Calculus Examples

$f (x) = \frac{1}{4} x^{4} - 2 x^{2}$

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 $\frac{1}{4} x^{4} - 2 x^{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{4} x^{4}] + \frac{d}{d x} [- 2 x^{2}]$ .

$f' (x) = \frac{d}{d x} (\frac{1}{4} x^{4}) + \frac{d}{d x} (- 2 x^{2})$

Step 1.1.2

Evaluate $\frac{d}{d x} [\frac{1}{4} x^{4}]$ .

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Step 1.1.2.1

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{1}{4} x^{4}$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [x^{4}]$ .

$f' (x) = \frac{1}{4} \cdot \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 2 x^{2})$

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 $n = 4$ .

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

Step 1.1.2.3

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

$f' (x) = \frac{4}{4} x^{3} + \frac{d}{d x} (- 2 x^{2})$

Step 1.1.2.4

Combine $\frac{4}{4}$ and $x^{3}$ .

$f' (x) = \frac{4 x^{3}}{4} + \frac{d}{d x} (- 2 x^{2})$

Step 1.1.2.5

Cancel the common factor of $4$ .

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Step 1.1.2.5.1

Cancel the common factor.

$f' (x) = \frac{4 x^{3}}{4} + \frac{d}{d x} (- 2 x^{2})$

Step 1.1.2.5.2

Divide $x^{3}$ by $1$ .

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 2$ .

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

Step 1.1.3.3

Multiply $2$ by $- 2$ .

$f' (x) = x^{3} - 4 x$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $x^{3} - 4 x$ .

$x^{3} - 4 x$

Step 2

Set the first derivative equal to $0$ then solve the equation $x^{3} - 4 x = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$x^{3} - 4 x = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $x$ out of $x^{3} - 4 x$ .

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Step 2.2.1.1

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 4 x = 0$

Step 2.2.1.2

Factor $x$ out of $- 4 x$ .

$x \cdot x^{2} + x \cdot - 4 = 0$

Step 2.2.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 4$ .

$x (x^{2} - 4) = 0$

Step 2.2.2

Rewrite $4$ as $2^{2}$ .

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

Step 2.2.3

Factor.

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Step 2.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 2.2.3.2

Remove unnecessary parentheses.

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

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 $0$ .

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

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

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

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Step 2.6.1

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

$x - 2 = 0$

Step 2.6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.7

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

$x = 0, - 2, 2$

Step 3

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

$0, - 2, 2$

Step 4

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

$(- \infty, - 2) \cup (- 2, 0) \cup (0, 2) \cup (2, \infty)$

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.

$f' (- 3) = {(- 3)}^{3} - 4 \cdot - 3$

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 $3$ .

$f' (- 3) = - 27 - 4 \cdot - 3$

Step 5.2.1.2

Multiply $- 4$ by $- 3$ .

$f' (- 3) = - 27 + 12$

Step 5.2.2

Add $- 27$ and $12$ .

$f' (- 3) = - 15$

Step 5.2.3

The final answer is $- 15$ .

$- 15$

Step 5.3

At $x = - 3$ the derivative is $- 15$ . Since this is negative, the function is decreasing on $(- \infty, - 2)$ .

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

Step 6

Substitute a value from the interval $(- 2, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

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

$f' (- 1) = {(- 1)}^{3} - 4 \cdot - 1$

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

$f' (- 1) = - 1 - 4 \cdot - 1$

Step 6.2.1.2

Multiply $- 4$ by $- 1$ .

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

Step 6.2.2

Add $- 1$ and $4$ .

$f' (- 1) = 3$

Step 6.2.3

The final answer is $3$ .

$3$

Step 6.3

At $x = - 1$ the derivative is $3$ . Since this is positive, the function is increasing on $(- 2, 0)$ .

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

Step 7

Substitute a value from the interval $(0, 2)$ 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.

$f' (1) = {(1)}^{3} - 4 \cdot 1$

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.

$f' (1) = 1 - 4 \cdot 1$

Step 7.2.1.2

Multiply $- 4$ by $1$ .

$f' (1) = 1 - 4$

Step 7.2.2

Subtract $4$ from $1$ .

$f' (1) = - 3$

Step 7.2.3

The final answer is $- 3$ .

$- 3$

Step 7.3

At $x = 1$ the derivative is $- 3$ . Since this is negative, the function is decreasing on $(0, 2)$ .

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

Step 8

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

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Step 8.1

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

$f' (3) = {(3)}^{3} - 4 \cdot 3$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

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

$f' (3) = 27 - 4 \cdot 3$

Step 8.2.1.2

Multiply $- 4$ by $3$ .

$f' (3) = 27 - 12$

Step 8.2.2

Subtract $12$ from $27$ .

$f' (3) = 15$

Step 8.2.3

The final answer is $15$ .

$15$

Step 8.3

At $x = 3$ the derivative is $15$ . Since this is positive, the function is increasing on $(2, \infty)$ .

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

Step 9

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

Increasing on: $(- 2, 0), (2, \infty)$

Decreasing on: $(- \infty, - 2), (0, 2)$

Step 10