Calculus Examples

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

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

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{3} - 3 x^{2} - 9 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [- 9 x]$ .

$f' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (- 9 x)$

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 $n = 3$ .

$f' (x) = 3 x^{2} + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (- 9 x)$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 3 x^{2}]$ .

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Step 1.1.2.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{2}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 3 x^{2} - 3 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 9 x)$

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 $n = 2$ .

$f' (x) = 3 x^{2} - 3 (2 x) + \frac{d}{d x} (- 9 x)$

Step 1.1.2.3

Multiply $2$ by $- 3$ .

$f' (x) = 3 x^{2} - 6 x + \frac{d}{d x} (- 9 x)$

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

$f' (x) = 3 x^{2} - 6 x - 9 \frac{d}{d x} x$

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 $n = 1$ .

$f' (x) = 3 x^{2} - 6 x - 9 \cdot 1$

Step 1.1.3.3

Multiply $- 9$ by $1$ .

$f' (x) = 3 x^{2} - 6 x - 9$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $3 x^{2} - 6 x - 9$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [- 9]$ .

$f'' (x) = \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (- 9)$

Step 1.2.2

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 1.2.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 3 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (- 9)$

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 $n = 2$ .

$f'' (x) = 3 (2 x) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (- 9)$

Step 1.2.2.3

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (- 9)$

Step 1.2.3

Evaluate $\frac{d}{d x} [- 6 x]$ .

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Step 1.2.3.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x$ with respect to $x$ is $- 6 \frac{d}{d x} [x]$ .

$f'' (x) = 6 x - 6 \frac{d}{d x} x + \frac{d}{d x} (- 9)$

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 $n = 1$ .

$f'' (x) = 6 x - 6 \cdot 1 + \frac{d}{d x} (- 9)$

Step 1.2.3.3

Multiply $- 6$ by $1$ .

$f'' (x) = 6 x - 6 + \frac{d}{d x} (- 9)$

Step 1.2.4

Differentiate using the Constant Rule.

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Step 1.2.4.1

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

$f'' (x) = 6 x - 6 + 0$

Step 1.2.4.2

Add $6 x - 6$ and $0$ .

$f'' (x) = 6 x - 6$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $6 x - 6$ .

$6 x - 6$

Step 2

Set the second derivative equal to $0$ then solve the equation $6 x - 6 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$6 x - 6 = 0$

Step 2.2

Add $6$ to both sides of the equation.

$6 x = 6$

Step 2.3

Divide each term in $6 x = 6$ by $6$ and simplify.

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Step 2.3.1

Divide each term in $6 x = 6$ by $6$ .

$\frac{6 x}{6} = \frac{6}{6}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $6$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{6}{6}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{6}{6}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $6$ by $6$ .

$x = 1$

Step 3

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

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Step 3.1

Substitute $1$ in $f (x) = x^{3} - 3 x^{2} - 9 x$ to find the value of $y$ .

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Step 3.1.1

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

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

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

One to any power is one.

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

Step 3.1.2.1.2

One to any power is one.

$f (1) = 1 - 3 \cdot 1 - 9 \cdot 1$

Step 3.1.2.1.3

Multiply $- 3$ by $1$ .

$f (1) = 1 - 3 - 9 \cdot 1$

Step 3.1.2.1.4

Multiply $- 9$ by $1$ .

$f (1) = 1 - 3 - 9$

Step 3.1.2.2

Simplify by subtracting numbers.

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Step 3.1.2.2.1

Subtract $3$ from $1$ .

$f (1) = - 2 - 9$

Step 3.1.2.2.2

Subtract $9$ from $- 2$ .

$f (1) = - 11$

Step 3.1.2.3

The final answer is $- 11$ .

$- 11$

Step 3.2

The point found by substituting $1$ in $f (x) = x^{3} - 3 x^{2} - 9 x$ is $(1, - 11)$ . This point can be an inflection point.

$(1, - 11)$

Step 4

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

$(- \infty, 1) \cup (1, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $0.9$ in the expression.

$f'' (0.9) = 6 (0.9) - 6$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $6$ by $0.9$ .

$f'' (0.9) = 5.4 - 6$

Step 5.2.2

Subtract $6$ from $5.4$ .

$f'' (0.9) = - 0.6$

Step 5.2.3

The final answer is $- 0.6$ .

$- 0.6$

Step 5.3

At $0.9$ , the second derivative is $- 0.6$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 1)$

Decreasing on $(- \infty, 1)$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(1, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1.1$ in the expression.

$f'' (1.1) = 6 (1.1) - 6$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $6$ by $1.1$ .

$f'' (1.1) = 6.6 - 6$

Step 6.2.2

Subtract $6$ from $6.6$ .

$f'' (1.1) = 0.6$

Step 6.2.3

The final answer is $0.6$ .

$0.6$

Step 6.3

At $1.1$ , the second derivative is $0.6$ . Since this is positive, the second derivative is increasing on the interval $(1, \infty)$ .

Increasing on $(1, \infty)$ 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 $(1, - 11)$ .

$(1, - 11)$

Step 8