1

Simplify the numerator.

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Step 2.4.1.1

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 3 \cdot - 20}}{2 \cdot 3}$

Step 2.4.1.2

Multiply $- 4 \cdot 3 \cdot - 20$ .

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Step 2.4.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{16 \pm \sqrt{256 - 12 \cdot - 20}}{2 \cdot 3}$

Step 2.4.1.2.2

Multiply $- 12$ by $- 20$ .

$x = \frac{16 \pm \sqrt{256 + 240}}{2 \cdot 3}$

Step 2.4.1.3

Add $256$ and $240$ .

$x = \frac{16 \pm \sqrt{496}}{2 \cdot 3}$

Step 2.4.1.4

Rewrite $496$ as $4^{2} \cdot 31$ .

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Step 2.4.1.4.1

Factor $16$ out of $496$ .

$x = \frac{16 \pm \sqrt{16 (31)}}{2 \cdot 3}$

Step 2.4.1.4.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{16 \pm \sqrt{4^{2} \cdot 31}}{2 \cdot 3}$

Step 2.4.1.5

Pull terms out from under the radical.

$x = \frac{16 \pm 4 \sqrt{31}}{2 \cdot 3}$

Step 2.4.2

Multiply $2$ by $3$ .

$x = \frac{16 \pm 4 \sqrt{31}}{6}$

Step 2.4.3

Simplify $\frac{16 \pm 4 \sqrt{31}}{6}$ .

$x = \frac{8 \pm 2 \sqrt{31}}{3}$

Step 2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 3 \cdot - 20}}{2 \cdot 3}$

Step 2.5.1.2

Multiply $- 4 \cdot 3 \cdot - 20$ .

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Step 2.5.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{16 \pm \sqrt{256 - 12 \cdot - 20}}{2 \cdot 3}$

Step 2.5.1.2.2

Multiply $- 12$ by $- 20$ .

$x = \frac{16 \pm \sqrt{256 + 240}}{2 \cdot 3}$

Step 2.5.1.3

Add $256$ and $240$ .

$x = \frac{16 \pm \sqrt{496}}{2 \cdot 3}$

Step 2.5.1.4

Rewrite $496$ as $4^{2} \cdot 31$ .

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Step 2.5.1.4.1

Factor $16$ out of $496$ .

$x = \frac{16 \pm \sqrt{16 (31)}}{2 \cdot 3}$

Step 2.5.1.4.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{16 \pm \sqrt{4^{2} \cdot 31}}{2 \cdot 3}$

Step 2.5.1.5

Pull terms out from under the radical.

$x = \frac{16 \pm 4 \sqrt{31}}{2 \cdot 3}$

Step 2.5.2

Multiply $2$ by $3$ .

$x = \frac{16 \pm 4 \sqrt{31}}{6}$

Step 2.5.3

Simplify $\frac{16 \pm 4 \sqrt{31}}{6}$ .

$x = \frac{8 \pm 2 \sqrt{31}}{3}$

Step 2.5.4

Change the $\pm$ to $+$ .

$x = \frac{8 + 2 \sqrt{31}}{3}$

Step 2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 3 \cdot - 20}}{2 \cdot 3}$

Step 2.6.1.2

Multiply $- 4 \cdot 3 \cdot - 20$ .

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Step 2.6.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{16 \pm \sqrt{256 - 12 \cdot - 20}}{2 \cdot 3}$

Step 2.6.1.2.2

Multiply $- 12$ by $- 20$ .

$x = \frac{16 \pm \sqrt{256 + 240}}{2 \cdot 3}$

Step 2.6.1.3

Add $256$ and $240$ .

$x = \frac{16 \pm \sqrt{496}}{2 \cdot 3}$

Step 2.6.1.4

Rewrite $496$ as $4^{2} \cdot 31$ .

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Step 2.6.1.4.1

Factor $16$ out of $496$ .

$x = \frac{16 \pm \sqrt{16 (31)}}{2 \cdot 3}$

Step 2.6.1.4.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{16 \pm \sqrt{4^{2} \cdot 31}}{2 \cdot 3}$

Step 2.6.1.5

Pull terms out from under the radical.

$x = \frac{16 \pm 4 \sqrt{31}}{2 \cdot 3}$

Step 2.6.2

Multiply $2$ by $3$ .

$x = \frac{16 \pm 4 \sqrt{31}}{6}$

Step 2.6.3

Simplify $\frac{16 \pm 4 \sqrt{31}}{6}$ .

$x = \frac{8 \pm 2 \sqrt{31}}{3}$

Step 2.6.4

Change the $\pm$ to $-$ .

$x = \frac{8 - 2 \sqrt{31}}{3}$

Step 2.7

The final answer is the combination of both solutions.

$x = \frac{8 + 2 \sqrt{31}}{3}, \frac{8 - 2 \sqrt{31}}{3}$

Step 3

The values which make the derivative equal to $0$ are $\frac{8 + 2 \sqrt{31}}{3}, \frac{8 - 2 \sqrt{31}}{3}$ .

$\frac{8 + 2 \sqrt{31}}{3}, \frac{8 - 2 \sqrt{31}}{3}$

Step 4

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

$(- \infty, \frac{8 - 2 \sqrt{31}}{3}) \cup (\frac{8 - 2 \sqrt{31}}{3}, \frac{8 + 2 \sqrt{31}}{3}) \cup (\frac{8 + 2 \sqrt{31}}{3}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{8 - 2 \sqrt{31}}{3})$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 2.0451763$ in the expression.

$f' (- 2.0451763) = 3 {(- 2.0451763)}^{2} - 16 \cdot - 2.0451763 - 20$

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 $- 2.0451763$ to the power of $2$ .

$f' (- 2.0451763) = 3 \cdot 4.18274609 - 16 \cdot - 2.0451763 - 20$

Step 5.2.1.2

Multiply $3$ by $4.18274609$ .

$f' (- 2.0451763) = 12.54823829 - 16 \cdot - 2.0451763 - 20$

Step 5.2.1.3

Multiply $- 16$ by $- 2.0451763$ .

$f' (- 2.0451763) = 12.54823829 + 32.7228208 - 20$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Add $12.54823829$ and $32.7228208$ .

$f' (- 2.0451763) = 45.27105909 - 20$

Step 5.2.2.2

Subtract $20$ from $45.27105909$ .

$f' (- 2.0451763) = 25.27105909$

Step 5.2.3

The final answer is $25.27105909$ .

$25.27105909$

Step 5.3

At $x = - 2.0451763$ the derivative is $25.27105909$ . Since this is positive, the function is increasing on $(- \infty, \frac{8 - 2 \sqrt{31}}{3})$ .

Increasing on $(- \infty, \frac{8 - 2 \sqrt{31}}{3})$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 1.0451763, \frac{8 + 2 \sqrt{31}}{3})$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $2.6666666$ in the expression.

$f' (2.6666666) = 3 {(2.6666666)}^{2} - 16 \cdot 2.6666666 - 20$

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 $2.6666666$ to the power of $2$ .

$f' (2.6666666) = 3 \cdot 7.11111075 - 16 \cdot 2.6666666 - 20$

Step 6.2.1.2

Multiply $3$ by $7.11111075$ .

$f' (2.6666666) = 21.33333226 - 16 \cdot 2.6666666 - 20$

Step 6.2.1.3

Multiply $- 16$ by $2.6666666$ .

$f' (2.6666666) = 21.33333226 - 42.6666656 - 20$

Step 6.2.2

Simplify by subtracting numbers.

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Step 6.2.2.1

Subtract $42.6666656$ from $21.33333226$ .

$f' (2.6666666) = - 21.\overline{3} - 20$

Step 6.2.2.2

Subtract $20$ from $- 21.\overline{3}$ .

$f' (2.6666666) = - 41.\overline{3}$

Step 6.2.3

The final answer is $- 41.\overline{3}$ .

$- 41.\overline{3}$

Step 6.3

At $x = 2.6666666$ the derivative is $- 41.\overline{3}$ . Since this is negative, the function is decreasing on $(- 1.0451763, \frac{8 + 2 \sqrt{31}}{3})$ .

Decreasing on $(\frac{8 - 2 \sqrt{31}}{3}, \frac{8 + 2 \sqrt{31}}{3})$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{8 + 2 \sqrt{31}}{3}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $7.3785095$ in the expression.

$f' (7.3785095) = 3 {(7.3785095)}^{2} - 16 \cdot 7.3785095 - 20$

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 $7.3785095$ to the power of $2$ .

$f' (7.3785095) = 3 \cdot 54.44240244 - 16 \cdot 7.3785095 - 20$

Step 7.2.1.2

Multiply $3$ by $54.44240244$ .

$f' (7.3785095) = 163.32720732 - 16 \cdot 7.3785095 - 20$

Step 7.2.1.3

Multiply $- 16$ by $7.3785095$ .

$f' (7.3785095) = 163.32720732 - 118.056152 - 20$

Step 7.2.2

Simplify by subtracting numbers.

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Step 7.2.2.1

Subtract $118.056152$ from $163.32720732$ .

$f' (7.3785095) = 45.27105532 - 20$

Step 7.2.2.2

Subtract $20$ from $45.27105532$ .

$f' (7.3785095) = 25.27105532$

Step 7.2.3

The final answer is $25.27105532$ .

$25.27105532$

Step 7.3

At $x = 7.3785095$ the derivative is $25.27105532$ . Since this is positive, the function is increasing on $(\frac{8 + 2 \sqrt{31}}{3}, \infty)$ .

Increasing on $(\frac{8 + 2 \sqrt{31}}{3}, \infty)$ since $f' (x) > 0$

Step 8

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

Increasing on: $(- \infty, \frac{8 - 2 \sqrt{31}}{3}), (\frac{8 + 2 \sqrt{31}}{3}, \infty)$

Decreasing on: $(\frac{8 - 2 \sqrt{31}}{3}, \frac{8 + 2 \sqrt{31}}{3})$

Step 9