Calculus Examples

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

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 $2 x^{3} + 3 x^{2} - 12 x$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 12 x]$ .

$f' (x) = \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 12 x)$

Step 1.1.2

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

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Step 1.1.2.1

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

$f' (x) = 2 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 12 x)$

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 $n = 3$ .

$f' (x) = 2 (3 x^{2}) + \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (- 12 x)$

Step 1.1.2.3

Multiply $3$ by $2$ .

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 2$ .

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

Step 1.1.3.3

Multiply $2$ by $3$ .

$f' (x) = 6 x^{2} + 6 x + \frac{d}{d x} (- 12 x)$

Step 1.1.4

Evaluate $\frac{d}{d x} [- 12 x]$ .

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Step 1.1.4.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x$ with respect to $x$ is $- 12 \frac{d}{d x} [x]$ .

$f' (x) = 6 x^{2} + 6 x - 12 \frac{d}{d x} x$

Step 1.1.4.2

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

$f' (x) = 6 x^{2} + 6 x - 12 \cdot 1$

Step 1.1.4.3

Multiply $- 12$ by $1$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $6 x^{2} + 6 x - 12$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [6 x] + \frac{d}{d x} [- 12]$ .

$f'' (x) = \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 12)$

Step 1.2.2

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Step 1.2.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 6 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 12)$

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 $n = 2$ .

$f'' (x) = 6 (2 x) + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 12)$

Step 1.2.2.3

Multiply $2$ by $6$ .

$f'' (x) = 12 x + \frac{d}{d x} (6 x) + \frac{d}{d x} (- 12)$

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

$f'' (x) = 12 x + 6 \frac{d}{d x} (x) + \frac{d}{d x} (- 12)$

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 $n = 1$ .

$f'' (x) = 12 x + 6 \cdot 1 + \frac{d}{d x} (- 12)$

Step 1.2.3.3

Multiply $6$ by $1$ .

$f'' (x) = 12 x + 6 + \frac{d}{d x} (- 12)$

Step 1.2.4

Differentiate using the Constant Rule.

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Step 1.2.4.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$f'' (x) = 12 x + 6 + 0$

Step 1.2.4.2

Add $12 x + 6$ and $0$ .

$f'' (x) = 12 x + 6$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $12 x + 6$ .

$12 x + 6$

Step 2

Set the second derivative equal to $0$ then solve the equation $12 x + 6 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$12 x + 6 = 0$

Step 2.2

Subtract $6$ from both sides of the equation.

$12 x = - 6$

Step 2.3

Divide each term in $12 x = - 6$ by $12$ and simplify.

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Step 2.3.1

Divide each term in $12 x = - 6$ by $12$ .

$\frac{12 x}{12} = \frac{- 6}{12}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $12$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{12 x}{12} = \frac{- 6}{12}$

Step 2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 6}{12}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $- 6$ and $12$ .

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Step 2.3.3.1.1

Factor $6$ out of $- 6$ .

$x = \frac{6 (- 1)}{12}$

Step 2.3.3.1.2

Cancel the common factors.

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Step 2.3.3.1.2.1

Factor $6$ out of $12$ .

$x = \frac{6 \cdot - 1}{6 \cdot 2}$

Step 2.3.3.1.2.2

Cancel the common factor.

$x = \frac{6 \cdot - 1}{6 \cdot 2}$

Step 2.3.3.1.2.3

Rewrite the expression.

$x = \frac{- 1}{2}$

Step 2.3.3.2

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 3

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

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Step 3.1

Substitute $- \frac{1}{2}$ in $f (x) = 2 x^{3} + 3 x^{2} - 12 x$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f (- \frac{1}{2}) = 2 {(- \frac{1}{2})}^{3} + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.1.2.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

$f (- \frac{1}{2}) = 2 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

$f (- \frac{1}{2}) = 2 ({(- 1)}^{3} (\frac{1^{3}}{2^{3}})) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.2

Raise $- 1$ to the power of $3$ .

$f (- \frac{1}{2}) = 2 (- \frac{1^{3}}{2^{3}}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.3

One to any power is one.

$f (- \frac{1}{2}) = 2 (- \frac{1}{2^{3}}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.4

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

$f (- \frac{1}{2}) = 2 (- \frac{1}{8}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.5

Cancel the common factor of $2$ .

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Step 3.1.2.1.5.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$f (- \frac{1}{2}) = 2 (\frac{- 1}{8}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.5.2

Factor $2$ out of $8$ .

$f (- \frac{1}{2}) = 2 (\frac{- 1}{2 (4)}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.5.3

Cancel the common factor.

$f (- \frac{1}{2}) = 2 (\frac{- 1}{2 \cdot 4}) + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.5.4

Rewrite the expression.

$f (- \frac{1}{2}) = \frac{- 1}{4} + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.6

Move the negative in front of the fraction.

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 {(- \frac{1}{2})}^{2} - 12 (- \frac{1}{2})$

Step 3.1.2.1.7

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.1.2.1.7.1

Apply the product rule to $- \frac{1}{2}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) - 12 (- \frac{1}{2})$

Step 3.1.2.1.7.2

Apply the product rule to $\frac{1}{2}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}})) - 12 (- \frac{1}{2})$

Step 3.1.2.1.8

Raise $- 1$ to the power of $2$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 (1 (\frac{1^{2}}{2^{2}})) - 12 (- \frac{1}{2})$

Step 3.1.2.1.9

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 (\frac{1^{2}}{2^{2}}) - 12 (- \frac{1}{2})$

Step 3.1.2.1.10

One to any power is one.

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 (\frac{1}{2^{2}}) - 12 (- \frac{1}{2})$

Step 3.1.2.1.11

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

$f (- \frac{1}{2}) = - \frac{1}{4} + 3 (\frac{1}{4}) - 12 (- \frac{1}{2})$

Step 3.1.2.1.12

Combine $3$ and $\frac{1}{4}$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} - 12 (- \frac{1}{2})$

Step 3.1.2.1.13

Cancel the common factor of $2$ .

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Step 3.1.2.1.13.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} - 12 (\frac{- 1}{2})$

Step 3.1.2.1.13.2

Factor $2$ out of $- 12$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} + 2 (- 6) (\frac{- 1}{2})$

Step 3.1.2.1.13.3

Cancel the common factor.

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} + 2 \cdot (- 6 (\frac{- 1}{2}))$

Step 3.1.2.1.13.4

Rewrite the expression.

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} - 6 \cdot - 1$

Step 3.1.2.1.14

Multiply $- 6$ by $- 1$ .

$f (- \frac{1}{2}) = - \frac{1}{4} + \frac{3}{4} + 6$

Step 3.1.2.2

Simplify terms.

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Step 3.1.2.2.1

Combine the numerators over the common denominator.

$f (- \frac{1}{2}) = 6 + \frac{- 1 + 3}{4}$

Step 3.1.2.2.2

Add $- 1$ and $3$ .

$f (- \frac{1}{2}) = 6 + \frac{2}{4}$

Step 3.1.2.2.3

Cancel the common factor of $2$ and $4$ .

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Step 3.1.2.2.3.1

Factor $2$ out of $2$ .

$f (- \frac{1}{2}) = 6 + \frac{2 (1)}{4}$

Step 3.1.2.2.3.2

Cancel the common factors.

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Step 3.1.2.2.3.2.1

Factor $2$ out of $4$ .

$f (- \frac{1}{2}) = 6 + \frac{2 \cdot 1}{2 \cdot 2}$

Step 3.1.2.2.3.2.2

Cancel the common factor.

$f (- \frac{1}{2}) = 6 + \frac{2 \cdot 1}{2 \cdot 2}$

Step 3.1.2.2.3.2.3

Rewrite the expression.

$f (- \frac{1}{2}) = 6 + \frac{1}{2}$

Step 3.1.2.3

To write $6$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (- \frac{1}{2}) = 6 \cdot \frac{2}{2} + \frac{1}{2}$

Step 3.1.2.4

Combine $6$ and $\frac{2}{2}$ .

$f (- \frac{1}{2}) = \frac{6 \cdot 2}{2} + \frac{1}{2}$

Step 3.1.2.5

Combine the numerators over the common denominator.

$f (- \frac{1}{2}) = \frac{6 \cdot 2 + 1}{2}$

Step 3.1.2.6

Simplify the numerator.

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Step 3.1.2.6.1

Multiply $6$ by $2$ .

$f (- \frac{1}{2}) = \frac{12 + 1}{2}$

Step 3.1.2.6.2

Add $12$ and $1$ .

$f (- \frac{1}{2}) = \frac{13}{2}$

Step 3.1.2.7

The final answer is $\frac{13}{2}$ .

$\frac{13}{2}$

Step 3.2

The point found by substituting $- \frac{1}{2}$ in $f (x) = 2 x^{3} + 3 x^{2} - 12 x$ is $(- \frac{1}{2}, \frac{13}{2})$ . This point can be an inflection point.

$(- \frac{1}{2}, \frac{13}{2})$

Step 4

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

$(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 0.6$ in the expression.

$f'' (- 0.6) = 12 (- 0.6) + 6$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $12$ by $- 0.6$ .

$f'' (- 0.6) = - 7.2 + 6$

Step 5.2.2

Add $- 7.2$ and $6$ .

$f'' (- 0.6) = - 1.2$

Step 5.2.3

The final answer is $- 1.2$ .

$- 1.2$

Step 5.3

At $- 0.6$ , the second derivative is $- 1.2$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - \frac{1}{2})$

Decreasing on $(- \infty, - \frac{1}{2})$ since $f'' (x) < 0$

Step 6

Substitute a value from the interval $(- \frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.4$ in the expression.

$f'' (- 0.4) = 12 (- 0.4) + 6$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $12$ by $- 0.4$ .

$f'' (- 0.4) = - 4.8 + 6$

Step 6.2.2

Add $- 4.8$ and $6$ .

$f'' (- 0.4) = 1.2$

Step 6.2.3

The final answer is $1.2$ .

$1.2$

Step 6.3

At $- 0.4$ , the second derivative is $1.2$ . Since this is positive, the second derivative is increasing on the interval $(- \frac{1}{2}, \infty)$ .

Increasing on $(- \frac{1}{2}, \infty)$ 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 $(- \frac{1}{2}, \frac{13}{2})$ .

$(- \frac{1}{2}, \frac{13}{2})$

Step 8