Calculus Examples

$f (x) = \frac{1}{3} x^{3} - x$

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 $\frac{1}{3} x^{3} - x$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [\frac{1}{3} x^{3}] + \frac{d}{d x} [- x]$

Step 1.1.2

Evaluate $\frac{d}{d x} [\frac{1}{3} x^{3}]$ .

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Step 1.1.2.1

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3} x^{3}$ with respect to $x$ is $\frac{1}{3} \frac{d}{d x} [x^{3}]$ .

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x]$

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 $n = 3$ .

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [- x]$

Step 1.1.2.3

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

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

Step 1.1.2.4

Combine $\frac{3}{3}$ and $x^{2}$ .

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

Step 1.1.2.5

Cancel the common factor of $3$ .

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Step 1.1.2.5.1

Cancel the common factor.

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

Step 1.1.2.5.2

Divide $x^{2}$ by $1$ .

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

Step 1.1.3

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

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Step 1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

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 $n = 1$ .

$x^{2} - 1 \cdot 1$

Step 1.1.3.3

Multiply $- 1$ by $1$ .

$f' (x) = x^{2} - 1$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $x^{2} - 1$ .

$x^{2} - 1$

Step 2

Set the first derivative equal to $0$ then solve the equation $x^{2} - 1 = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$x^{2} - 1 = 0$

Step 2.2

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Step 2.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 2.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 2.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 2.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 3

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

$1, - 1$

Step 4

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

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

Step 5

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

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Step 5.1

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

$f' (- 2) = {(- 2)}^{2} - 1$

Step 5.2

Simplify the result.

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Step 5.2.1

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

$f' (- 2) = 4 - 1$

Step 5.2.2

Subtract $1$ from $4$ .

$f' (- 2) = 3$

Step 5.2.3

The final answer is $3$ .

$3$

Step 5.3

At $x = - 2$ the derivative is $3$ . Since this is positive, the function is increasing on $(- \infty, - 1)$ .

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $0$ in the expression.

$f' (0) = {(0)}^{2} - 1$

Step 6.2

Simplify the result.

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Step 6.2.1

Raising $0$ to any positive power yields $0$ .

$f' (0) = 0 - 1$

Step 6.2.2

Subtract $1$ from $0$ .

$f' (0) = - 1$

Step 6.2.3

The final answer is $- 1$ .

$- 1$

Step 6.3

At $x = 0$ the derivative is $- 1$ . Since this is negative, the function is decreasing on $(- 1, 1)$ .

Decreasing on $(- 1, 1)$ 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.

$f' (2) = {(2)}^{2} - 1$

Step 7.2

Simplify the result.

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Step 7.2.1

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

$f' (2) = 4 - 1$

Step 7.2.2

Subtract $1$ from $4$ .

$f' (2) = 3$

Step 7.2.3

The final answer is $3$ .

$3$

Step 7.3

At $x = 2$ the derivative is $3$ . 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: $(- \infty, - 1), (1, \infty)$

Decreasing on: $(- 1, 1)$

Step 9