Calculus Examples

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

Step 1

Find the first 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

The first derivative of $f (x)$ with respect to $x$ is $6 x^{2} + 6 x - 12$ .

$6 x^{2} + 6 x - 12$

Step 2

Set the first derivative equal to $0$ then solve the equation $6 x^{2} + 6 x - 12 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$6 x^{2} + 6 x - 12 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $6$ out of $6 x^{2} + 6 x - 12$ .

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Step 2.2.1.1

Factor $6$ out of $6 x^{2}$ .

$6 (x^{2}) + 6 x - 12 = 0$

Step 2.2.1.2

Factor $6$ out of $6 x$ .

$6 (x^{2}) + 6 (x) - 12 = 0$

Step 2.2.1.3

Factor $6$ out of $- 12$ .

$6 (x^{2}) + 6 x + 6 \cdot - 2 = 0$

Step 2.2.1.4

Factor $6$ out of $6 (x^{2}) + 6 x$ .

$6 (x^{2} + x) + 6 \cdot - 2 = 0$

Step 2.2.1.5

Factor $6$ out of $6 (x^{2} + x) + 6 \cdot - 2$ .

$6 (x^{2} + x - 2) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $x^{2} + x - 2$ using the AC method.

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Step 2.2.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 2.2.2.1.2

Write the factored form using these integers.

$6 ((x - 1) (x + 2)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$6 (x - 1) (x + 2) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 2 = 0$

Step 2.4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

The final solution is all the values that make $6 (x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 3

The values which make the derivative equal to $0$ are $1, - 2$ .

$1, - 2$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 2) \cup (- 2, 1) \cup (1, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 2)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

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

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

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

$f' (- 3) = 6 \cdot 9 + 6 (- 3) - 12$

Step 5.2.1.2

Multiply $6$ by $9$ .

$f' (- 3) = 54 + 6 (- 3) - 12$

Step 5.2.1.3

Multiply $6$ by $- 3$ .

$f' (- 3) = 54 - 18 - 12$

Step 5.2.2

Simplify by subtracting numbers.

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Step 5.2.2.1

Subtract $18$ from $54$ .

$f' (- 3) = 36 - 12$

Step 5.2.2.2

Subtract $12$ from $36$ .

$f' (- 3) = 24$

Step 5.2.3

The final answer is $24$ .

$24$

Step 5.3

At $x = - 3$ the derivative is $24$ . Since this is positive, the function is increasing on $(- \infty, - 2)$ .

Increasing on $(- \infty, - 2)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 2, 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

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

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Step 6.2.1.1.1

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

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

Step 6.2.1.1.2

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

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

Step 6.2.1.2

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

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

Step 6.2.1.3

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

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

Step 6.2.1.4

One to any power is one.

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

Step 6.2.1.5

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

$f' (- \frac{1}{2}) = 6 (\frac{1}{4}) + 6 (- \frac{1}{2}) - 12$

Step 6.2.1.6

Cancel the common factor of $2$ .

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Step 6.2.1.6.1

Factor $2$ out of $6$ .

$f' (- \frac{1}{2}) = 2 (3) (\frac{1}{4}) + 6 (- \frac{1}{2}) - 12$

Step 6.2.1.6.2

Factor $2$ out of $4$ .

$f' (- \frac{1}{2}) = 2 \cdot (3 (\frac{1}{2 \cdot 2})) + 6 (- \frac{1}{2}) - 12$

Step 6.2.1.6.3

Cancel the common factor.

$f' (- \frac{1}{2}) = 2 \cdot (3 (\frac{1}{2 \cdot 2})) + 6 (- \frac{1}{2}) - 12$

Step 6.2.1.6.4

Rewrite the expression.

$f' (- \frac{1}{2}) = 3 (\frac{1}{2}) + 6 (- \frac{1}{2}) - 12$

Step 6.2.1.7

Combine $3$ and $\frac{1}{2}$ .

$f' (- \frac{1}{2}) = \frac{3}{2} + 6 (- \frac{1}{2}) - 12$

Step 6.2.1.8

Cancel the common factor of $2$ .

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Step 6.2.1.8.1

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

$f' (- \frac{1}{2}) = \frac{3}{2} + 6 (\frac{- 1}{2}) - 12$

Step 6.2.1.8.2

Factor $2$ out of $6$ .

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

Step 6.2.1.8.3

Cancel the common factor.

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

Step 6.2.1.8.4

Rewrite the expression.

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

Step 6.2.1.9

Multiply $3$ by $- 1$ .

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

Step 6.2.2

Find the common denominator.

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Step 6.2.2.1

Write $- 3$ as a fraction with denominator $1$ .

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

Step 6.2.2.2

Multiply $\frac{- 3}{1}$ by $\frac{2}{2}$ .

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

Step 6.2.2.3

Multiply $\frac{- 3}{1}$ by $\frac{2}{2}$ .

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

Step 6.2.2.4

Write $- 12$ as a fraction with denominator $1$ .

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

Step 6.2.2.5

Multiply $\frac{- 12}{1}$ by $\frac{2}{2}$ .

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

Step 6.2.2.6

Multiply $\frac{- 12}{1}$ by $\frac{2}{2}$ .

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

Step 6.2.3

Combine the numerators over the common denominator.

$f' (- \frac{1}{2}) = \frac{3 - 3 \cdot 2 - 12 \cdot 2}{2}$

Step 6.2.4

Simplify each term.

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Step 6.2.4.1

Multiply $- 3$ by $2$ .

$f' (- \frac{1}{2}) = \frac{3 - 6 - 12 \cdot 2}{2}$

Step 6.2.4.2

Multiply $- 12$ by $2$ .

$f' (- \frac{1}{2}) = \frac{3 - 6 - 24}{2}$

Step 6.2.5

Simplify the expression.

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Step 6.2.5.1

Subtract $6$ from $3$ .

$f' (- \frac{1}{2}) = \frac{- 3 - 24}{2}$

Step 6.2.5.2

Subtract $24$ from $- 3$ .

$f' (- \frac{1}{2}) = \frac{- 27}{2}$

Step 6.2.5.3

Move the negative in front of the fraction.

$f' (- \frac{1}{2}) = - \frac{27}{2}$

Step 6.2.6

The final answer is $- \frac{27}{2}$ .

$- \frac{27}{2}$

Step 6.3

At $x = - \frac{1}{2}$ the derivative is $- \frac{27}{2}$ . Since this is negative, the function is decreasing on $(- 2, 1)$ .

Decreasing on $(- 2, 1)$ since $f' (x) < 0$

Step 7

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

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Step 7.1

Replace the variable $x$ with $2$ in the expression.

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

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

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

$f' (2) = 6 \cdot 4 + 6 (2) - 12$

Step 7.2.1.2

Multiply $6$ by $4$ .

$f' (2) = 24 + 6 (2) - 12$

Step 7.2.1.3

Multiply $6$ by $2$ .

$f' (2) = 24 + 12 - 12$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Add $24$ and $12$ .

$f' (2) = 36 - 12$

Step 7.2.2.2

Subtract $12$ from $36$ .

$f' (2) = 24$

Step 7.2.3

The final answer is $24$ .

$24$

Step 7.3

At $x = 2$ the derivative is $24$ . Since this is positive, the function is increasing on $(1, \infty)$ .

Increasing on $(1, \infty)$ since $f' (x) > 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, - 2), (1, \infty)$

Decreasing on: $(- 2, 1)$

Step 9