Calculus Examples

$\frac{- x^{2} - 16}{x}$

Step 1

Write $\frac{- x^{2} - 16}{x}$ as a function.

$f (x) = \frac{- x^{2} - 16}{x}$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = - x^{2} - 16$ and $g (x) = x$ .

$\frac{x \frac{d}{d x} [- x^{2} - 16] - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $- x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 16]$ .

$\frac{x (\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [- 16]) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.2.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{x (- \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{x (- (2 x) + \frac{d}{d x} [- 16]) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.2.4

Multiply $2$ by $- 1$ .

$\frac{x (- 2 x + \frac{d}{d x} [- 16]) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.2.5

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$\frac{x (- 2 x + 0) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.2.6

Add $- 2 x$ and $0$ .

$\frac{x (- 2 x) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.3

Raise $x$ to the power of $1$ .

$\frac{- 2 (x^{1} x) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.4

Raise $x$ to the power of $1$ .

$\frac{- 2 (x^{1} x^{1}) - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 2 x^{1 + 1} - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.6

Add $1$ and $1$ .

$\frac{- 2 x^{2} - (- x^{2} - 16) \frac{d}{d x} [x]}{x^{2}}$

Step 2.1.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{- 2 x^{2} - (- x^{2} - 16) \cdot 1}{x^{2}}$

Step 2.1.8

Multiply $- 1$ by $1$ .

$\frac{- 2 x^{2} - (- x^{2} - 16)}{x^{2}}$

Step 2.1.9

Simplify.

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Step 2.1.9.1

Apply the distributive property.

$\frac{- 2 x^{2} - - x^{2} - - 16}{x^{2}}$

Step 2.1.9.2

Simplify the numerator.

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Step 2.1.9.2.1

Simplify each term.

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Step 2.1.9.2.1.1

Multiply $- - x^{2}$ .

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Step 2.1.9.2.1.1.1

Multiply $- 1$ by $- 1$ .

$\frac{- 2 x^{2} + 1 x^{2} - - 16}{x^{2}}$

Step 2.1.9.2.1.1.2

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

$\frac{- 2 x^{2} + x^{2} - - 16}{x^{2}}$

Step 2.1.9.2.1.2

Multiply $- 1$ by $- 16$ .

$\frac{- 2 x^{2} + x^{2} + 16}{x^{2}}$

Step 2.1.9.2.2

Add $- 2 x^{2}$ and $x^{2}$ .

$\frac{- x^{2} + 16}{x^{2}}$

Step 2.1.9.3

Simplify the numerator.

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Step 2.1.9.3.1

Rewrite $16$ as $4^{2}$ .

$\frac{- x^{2} + 4^{2}}{x^{2}}$

Step 2.1.9.3.2

Reorder $- x^{2}$ and $4^{2}$ .

$\frac{4^{2} - x^{2}}{x^{2}}$

Step 2.1.9.3.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = x$ .

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

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{(4 + x) (4 - x)}{x^{2}}$ .

$\frac{(4 + x) (4 - x)}{x^{2}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{(4 + x) (4 - x)}{x^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

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

Step 3.2

Set the numerator equal to zero.

$(4 + x) (4 - x) = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

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

$4 + x = 0$

$4 - x = 0$

Step 3.3.2

Set $4 + x$ equal to $0$ and solve for $x$ .

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Step 3.3.2.1

Set $4 + x$ equal to $0$ .

$4 + x = 0$

Step 3.3.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 3.3.3

Set $4 - x$ equal to $0$ and solve for $x$ .

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Step 3.3.3.1

Set $4 - x$ equal to $0$ .

$4 - x = 0$

Step 3.3.3.2

Solve $4 - x = 0$ for $x$ .

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Step 3.3.3.2.1

Subtract $4$ from both sides of the equation.

$- x = - 4$

Step 3.3.3.2.2

Divide each term in $- x = - 4$ by $- 1$ and simplify.

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Step 3.3.3.2.2.1

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Step 3.3.3.2.2.2

Simplify the left side.

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Step 3.3.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Step 3.3.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Step 3.3.3.2.2.3

Simplify the right side.

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Step 3.3.3.2.2.3.1

Divide $- 4$ by $- 1$ .

$x = 4$

Step 3.3.4

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

$x = - 4, 4$

Step 4

The values which make the derivative equal to $0$ are $- 4, 4$ .

$- 4, 4$

Step 5

Find where the derivative is undefined.

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Step 5.1

Set the denominator in $\frac{(4 + x) (4 - x)}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 5.2

Solve for $x$ .

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Step 5.2.1

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

$x = \pm \sqrt{0}$

Step 5.2.2

Simplify $\pm \sqrt{0}$ .

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Step 5.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 5.2.2.2

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

$x = \pm 0$

Step 5.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 6

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

$(- \infty, - 4) \cup (- 4, 0) \cup (0, 4) \cup (4, \infty)$

Step 7

Substitute a value from the interval $(- \infty, - 4)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- 5$ in the expression.

$f' (- 5) = \frac{(4 - 5) (4 - (- 5))}{{(- 5)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Remove parentheses.

$f' (- 5) = \frac{(4 - 5) (4 - (- 5))}{{(- 5)}^{2}}$

Step 7.2.2

Simplify the numerator.

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Step 7.2.2.1

Subtract $5$ from $4$ .

$f' (- 5) = \frac{- 1 (4 - (- 5))}{{(- 5)}^{2}}$

Step 7.2.2.2

Multiply $- 1$ by $- 5$ .

$f' (- 5) = \frac{- 1 (4 + 5)}{{(- 5)}^{2}}$

Step 7.2.2.3

Add $4$ and $5$ .

$f' (- 5) = \frac{- 1 \cdot 9}{{(- 5)}^{2}}$

Step 7.2.3

Simplify the expression.

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Step 7.2.3.1

Raise $- 5$ to the power of $2$ .

$f' (- 5) = \frac{- 1 \cdot 9}{25}$

Step 7.2.3.2

Multiply $- 1$ by $9$ .

$f' (- 5) = \frac{- 9}{25}$

Step 7.2.3.3

Move the negative in front of the fraction.

$f' (- 5) = - \frac{9}{25}$

Step 7.2.4

The final answer is $- \frac{9}{25}$ .

$- \frac{9}{25}$

Step 7.3

At $x = - 5$ the derivative is $- \frac{9}{25}$ . Since this is negative, the function is decreasing on $(- \infty, - 4)$ .

Decreasing on $(- \infty, - 4)$ since $f' (x) < 0$

Step 8

Substitute a value from the interval $(- 4, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

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

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

Step 8.2

Simplify the result.

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Step 8.2.1

Remove parentheses.

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

Step 8.2.2

Simplify the numerator.

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Step 8.2.2.1

Subtract $2$ from $4$ .

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

Step 8.2.2.2

Multiply $- 1$ by $- 2$ .

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

Step 8.2.2.3

Add $4$ and $2$ .

$f' (- 2) = \frac{2 \cdot 6}{{(- 2)}^{2}}$

Step 8.2.3

Simplify the expression.

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Step 8.2.3.1

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

$f' (- 2) = \frac{2 \cdot 6}{4}$

Step 8.2.3.2

Multiply $2$ by $6$ .

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

Step 8.2.3.3

Divide $12$ by $4$ .

$f' (- 2) = 3$

Step 8.2.4

The final answer is $3$ .

$3$

Step 8.3

At $x = - 2$ the derivative is $3$ . Since this is positive, the function is increasing on $(- 4, 0)$ .

Increasing on $(- 4, 0)$ since $f' (x) > 0$

Step 9

Substitute a value from the interval $(0, 4)$ into the derivative to determine if the function is increasing or decreasing.

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Step 9.1

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

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

Step 9.2

Simplify the result.

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Step 9.2.1

Remove parentheses.

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

Step 9.2.2

Simplify the numerator.

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Step 9.2.2.1

Add $4$ and $2$ .

$f' (2) = \frac{6 (4 - (2))}{2^{2}}$

Step 9.2.2.2

Multiply $- 1$ by $2$ .

$f' (2) = \frac{6 (4 - 2)}{2^{2}}$

Step 9.2.2.3

Subtract $2$ from $4$ .

$f' (2) = \frac{6 \cdot 2}{2^{2}}$

Step 9.2.3

Simplify the expression.

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Step 9.2.3.1

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

$f' (2) = \frac{6 \cdot 2}{4}$

Step 9.2.3.2

Multiply $6$ by $2$ .

$f' (2) = \frac{12}{4}$

Step 9.2.3.3

Divide $12$ by $4$ .

$f' (2) = 3$

Step 9.2.4

The final answer is $3$ .

$3$

Step 9.3

At $x = 2$ the derivative is $3$ . Since this is positive, the function is increasing on $(0, 4)$ .

Increasing on $(0, 4)$ since $f' (x) > 0$

Step 10

Substitute a value from the interval $(4, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 10.1

Replace the variable $x$ with $5$ in the expression.

$f' (5) = \frac{(4 + 5) (4 - (5))}{{(5)}^{2}}$

Step 10.2

Simplify the result.

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Step 10.2.1

Remove parentheses.

$f' (5) = \frac{(4 + 5) (4 - (5))}{{(5)}^{2}}$

Step 10.2.2

Simplify the numerator.

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Step 10.2.2.1

Add $4$ and $5$ .

$f' (5) = \frac{9 (4 - (5))}{5^{2}}$

Step 10.2.2.2

Multiply $- 1$ by $5$ .

$f' (5) = \frac{9 (4 - 5)}{5^{2}}$

Step 10.2.2.3

Subtract $5$ from $4$ .

$f' (5) = \frac{9 \cdot - 1}{5^{2}}$

Step 10.2.3

Simplify the expression.

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Step 10.2.3.1

Raise $5$ to the power of $2$ .

$f' (5) = \frac{9 \cdot - 1}{25}$

Step 10.2.3.2

Multiply $9$ by $- 1$ .

$f' (5) = \frac{- 9}{25}$

Step 10.2.3.3

Move the negative in front of the fraction.

$f' (5) = - \frac{9}{25}$

Step 10.2.4

The final answer is $- \frac{9}{25}$ .

$- \frac{9}{25}$

Step 10.3

At $x = 5$ the derivative is $- \frac{9}{25}$ . Since this is negative, the function is decreasing on $(4, \infty)$ .

Decreasing on $(4, \infty)$ since $f' (x) < 0$

Step 11

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

Increasing on: $(- 4, 0), (0, 4)$

Decreasing on: $(- \infty, - 4), (4, \infty)$

Step 12