Calculus Examples

$f (x) = (x - 3) (x^{2} - 6 x - 18)$

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 using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x - 3$ and $g (x) = x^{2} - 6 x - 18$ .

$f' (x) = (x - 3) \frac{d}{d x} (x^{2} - 6 x - 18) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2

Differentiate.

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Step 1.1.2.1

By the Sum Rule, the derivative of $x^{2} - 6 x - 18$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [- 18]$ .

$f' (x) = (x - 3) (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (- 18)) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

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 $n = 2$ .

$f' (x) = (x - 3) (2 x + \frac{d}{d x} (- 6 x) + \frac{d}{d x} (- 18)) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2.3

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

$f' (x) = (x - 3) (2 x - 6 \frac{d}{d x} x + \frac{d}{d x} (- 18)) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2.4

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

$f' (x) = (x - 3) (2 x - 6 \cdot 1 + \frac{d}{d x} (- 18)) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2.5

Multiply $- 6$ by $1$ .

$f' (x) = (x - 3) (2 x - 6 + \frac{d}{d x} (- 18)) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2.6

Since $- 18$ is constant with respect to $x$ , the derivative of $- 18$ with respect to $x$ is $0$ .

$f' (x) = (x - 3) (2 x - 6 + 0) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2.7

Add $2 x - 6$ and $0$ .

$f' (x) = (x - 3) (2 x - 6) + (x^{2} - 6 x - 18) \frac{d}{d x} (x - 3)$

Step 1.1.2.8

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

$f' (x) = (x - 3) (2 x - 6) + (x^{2} - 6 x - 18) (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))$

Step 1.1.2.9

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

$f' (x) = (x - 3) (2 x - 6) + (x^{2} - 6 x - 18) (1 + \frac{d}{d x} (- 3))$

Step 1.1.2.10

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f' (x) = (x - 3) (2 x - 6) + (x^{2} - 6 x - 18) (1 + 0)$

Step 1.1.2.11

Simplify the expression.

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Step 1.1.2.11.1

Add $1$ and $0$ .

$f' (x) = (x - 3) (2 x - 6) + (x^{2} - 6 x - 18) \cdot 1$

Step 1.1.2.11.2

Multiply $x^{2} - 6 x - 18$ by $1$ .

$f' (x) = (x - 3) (2 x - 6) + x^{2} - 6 x - 18$

Step 1.1.3

Simplify.

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Step 1.1.3.1

Apply the distributive property.

$f' (x) = (x - 3) (2 x) + (x - 3) \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.2

Apply the distributive property.

$f' (x) = x (2 x) - 3 (2 x) + (x - 3) \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.3

Apply the distributive property.

$f' (x) = x (2 x) - 3 (2 x) + x \cdot - 6 - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4

Combine terms.

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Step 1.1.3.4.1

Raise $x$ to the power of $1$ .

$f' (x) = 2 (x \cdot x) - 3 (2 x) + x \cdot - 6 - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4.2

Raise $x$ to the power of $1$ .

$f' (x) = 2 (x \cdot x) - 3 (2 x) + x \cdot - 6 - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (x) = 2 x^{1 + 1} - 3 (2 x) + x \cdot - 6 - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4.4

Add $1$ and $1$ .

$f' (x) = 2 x^{2} - 3 (2 x) + x \cdot - 6 - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4.5

Multiply $2$ by $- 3$ .

$f' (x) = 2 x^{2} - 6 x + x \cdot - 6 - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4.6

Move $- 6$ to the left of $x$ .

$f' (x) = 2 x^{2} - 6 x - 6 \cdot x - 3 \cdot - 6 + x^{2} - 6 x - 18$

Step 1.1.3.4.7

Multiply $- 3$ by $- 6$ .

$f' (x) = 2 x^{2} - 6 x - 6 x + 18 + x^{2} - 6 x - 18$

Step 1.1.3.4.8

Subtract $6 x$ from $- 6 x$ .

$f' (x) = 2 x^{2} - 12 x + 18 + x^{2} - 6 x - 18$

Step 1.1.3.4.9

Add $2 x^{2}$ and $x^{2}$ .

$f' (x) = 3 x^{2} - 12 x + 18 - 6 x - 18$

Step 1.1.3.4.10

Subtract $6 x$ from $- 12 x$ .

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

Step 1.1.3.4.11

Subtract $18$ from $18$ .

$f' (x) = 3 x^{2} - 18 x + 0$

Step 1.1.3.4.12

Add $3 x^{2} - 18 x$ and $0$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $3 x^{2} - 18 x$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 18 x]$ .

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

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

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

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 $n = 2$ .

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

Step 1.2.2.3

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (- 18 x)$

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

$f'' (x) = 6 x - 18 \frac{d}{d x} x$

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 $n = 1$ .

$f'' (x) = 6 x - 18 \cdot 1$

Step 1.2.3.3

Multiply $- 18$ by $1$ .

$f'' (x) = 6 x - 18$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $6 x - 18$ .

$6 x - 18$

Step 2

Set the second derivative equal to $0$ then solve the equation $6 x - 18 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$6 x - 18 = 0$

Step 2.2

Add $18$ to both sides of the equation.

$6 x = 18$

Step 2.3

Divide each term in $6 x = 18$ by $6$ and simplify.

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Step 2.3.1

Divide each term in $6 x = 18$ by $6$ .

$\frac{6 x}{6} = \frac{18}{6}$

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.

$\frac{6 x}{6} = \frac{18}{6}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{18}{6}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $18$ by $6$ .

$x = 3$

Step 3

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

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Step 3.1

Substitute $3$ in $f (x) = (x - 3) (x^{2} - 6 x - 18)$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = ((3) - 3) ({(3)}^{2} - 6 \cdot 3 - 18)$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Subtract $3$ from $3$ .

$f (3) = 0 ({(3)}^{2} - 6 \cdot 3 - 18)$

Step 3.1.2.2

Simplify each term.

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Step 3.1.2.2.1

Raise $3$ to the power of $2$ .

$f (3) = 0 (9 - 6 \cdot 3 - 18)$

Step 3.1.2.2.2

Multiply $- 6$ by $3$ .

$f (3) = 0 (9 - 18 - 18)$

Step 3.1.2.3

Simplify the expression.

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Step 3.1.2.3.1

Subtract $18$ from $9$ .

$f (3) = 0 (- 9 - 18)$

Step 3.1.2.3.2

Subtract $18$ from $- 9$ .

$f (3) = 0 \cdot - 27$

Step 3.1.2.3.3

Multiply $0$ by $- 27$ .

$f (3) = 0$

Step 3.1.2.4

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $3$ in $f (x) = (x - 3) (x^{2} - 6 x - 18)$ is $(3, 0)$ . This point can be an inflection point.

$(3, 0)$

Step 4

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

$(- \infty, 3) \cup (3, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 3)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $2.9$ in the expression.

$f'' (2.9) = 6 (2.9) - 18$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $6$ by $2.9$ .

$f'' (2.9) = 17.4 - 18$

Step 5.2.2

Subtract $18$ from $17.4$ .

$f'' (2.9) = - 0.6$

Step 5.2.3

The final answer is $- 0.6$ .

$- 0.6$

Step 5.3

At $2.9$ , the second derivative is $- 0.6$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 3)$

Decreasing on $(- \infty, 3)$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(3, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $3.1$ in the expression.

$f'' (3.1) = 6 (3.1) - 18$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $6$ by $3.1$ .

$f'' (3.1) = 18.6 - 18$

Step 6.2.2

Subtract $18$ from $18.6$ .

$f'' (3.1) = 0.6$

Step 6.2.3

The final answer is $0.6$ .

$0.6$

Step 6.3

At $3.1$ , the second derivative is $0.6$ . Since this is positive, the second derivative is increasing on the interval $(3, \infty)$ .

Increasing on $(3, \infty)$ 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 $(3, 0)$ .

$(3, 0)$

Step 8