Calculus Examples

$f (x) = x^{2} + \frac{8}{x}$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{2} + \frac{8}{x}$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [\frac{8}{x}]$ .

$f' (x) = \frac{d}{d x} (x^{2}) + \frac{d}{d x} (\frac{8}{x})$

Step 1.1.1.2

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

$f' (x) = 2 x + \frac{d}{d x} (\frac{8}{x})$

Step 1.1.2

Evaluate $\frac{d}{d x} [\frac{8}{x}]$ .

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Step 1.1.2.1

Since $8$ is constant with respect to $x$ , the derivative of $\frac{8}{x}$ with respect to $x$ is $8 \frac{d}{d x} [\frac{1}{x}]$ .

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

Step 1.1.2.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 1.1.2.3

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

$f' (x) = 2 x + 8 (- x^{- 2})$

Step 1.1.2.4

Multiply $- 1$ by $8$ .

$f' (x) = 2 x - 8 x^{- 2}$

Step 1.1.3

Simplify.

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Step 1.1.3.1

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

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

Step 1.1.3.2

Combine terms.

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Step 1.1.3.2.1

Combine $- 8$ and $\frac{1}{x^{2}}$ .

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

Step 1.1.3.2.2

Move the negative in front of the fraction.

$f' (x) = 2 x - \frac{8}{x^{2}}$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $2 x - \frac{8}{x^{2}}$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- \frac{8}{x^{2}}]$ .

$f'' (x) = \frac{d}{d x} (2 x) + \frac{d}{d x} (- \frac{8}{x^{2}})$

Step 1.2.2

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

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Step 1.2.2.1

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

$f'' (x) = 2 \frac{d}{d x} (x) + \frac{d}{d x} (- \frac{8}{x^{2}})$

Step 1.2.2.2

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

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

Step 1.2.2.3

Multiply $2$ by $1$ .

$f'' (x) = 2 + \frac{d}{d x} (- \frac{8}{x^{2}})$

Step 1.2.3

Evaluate $\frac{d}{d x} [- \frac{8}{x^{2}}]$ .

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Step 1.2.3.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- \frac{8}{x^{2}}$ with respect to $x$ is $- 8 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f'' (x) = 2 - 8 \frac{d}{d x} \frac{1}{x^{2}}$

Step 1.2.3.2

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$f'' (x) = 2 - 8 \frac{d}{d x} {(x^{2})}^{- 1}$

Step 1.2.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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Step 1.2.3.3.1

To apply the Chain Rule, set $u$ as $x^{2}$ .

$f'' (x) = 2 - 8 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2}))$

Step 1.2.3.3.2

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

$f'' (x) = 2 - 8 (- u^{- 2} \frac{d}{d x} x^{2})$

Step 1.2.3.3.3

Replace all occurrences of $u$ with $x^{2}$ .

$f'' (x) = 2 - 8 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2})$

Step 1.2.3.4

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

$f'' (x) = 2 - 8 (- {(x^{2})}^{- 2} (2 x))$

Step 1.2.3.5

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Step 1.2.3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = 2 - 8 (- x^{2 \cdot - 2} (2 x))$

Step 1.2.3.5.2

Multiply $2$ by $- 2$ .

$f'' (x) = 2 - 8 (- x^{- 4} (2 x))$

Step 1.2.3.6

Multiply $2$ by $- 1$ .

$f'' (x) = 2 - 8 (- 2 x^{- 4} x)$

Step 1.2.3.7

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

$f'' (x) = 2 - 8 (- 2 (x \cdot x^{- 4}))$

Step 1.2.3.8

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

$f'' (x) = 2 - 8 (- 2 x^{1 - 4})$

Step 1.2.3.9

Subtract $4$ from $1$ .

$f'' (x) = 2 - 8 (- 2 x^{- 3})$

Step 1.2.3.10

Multiply $- 2$ by $- 8$ .

$f'' (x) = 2 + 16 x^{- 3}$

Step 1.2.4

Simplify.

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Step 1.2.4.1

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

$f'' (x) = 2 + 16 (\frac{1}{x^{3}})$

Step 1.2.4.2

Combine $16$ and $\frac{1}{x^{3}}$ .

$f'' (x) = 2 + \frac{16}{x^{3}}$

Step 1.2.4.3

Reorder terms.

$f'' (x) = \frac{16}{x^{3}} + 2$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{16}{x^{3}} + 2$ .

$\frac{16}{x^{3}} + 2$

Step 2

Set the second derivative equal to $0$ then solve the equation $\frac{16}{x^{3}} + 2 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$\frac{16}{x^{3}} + 2 = 0$

Step 2.2

Subtract $2$ from both sides of the equation.

$\frac{16}{x^{3}} = - 2$

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 $\frac{16}{x^{3}} = - 2$ by $x^{3}$ to eliminate the fractions.

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Step 2.4.1

Multiply each term in $\frac{16}{x^{3}} = - 2$ by $x^{3}$ .

$\frac{16}{x^{3}} x^{3} = - 2 x^{3}$

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.

$\frac{16}{x^{3}} x^{3} = - 2 x^{3}$

Step 2.4.2.1.2

Rewrite the expression.

$16 = - 2 x^{3}$

Step 2.5

Solve the equation.

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Step 2.5.1

Rewrite the equation as $- 2 x^{3} = 16$ .

$- 2 x^{3} = 16$

Step 2.5.2

Subtract $16$ from both sides of the equation.

$- 2 x^{3} - 16 = 0$

Step 2.5.3

Factor the left side of the equation.

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Step 2.5.3.1

Factor $- 2$ out of $- 2 x^{3} - 16$ .

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Step 2.5.3.1.1

Factor $- 2$ out of $- 2 x^{3}$ .

$- 2 x^{3} - 16 = 0$

Step 2.5.3.1.2

Factor $- 2$ out of $- 16$ .

$- 2 x^{3} - 2 \cdot 8 = 0$

Step 2.5.3.1.3

Factor $- 2$ out of $- 2 (x^{3}) - 2 (8)$ .

$- 2 (x^{3} + 8) = 0$

Step 2.5.3.2

Rewrite $8$ as $2^{3}$ .

$- 2 (x^{3} + 2^{3}) = 0$

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 $b = 2$ .

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

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 $2$ by $- 1$ .

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

Step 2.5.3.4.1.2

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

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

Step 2.5.3.4.2

Remove unnecessary parentheses.

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

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 $0$ .

$x + 2 = 0$

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

Step 2.5.5

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

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Step 2.5.5.1

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

$x + 2 = 0$

Step 2.5.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.5.6

Set $x^{2} - 2 x + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.6.1

Set $x^{2} - 2 x + 4$ equal to $0$ .

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

Step 2.5.6.2

Solve $x^{2} - 2 x + 4 = 0$ 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 $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

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

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

Step 2.5.6.2.3.1.2

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

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Step 2.5.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.6.2.3.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 2.5.6.2.3.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 2.5.6.2.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 2.5.6.2.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.6.2.3.1.6

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.6.2.3.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.5.6.2.3.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.5.6.2.3.1.7.2

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

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

Step 2.5.6.2.3.1.8

Pull terms out from under the radical.

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

Step 2.5.6.2.3.1.9

Move $2$ to the left of $i$ .

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

Step 2.5.6.2.3.2

Multiply $2$ by $1$ .

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

Step 2.5.6.2.3.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

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

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

Step 2.5.6.2.4.1.2

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

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Step 2.5.6.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.6.2.4.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 2.5.6.2.4.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 2.5.6.2.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 2.5.6.2.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.6.2.4.1.6

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.6.2.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.5.6.2.4.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.5.6.2.4.1.7.2

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

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

Step 2.5.6.2.4.1.8

Pull terms out from under the radical.

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

Step 2.5.6.2.4.1.9

Move $2$ to the left of $i$ .

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

Step 2.5.6.2.4.2

Multiply $2$ by $1$ .

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

Step 2.5.6.2.4.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

Step 2.5.6.2.4.4

Change the $\pm$ to $+$ .

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

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

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

Step 2.5.6.2.5.1.2

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

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Step 2.5.6.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.6.2.5.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 2.5.6.2.5.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 2.5.6.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 2.5.6.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.6.2.5.1.6

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.6.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.5.6.2.5.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.5.6.2.5.1.7.2

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

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

Step 2.5.6.2.5.1.8

Pull terms out from under the radical.

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

Step 2.5.6.2.5.1.9

Move $2$ to the left of $i$ .

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

Step 2.5.6.2.5.2

Multiply $2$ by $1$ .

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

Step 2.5.6.2.5.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

Step 2.5.6.2.5.4

Change the $\pm$ to $-$ .

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

Step 2.5.6.2.6

The final answer is the combination of both solutions.

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

Step 2.5.7

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

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

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $- 2$ in $f (x) = x^{2} + \frac{8}{x}$ to find the value of $y$ .

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Step 3.1.1

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

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

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

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

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

Step 3.1.2.1.2

Divide $8$ by $- 2$ .

$f (- 2) = 4 - 4$

Step 3.1.2.2

Subtract $4$ from $4$ .

$f (- 2) = 0$

Step 3.1.2.3

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $- 2$ in $f (x) = x^{2} + \frac{8}{x}$ is $(- 2, 0)$ . This point can be an inflection point.

$(- 2, 0)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Step 5

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

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Step 5.1

Replace the variable $x$ with $- 2.1$ in the expression.

$f'' (- 2.1) = \frac{16}{{(- 2.1)}^{3}} + 2$

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 $- 2.1$ to the power of $3$ .

$f'' (- 2.1) = \frac{16}{- 9.261} + 2$

Step 5.2.1.2

Divide $16$ by $- 9.261$ .

$f'' (- 2.1) = - 1.72767519 + 2$

Step 5.2.2

Add $- 1.72767519$ and $2$ .

$f'' (- 2.1) = 0.2723248$

Step 5.2.3

The final answer is $0.2723248$ .

$0.2723248$

Step 5.3

At $- 2.1$ , the second derivative is $0.2723248$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 2)$ .

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $- 1.9$ in the expression.

$f'' (- 1.9) = \frac{16}{{(- 1.9)}^{3}} + 2$

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 $- 1.9$ to the power of $3$ .

$f'' (- 1.9) = \frac{16}{- 6.859} + 2$

Step 6.2.1.2

Divide $16$ by $- 6.859$ .

$f'' (- 1.9) = - 2.33270155 + 2$

Step 6.2.2

Add $- 2.33270155$ and $2$ .

$f'' (- 1.9) = - 0.33270155$

Step 6.2.3

The final answer is $- 0.33270155$ .

$- 0.33270155$

Step 6.3

At $- 1.9$ , the second derivative is $- 0.33270155$ . Since this is negative, the second derivative is decreasing on the interval $(- 2, \infty)$

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

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(- 2, 0)$ .

$(- 2, 0)$

Step 8