Calculus Examples

$f (x) = 3 x^{3} - 9 x$

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 $3 x^{3} - 9 x$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [- 9 x]$ .

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

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 $n = 3$ .

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

Step 1.1.2.3

Multiply $3$ by $3$ .

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

Step 1.1.3

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

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Step 1.1.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

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

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 $n = 1$ .

$9 x^{2} - 9 \cdot 1$

Step 1.1.3.3

Multiply $- 9$ by $1$ .

$f' (x) = 9 x^{2} - 9$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $9 x^{2} - 9$ with respect to $x$ is $\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- 9]$ .

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

Step 1.2.2

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

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Step 1.2.2.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

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

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 $n = 2$ .

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

Step 1.2.2.3

Multiply $2$ by $9$ .

$18 x + \frac{d}{d x} [- 9]$

Step 1.2.3

Differentiate using the Constant Rule.

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Step 1.2.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$18 x + 0$

Step 1.2.3.2

Add $18 x$ and $0$ .

$f'' (x) = 18 x$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $18 x$ .

$18 x$

Step 2

Set the second derivative equal to $0$ then solve the equation $18 x = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$18 x = 0$

Step 2.2

Divide each term in $18 x = 0$ by $18$ and simplify.

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Step 2.2.1

Divide each term in $18 x = 0$ by $18$ .

$\frac{18 x}{18} = \frac{0}{18}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $18$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{18 x}{18} = \frac{0}{18}$

Step 2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{18}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $0$ by $18$ .

$x = 0$

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $0$ in $f (x) = 3 x^{3} - 9 x$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 3 {(0)}^{3} - 9 \cdot 0$

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

Raising $0$ to any positive power yields $0$ .

$f (0) = 3 \cdot 0 - 9 \cdot 0$

Step 3.1.2.1.2

Multiply $3$ by $0$ .

$f (0) = 0 - 9 \cdot 0$

Step 3.1.2.1.3

Multiply $- 9$ by $0$ .

$f (0) = 0 + 0$

Step 3.1.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 3.1.2.3

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $f (x) = 3 x^{3} - 9 x$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4

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

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

Step 5

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

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Step 5.1

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = 18 (- 0.1)$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $18$ by $- 0.1$ .

$f'' (- 0.1) = - 1.8$

Step 5.2.2

The final answer is $- 1.8$ .

$- 1.8$

Step 5.3

At $- 0.1$ , the second derivative is $- 1.8$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = 18 (0.1)$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $18$ by $0.1$ .

$f'' (0.1) = 1.8$

Step 6.2.2

The final answer is $1.8$ .

$1.8$

Step 6.3

At $0.1$ , the second derivative is $1.8$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

Increasing on $(0, \infty)$ 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 $(0, 0)$ .

$(0, 0)$

Step 8