Calculus Examples

f (x) = - \frac{1}{3} x^{3} - 9 x^{2}

$f (x) = - \frac{1}{3} x^{3} - 9 x^{2}$

Step 1

Find the second 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

- \frac{1}{3} x^{3} - 9 x^{2}

$- \frac{1}{3} x^{3} - 9 x^{2}$ with respect to

x

$x$ is

\frac{d}{d x} [- \frac{1}{3} x^{3}] + \frac{d}{d x} [- 9 x^{2}]

$\frac{d}{d x} [- \frac{1}{3} x^{3}] + \frac{d}{d x} [- 9 x^{2}]$ .

\frac{d}{d x} [- \frac{1}{3} x^{3}] + \frac{d}{d x} [- 9 x^{2}]

$\frac{d}{d x} [- \frac{1}{3} x^{3}] + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2

Evaluate

\frac{d}{d x} [- \frac{1}{3} x^{3}]

$\frac{d}{d x} [- \frac{1}{3} x^{3}]$ .

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Step 1.1.2.1

Since

- \frac{1}{3}

$- \frac{1}{3}$ is constant with respect to

x

$x$ , the derivative of

- \frac{1}{3} x^{3}

$- \frac{1}{3} x^{3}$ with respect to

x

$x$ is

- \frac{1}{3} \frac{d}{d x} [x^{3}]

$- \frac{1}{3} \frac{d}{d x} [x^{3}]$ .

- \frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 9 x^{2}]

$- \frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 9 x^{2}]$

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 = 3

$n = 3$ .

- \frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- 9 x^{2}]

$- \frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.3

Multiply

3

$3$ by

- 1

$- 1$ .

- 3 (\frac{1}{3}) x^{2} + \frac{d}{d x} [- 9 x^{2}]

$- 3 (\frac{1}{3}) x^{2} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.4

Combine

- 3

$- 3$ and

\frac{1}{3}

$\frac{1}{3}$ .

\frac{- 3}{3} x^{2} + \frac{d}{d x} [- 9 x^{2}]

$\frac{- 3}{3} x^{2} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.5

Combine

\frac{- 3}{3}

$\frac{- 3}{3}$ and

x^{2}

$x^{2}$ .

\frac{- 3 x^{2}}{3} + \frac{d}{d x} [- 9 x^{2}]

$\frac{- 3 x^{2}}{3} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.6

Cancel the common factor of

- 3

$- 3$ and

3

$3$ .

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Step 1.1.2.6.1

Factor

3

$3$ out of

- 3 x^{2}

$- 3 x^{2}$ .

\frac{3 (- x^{2})}{3} + \frac{d}{d x} [- 9 x^{2}]

$\frac{3 (- x^{2})}{3} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.6.2

Cancel the common factors.

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Step 1.1.2.6.2.1

Factor

3

$3$ out of

3

$3$ .

\frac{3 (- x^{2})}{3 (1)} + \frac{d}{d x} [- 9 x^{2}]

$\frac{3 (- x^{2})}{3 (1)} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.6.2.2

Cancel the common factor.

\frac{3 (- x^{2})}{3 \cdot 1} + \frac{d}{d x} [- 9 x^{2}]

$\frac{3 (- x^{2})}{3 \cdot 1} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.6.2.3

Rewrite the expression.

\frac{- x^{2}}{1} + \frac{d}{d x} [- 9 x^{2}]

$\frac{- x^{2}}{1} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.2.6.2.4

Divide

- x^{2}

$- x^{2}$ by

1

$1$ .

- x^{2} + \frac{d}{d x} [- 9 x^{2}]

$- x^{2} + \frac{d}{d x} [- 9 x^{2}]$

- x^{2} + \frac{d}{d x} [- 9 x^{2}]

$- x^{2} + \frac{d}{d x} [- 9 x^{2}]$

- x^{2} + \frac{d}{d x} [- 9 x^{2}]

$- x^{2} + \frac{d}{d x} [- 9 x^{2}]$

- x^{2} + \frac{d}{d x} [- 9 x^{2}]

$- x^{2} + \frac{d}{d x} [- 9 x^{2}]$

Step 1.1.3

Evaluate

\frac{d}{d x} [- 9 x^{2}]

$\frac{d}{d x} [- 9 x^{2}]$ .

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Step 1.1.3.1

Since

- 9

$- 9$ is constant with respect to

x

$x$ , the derivative of

- 9 x^{2}

$- 9 x^{2}$ with respect to

x

$x$ is

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

$- 9 \frac{d}{d x} [x^{2}]$ .

- x^{2} - 9 \frac{d}{d x} [x^{2}]

$- x^{2} - 9 \frac{d}{d x} [x^{2}]$

Step 1.1.3.2

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

$n = 2$ .

- x^{2} - 9 (2 x)

$- x^{2} - 9 (2 x)$

Step 1.1.3.3

Multiply

2

$2$ by

- 9

$- 9$ .

$f' (x) = - x^{2} - 18 x$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of

$- x^{2} - 18 x$ with respect to

$x$ is

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

$\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 18 x]$

Step 1.2.2

Evaluate

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

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Step 1.2.2.1

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- x^{2}$ with respect to

$x$ is

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

$- \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 18 x]$

Step 1.2.2.2

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 2$ .

$- (2 x) + \frac{d}{d x} [- 18 x]$

Step 1.2.2.3

Multiply

$2$ by

$- 1$ .

$- 2 x + \frac{d}{d x} [- 18 x]$

Step 1.2.3

Evaluate

$\frac{d}{d x} [- 18 x]$ .

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Step 1.2.3.1

Since

$- 18$ is constant with respect to

$x$ , the derivative of

$- 18 x$ with respect to

$x$ is

$- 18 \frac{d}{d x} [x]$ .

$- 2 x - 18 \frac{d}{d x} [x]$

Step 1.2.3.2

Differentiate using the Power Rule which states that

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

$n x^{n - 1}$ where

$n = 1$ .

$- 2 x - 18 \cdot 1$

Step 1.2.3.3

Multiply

$- 18$ by

$1$ .

$f'' (x) = - 2 x - 18$

Step 1.3

The second derivative of

$f (x)$ with respect to

$x$ is

$- 2 x - 18$ .

$- 2 x - 18$

Step 2

Set the second derivative equal to

$0$ then solve the equation

$- 2 x - 18 = 0$ .

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Step 2.1

Set the second derivative equal to

$0$ .

$- 2 x - 18 = 0$

Step 2.2

Add

$18$ to both sides of the equation.

$- 2 x = 18$

Step 2.3

Divide each term in

$- 2 x = 18$ by

$- 2$ and simplify.

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Step 2.3.1

Divide each term in

$- 2 x = 18$ by

$- 2$ .

$\frac{- 2 x}{- 2} = \frac{18}{- 2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of

$- 2$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{18}{- 2}$

Step 2.3.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{18}{- 2}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide

$18$ by

$- 2$ .

$x = - 9$

Step 3

Find the points where the second derivative is

$0$ .

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Step 3.1

Substitute

$- 9$ in

$f (x) = - \frac{1}{3} x^{3} - 9 x^{2}$ to find the value of

$y$ .

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Step 3.1.1

Replace the variable

$x$ with

$- 9$ in the expression.

$f (- 9) = - \frac{1}{3} \cdot {(- 9)}^{3} - 9 {(- 9)}^{2}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Raise

$- 9$ to the power of

$3$ .

$f (- 9) = - \frac{1}{3} \cdot - 729 - 9 {(- 9)}^{2}$

Step 3.1.2.1.2

Cancel the common factor of

$3$ .

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Step 3.1.2.1.2.1

Move the leading negative in

$- \frac{1}{3}$ into the numerator.

$f (- 9) = \frac{- 1}{3} \cdot - 729 - 9 {(- 9)}^{2}$

Step 3.1.2.1.2.2

Factor

$3$ out of

$- 729$ .

$f (- 9) = \frac{- 1}{3} \cdot (3 (- 243)) - 9 {(- 9)}^{2}$

Step 3.1.2.1.2.3

Cancel the common factor.

$f (- 9) = \frac{- 1}{3} \cdot (3 \cdot - 243) - 9 {(- 9)}^{2}$

Step 3.1.2.1.2.4

Rewrite the expression.

$f (- 9) = - 1 \cdot - 243 - 9 {(- 9)}^{2}$

Step 3.1.2.1.3

Multiply

$- 1$ by

$- 243$ .

$f (- 9) = 243 - 9 {(- 9)}^{2}$

Step 3.1.2.1.4

Multiply

$- 9$ by

${(- 9)}^{2}$ by adding the exponents.

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Step 3.1.2.1.4.1

Multiply

$- 9$ by

${(- 9)}^{2}$ .

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Step 3.1.2.1.4.1.1

Raise

$- 9$ to the power of

$1$ .

$f (- 9) = 243 + (- 9) {(- 9)}^{2}$

Step 3.1.2.1.4.1.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (- 9) = 243 + {(- 9)}^{1 + 2}$

Step 3.1.2.1.4.2

Add

$1$ and

$2$ .

$f (- 9) = 243 + {(- 9)}^{3}$

Step 3.1.2.1.5

Raise

$- 9$ to the power of

$3$ .

$f (- 9) = 243 - 729$

Step 3.1.2.2

Subtract

$729$ from

$243$ .

$f (- 9) = - 486$

Step 3.1.2.3

The final answer is

$- 486$ .

$- 486$

Step 3.2

The point found by substituting

$- 9$ in

$f (x) = - \frac{1}{3} x^{3} - 9 x^{2}$ is

$(- 9, - 486)$ . This point can be an inflection point.

$(- 9, - 486)$

Step 4

Split

$(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 9) \cup (- 9, \infty)$

Step 5

Substitute a value from the interval

$(- \infty, - 9)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable

$x$ with

$- 9.1$ in the expression.

$f'' (- 9.1) = - 2 \cdot - 9.1 - 18$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply

$- 2$ by

$- 9.1$ .

$f'' (- 9.1) = 18.2 - 18$

Step 5.2.2

Subtract

$18$ from

$18.2$ .

$f'' (- 9.1) = 0.2$

Step 5.2.3

The final answer is

$0.2$ .

$0.2$

Step 5.3

$- 9.1$ , the second derivative is

$0.2$ . Since this is positive, the second derivative is increasing on the interval

$(- \infty, - 9)$ .

Increasing on

$(- \infty, - 9)$ since

$f'' (x) > 0$

Increasing on

$(- \infty, - 9)$ since

$f'' (x) > 0$

Step 6

Substitute a value from the interval

$(- 9, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable

$x$ with

$- 8.9$ in the expression.

$f'' (- 8.9) = - 2 \cdot - 8.9 - 18$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply

$- 2$ by

$- 8.9$ .

$f'' (- 8.9) = 17.8 - 18$

Step 6.2.2

Subtract

$18$ from

$17.8$ .

$f'' (- 8.9) = - 0.2$

Step 6.2.3

The final answer is

$- 0.2$ .

$- 0.2$

Step 6.3

$- 8.9$ , the second derivative is

$- 0.2$ . Since this is negative, the second derivative is decreasing on the interval

$(- 9, \infty)$

Decreasing on

$(- 9, \infty)$ since

$f'' (x) < 0$

Decreasing on

$(- 9, \infty)$ since

$f'' (x) < 0$

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is

$(- 9, - 486)$ .

$(- 9, - 486)$

Step 8