Calculus Examples

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

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

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

Step 1.1.2

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

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Step 1.1.2.1

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

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

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 $n = 1$ .

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

Step 1.1.2.3

Multiply $4$ by $1$ .

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

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

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

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 $n = - 2$ .

$f' (x) = 4 - 2 (- 2 x^{- 3})$

Step 1.1.3.3

Multiply $- 2$ by $- 2$ .

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

Step 1.1.4

Simplify.

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Step 1.1.4.1

Reorder terms.

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

Step 1.1.4.2

Simplify each term.

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Step 1.1.4.2.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.1.4.2.2

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

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

Step 1.2

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

$\frac{4}{x^{3}} + 4$

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{4}{x^{3}} + 4 = 0$

Step 2.2

Subtract $4$ from both sides of the equation.

$\frac{4}{x^{3}} = - 4$

Step 2.3

Find the LCD of the terms in the equation.

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Step 2.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x^{3}, 1$

Step 2.3.2

The LCM of one and any expression is the expression.

$x^{3}$

Step 2.4

Multiply each term in $\frac{4}{x^{3}} = - 4$ by $x^{3}$ to eliminate the fractions.

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Step 2.4.1

Multiply each term in $\frac{4}{x^{3}} = - 4$ by $x^{3}$ .

$\frac{4}{x^{3}} x^{3} = - 4 x^{3}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $x^{3}$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{4}{x^{3}} x^{3} = - 4 x^{3}$

Step 2.4.2.1.2

Rewrite the expression.

$4 = - 4 x^{3}$

Step 2.5

Solve the equation.

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Step 2.5.1

Rewrite the equation as $- 4 x^{3} = 4$ .

$- 4 x^{3} = 4$

Step 2.5.2

Subtract $4$ from both sides of the equation.

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

Step 2.5.3

Factor the left side of the equation.

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Step 2.5.3.1

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

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Step 2.5.3.1.1

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

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

Step 2.5.3.1.2

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

$- 4 x^{3} - 4 \cdot 1 = 0$

Step 2.5.3.1.3

Factor $- 4$ out of $- 4 (x^{3}) - 4 (1)$ .

$- 4 (x^{3} + 1) = 0$

Step 2.5.3.2

Rewrite $1$ as $1^{3}$ .

$- 4 (x^{3} + 1^{3}) = 0$

Step 2.5.3.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

$- 4 ((x + 1) (x^{2} - x \cdot 1 + 1^{2})) = 0$

Step 2.5.3.4

Factor.

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Step 2.5.3.4.1

Simplify.

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Step 2.5.3.4.1.1

Multiply $- 1$ by $1$ .

$- 4 ((x + 1) (x^{2} - x + 1^{2})) = 0$

Step 2.5.3.4.1.2

One to any power is one.

$- 4 ((x + 1) (x^{2} - x + 1)) = 0$

Step 2.5.3.4.2

Remove unnecessary parentheses.

$- 4 (x + 1) (x^{2} - x + 1) = 0$

Step 2.5.4

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

$x + 1 = 0$

$x^{2} - x + 1 = 0$

Step 2.5.5

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

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Step 2.5.5.1

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

$x + 1 = 0$

Step 2.5.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.5.6

Set $x^{2} - x + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.6.1

Set $x^{2} - x + 1$ equal to $0$ .

$x^{2} - x + 1 = 0$

Step 2.5.6.2

Solve $x^{2} - x + 1 = 0$ for $x$ .

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Step 2.5.6.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.5.6.2.2

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 2.5.6.2.3

Simplify.

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Step 2.5.6.2.3.1

Simplify the numerator.

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Step 2.5.6.2.3.1.1

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.5.6.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.5.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.5.6.2.3.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.5.6.2.3.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.5.6.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.5.6.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.5.6.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.5.6.2.3.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 2.5.6.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.5.6.2.4.1

Simplify the numerator.

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Step 2.5.6.2.4.1.1

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.5.6.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.5.6.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.5.6.2.4.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.5.6.2.4.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.5.6.2.4.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.5.6.2.4.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.5.6.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.5.6.2.4.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 2.5.6.2.4.3

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{3}}{2}$

Step 2.5.6.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.5.6.2.5.1

Simplify the numerator.

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Step 2.5.6.2.5.1.1

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

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.5.6.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.5.6.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.5.6.2.5.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.5.6.2.5.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.5.6.2.5.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.5.6.2.5.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.5.6.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.5.6.2.5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 2.5.6.2.5.3

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{3}}{2}$

Step 2.5.6.2.6

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 2.5.7

The final solution is all the values that make $- 4 (x + 1) (x^{2} - x + 1) = 0$ true.

$x = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 3

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

$- 1$

Step 4

Find where the derivative is undefined.

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Step 4.1

Set the denominator in $\frac{4}{x^{3}}$ equal to $0$ to find where the expression is undefined.

$x^{3} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

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

$x = \sqrt[3]{0}$

Step 4.2.2

Simplify $\sqrt[3]{0}$ .

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Step 4.2.2.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 4.2.2.2

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 5

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 6

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 6.1

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

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

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

$f' (- 2) = \frac{4}{- 8} + 4$

Step 6.2.1.2

Cancel the common factor of $4$ and $- 8$ .

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Step 6.2.1.2.1

Factor $4$ out of $4$ .

$f' (- 2) = \frac{4 (1)}{- 8} + 4$

Step 6.2.1.2.2

Cancel the common factors.

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Step 6.2.1.2.2.1

Factor $4$ out of $- 8$ .

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

Step 6.2.1.2.2.2

Cancel the common factor.

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

Step 6.2.1.2.2.3

Rewrite the expression.

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

Step 6.2.1.3

Move the negative in front of the fraction.

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

Step 6.2.2

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.2.3

Combine $4$ and $\frac{2}{2}$ .

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

Step 6.2.4

Combine the numerators over the common denominator.

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

Step 6.2.5

Simplify the numerator.

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Step 6.2.5.1

Multiply $4$ by $2$ .

$f' (- 2) = \frac{- 1 + 8}{2}$

Step 6.2.5.2

Add $- 1$ and $8$ .

$f' (- 2) = \frac{7}{2}$

Step 6.2.6

The final answer is $\frac{7}{2}$ .

$\frac{7}{2}$

Step 6.3

At $x = - 2$ the derivative is $\frac{7}{2}$ . Since this is positive, the function is increasing on $(- \infty, - 1)$ .

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

Step 7

Substitute a value from the interval $(- 1, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

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

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

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

Simplify the denominator.

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Step 7.2.1.1.1

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

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

Step 7.2.1.1.2

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

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

Step 7.2.1.1.3

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

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

Step 7.2.1.1.4

One to any power is one.

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

Step 7.2.1.1.5

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

$f' (- \frac{1}{2}) = \frac{4}{- \frac{1}{8}} + 4$

Step 7.2.1.2

Multiply the numerator by the reciprocal of the denominator.

$f' (- \frac{1}{2}) = 4 (- 1 \cdot 8) + 4$

Step 7.2.1.3

Multiply $4 (- 1 \cdot 8)$ .

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Step 7.2.1.3.1

Multiply $- 1$ by $8$ .

$f' (- \frac{1}{2}) = 4 \cdot - 8 + 4$

Step 7.2.1.3.2

Multiply $4$ by $- 8$ .

$f' (- \frac{1}{2}) = - 32 + 4$

Step 7.2.2

Add $- 32$ and $4$ .

$f' (- \frac{1}{2}) = - 28$

Step 7.2.3

The final answer is $- 28$ .

$- 28$

Step 7.3

At $x = - \frac{1}{2}$ the derivative is $- 28$ . Since this is negative, the function is decreasing on $(- 1, 0)$ .

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

Step 8

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 8.1

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

$f' (1) = \frac{4}{{(1)}^{3}} + 4$

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

One to any power is one.

$f' (1) = \frac{4}{1} + 4$

Step 8.2.1.2

Divide $4$ by $1$ .

$f' (1) = 4 + 4$

Step 8.2.2

Add $4$ and $4$ .

$f' (1) = 8$

Step 8.2.3

The final answer is $8$ .

$8$

Step 8.3

At $x = 1$ the derivative is $8$ . Since this is positive, the function is increasing on $(0, \infty)$ .

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

Step 9

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

Increasing on: $(- \infty, - 1), (0, \infty)$

Decreasing on: $(- 1, 0)$

Step 10