Calculus Examples

$f (x) = 2 x^{3} + 5 x^{2} - 12$

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 $2 x^{3} + 5 x^{2} - 12$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 12]$ .

$\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 12]$

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

$2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 12]$

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 $n = 3$ .

$2 (3 x^{2}) + \frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 12]$

Step 1.1.2.3

Multiply $3$ by $2$ .

$6 x^{2} + \frac{d}{d x} [5 x^{2}] + \frac{d}{d x} [- 12]$

Step 1.1.3

Evaluate $\frac{d}{d x} [5 x^{2}]$ .

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Step 1.1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$6 x^{2} + 5 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 12]$

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 $n = 2$ .

$6 x^{2} + 5 (2 x) + \frac{d}{d x} [- 12]$

Step 1.1.3.3

Multiply $2$ by $5$ .

$6 x^{2} + 10 x + \frac{d}{d x} [- 12]$

Step 1.1.4

Differentiate using the Constant Rule.

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Step 1.1.4.1

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

$6 x^{2} + 10 x + 0$

Step 1.1.4.2

Add $6 x^{2} + 10 x$ and $0$ .

$f' (x) = 6 x^{2} + 10 x$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $6 x^{2} + 10 x$ .

$6 x^{2} + 10 x$

Step 2

Set the first derivative equal to $0$ then solve the equation $6 x^{2} + 10 x = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$6 x^{2} + 10 x = 0$

Step 2.2

Factor $2 x$ out of $6 x^{2} + 10 x$ .

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Step 2.2.1

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

$2 x (3 x) + 10 x = 0$

Step 2.2.2

Factor $2 x$ out of $10 x$ .

$2 x (3 x) + 2 x (5) = 0$

Step 2.2.3

Factor $2 x$ out of $2 x (3 x) + 2 x (5)$ .

$2 x (3 x + 5) = 0$

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 $0$ .

$x = 0$

$3 x + 5 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $3 x + 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $3 x + 5$ equal to $0$ .

$3 x + 5 = 0$

Step 2.5.2

Solve $3 x + 5 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $5$ from both sides of the equation.

$3 x = - 5$

Step 2.5.2.2

Divide each term in $3 x = - 5$ by $3$ and simplify.

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Step 2.5.2.2.1

Divide each term in $3 x = - 5$ by $3$ .

$\frac{3 x}{3} = \frac{- 5}{3}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 5}{3}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{3}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{3}$

Step 2.6

The final solution is all the values that make $2 x (3 x + 5) = 0$ true.

$x = 0, - \frac{5}{3}$

Step 3

The values which make the derivative equal to $0$ are $0, - \frac{5}{3}$ .

$0, - \frac{5}{3}$

Step 4

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

$(- \infty, - \frac{5}{3}) \cup (- \frac{5}{3}, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \frac{5}{3})$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 2.6666666$ in the expression.

$f' (- 2.6666666) = 6 {(- 2.6666666)}^{2} + 10 (- 2.6666666)$

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 $- 2.6666666$ to the power of $2$ .

$f' (- 2.6666666) = 6 \cdot 7.11111075 + 10 (- 2.6666666)$

Step 5.2.1.2

Multiply $6$ by $7.11111075$ .

$f' (- 2.6666666) = 42.66666453 + 10 (- 2.6666666)$

Step 5.2.1.3

Multiply $10$ by $- 2.6666666$ .

$f' (- 2.6666666) = 42.66666453 - 26.666666$

Step 5.2.2

Subtract $26.666666$ from $42.66666453$ .

$f' (- 2.6666666) = 15.99999853$

Step 5.2.3

The final answer is $15.99999853$ .

$15.99999853$

Step 5.3

At $x = - 2.6666666$ the derivative is $15.99999853$ . Since this is positive, the function is increasing on $(- \infty, - \frac{5}{3})$ .

Increasing on $(- \infty, - \frac{5}{3})$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 1.6666666, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.8333333$ in the expression.

$f' (- 0.8333333) = 6 {(- 0.8333333)}^{2} + 10 (- 0.8333333)$

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

Raise $- 0.8333333$ to the power of $2$ .

$f' (- 0.8333333) = 6 \cdot 0.69444438 + 10 (- 0.8333333)$

Step 6.2.1.2

Multiply $6$ by $0.69444438$ .

$f' (- 0.8333333) = 4.166666\overline{3} + 10 (- 0.8333333)$

Step 6.2.1.3

Multiply $10$ by $- 0.8333333$ .

$f' (- 0.8333333) = 4.166666\overline{3} - 8.333333$

Step 6.2.2

Subtract $8.333333$ from $4.166666\overline{3}$ .

$f' (- 0.8333333) = - 4.1\overline{6}$

Step 6.2.3

The final answer is $- 4.1\overline{6}$ .

$- 4.1\overline{6}$

Step 6.3

At $x = - 0.8333333$ the derivative is $- 4.1\overline{6}$ . Since this is negative, the function is decreasing on $(- 1.6666666, 0)$ .

Decreasing on $(- \frac{5}{3}, 0)$ since $f' (x) < 0$

Step 7

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

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Step 7.1

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

$f' (1) = 6 {(1)}^{2} + 10 (1)$

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

One to any power is one.

$f' (1) = 6 \cdot 1 + 10 (1)$

Step 7.2.1.2

Multiply $6$ by $1$ .

$f' (1) = 6 + 10 (1)$

Step 7.2.1.3

Multiply $10$ by $1$ .

$f' (1) = 6 + 10$

Step 7.2.2

Add $6$ and $10$ .

$f' (1) = 16$

Step 7.2.3

The final answer is $16$ .

$16$

Step 7.3

At $x = 1$ the derivative is $16$ . Since this is positive, the function is increasing on $(0, \infty)$ .

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

Step 8

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

Increasing on: $(- \infty, - \frac{5}{3}), (0, \infty)$

Decreasing on: $(- \frac{5}{3}, 0)$

Step 9