Calculus Examples

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

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.1.1.1

Move the negative in front of the fraction.

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

Step 1.1.1.1.2

Since $- 2$ is constant with respect to $x$ , the derivative of $- \frac{2}{x^{2} + 5}$ with respect to $x$ is $- 2 \frac{d}{d x} [\frac{1}{x^{2} + 5}]$ .

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

Step 1.1.1.1.3

Rewrite $\frac{1}{x^{2} + 5}$ as ${(x^{2} + 5)}^{- 1}$ .

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

Step 1.1.1.2

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 $g (x) = x^{2} + 5$ .

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Step 1.1.1.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 5$ .

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

Step 1.1.1.2.2

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

$f' (x) = - 2 (- u^{- 2} \frac{d}{d x} (x^{2} + 5))$

Step 1.1.1.2.3

Replace all occurrences of $u$ with $x^{2} + 5$ .

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

Step 1.1.1.3

Differentiate.

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Step 1.1.1.3.1

Multiply $- 1$ by $- 2$ .

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

Step 1.1.1.3.2

By the Sum Rule, the derivative of $x^{2} + 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5]$ .

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

Step 1.1.1.3.3

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

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

Step 1.1.1.3.4

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$f' (x) = 2 {(x^{2} + 5)}^{- 2} (2 x + 0)$

Step 1.1.1.3.5

Simplify the expression.

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Step 1.1.1.3.5.1

Add $2 x$ and $0$ .

$f' (x) = 2 {(x^{2} + 5)}^{- 2} (2 x)$

Step 1.1.1.3.5.2

Multiply $2$ by $2$ .

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

Step 1.1.1.4

Simplify.

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Step 1.1.1.4.1

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

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

Step 1.1.1.4.2

Combine terms.

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Step 1.1.1.4.2.1

Combine $4$ and $\frac{1}{{(x^{2} + 5)}^{2}}$ .

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

Step 1.1.1.4.2.2

Combine $\frac{4}{{(x^{2} + 5)}^{2}}$ and $x$ .

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $\frac{4 x}{{(x^{2} + 5)}^{2}}$ with respect to $x$ is $4 \frac{d}{d x} [\frac{x}{{(x^{2} + 5)}^{2}}]$ .

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

Step 1.1.2.2

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 $f (x) = x$ and $g (x) = {(x^{2} + 5)}^{2}$ .

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

Step 1.1.2.3

Differentiate using the Power Rule.

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Step 1.1.2.3.1

Multiply the exponents in ${({(x^{2} + 5)}^{2})}^{2}$ .

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Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Multiply $2$ by $2$ .

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

Step 1.1.2.3.2

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

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

Step 1.1.2.3.3

Multiply ${(x^{2} + 5)}^{2}$ by $1$ .

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

Step 1.1.2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = x^{2} + 5$ .

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Step 1.1.2.4.1

To apply the Chain Rule, set $u$ as $x^{2} + 5$ .

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

Step 1.1.2.4.2

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

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

Step 1.1.2.4.3

Replace all occurrences of $u$ with $x^{2} + 5$ .

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

Step 1.1.2.5

Simplify with factoring out.

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Step 1.1.2.5.1

Multiply $2$ by $- 1$ .

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

Step 1.1.2.5.2

Factor $x^{2} + 5$ out of ${(x^{2} + 5)}^{2} - 2 x ((x^{2} + 5) \frac{d}{d x} [x^{2} + 5])$ .

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Step 1.1.2.5.2.1

Factor $x^{2} + 5$ out of ${(x^{2} + 5)}^{2}$ .

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

Step 1.1.2.5.2.2

Factor $x^{2} + 5$ out of $- 2 x ((x^{2} + 5) \frac{d}{d x} [x^{2} + 5])$ .

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

Step 1.1.2.5.2.3

Factor $x^{2} + 5$ out of $(x^{2} + 5) (x^{2} + 5) + (x^{2} + 5) (- 2 x (\frac{d}{d x} [x^{2} + 5]))$ .

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

Step 1.1.2.6

Cancel the common factors.

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Step 1.1.2.6.1

Factor $x^{2} + 5$ out of ${(x^{2} + 5)}^{4}$ .

$f'' (x) = 4 (\frac{(x^{2} + 5) (x^{2} + 5 - 2 x (\frac{d}{d x} (x^{2} + 5)))}{(x^{2} + 5) {(x^{2} + 5)}^{3}})$

Step 1.1.2.6.2

Cancel the common factor.

$f'' (x) = 4 (\frac{(x^{2} + 5) (x^{2} + 5 - 2 x (\frac{d}{d x} (x^{2} + 5)))}{(x^{2} + 5) {(x^{2} + 5)}^{3}})$

Step 1.1.2.6.3

Rewrite the expression.

$f'' (x) = 4 (\frac{x^{2} + 5 - 2 x (\frac{d}{d x} (x^{2} + 5))}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.7

By the Sum Rule, the derivative of $x^{2} + 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5]$ .

$f'' (x) = 4 (\frac{x^{2} + 5 - 2 x (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (5))}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.8

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

$f'' (x) = 4 (\frac{x^{2} + 5 - 2 x (2 x + \frac{d}{d x} (5))}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.9

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$f'' (x) = 4 (\frac{x^{2} + 5 - 2 x (2 x + 0)}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.10

Simplify the expression.

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Step 1.1.2.10.1

Add $2 x$ and $0$ .

$f'' (x) = 4 (\frac{x^{2} + 5 - 2 x (2 x)}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.10.2

Multiply $2$ by $- 2$ .

$f'' (x) = 4 (\frac{x^{2} + 5 - 4 x \cdot x}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.11

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

$f'' (x) = 4 (\frac{x^{2} + 5 - 4 (x \cdot x)}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.12

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

$f'' (x) = 4 (\frac{x^{2} + 5 - 4 (x \cdot x)}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.13

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

$f'' (x) = 4 (\frac{x^{2} + 5 - 4 x^{1 + 1}}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.14

Add $1$ and $1$ .

$f'' (x) = 4 (\frac{x^{2} + 5 - 4 x^{2}}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.15

Subtract $4 x^{2}$ from $x^{2}$ .

$f'' (x) = 4 (\frac{- 3 x^{2} + 5}{{(x^{2} + 5)}^{3}})$

Step 1.1.2.16

Combine $4$ and $\frac{- 3 x^{2} + 5}{{(x^{2} + 5)}^{3}}$ .

$f'' (x) = \frac{4 (- 3 x^{2} + 5)}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17

Simplify.

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Step 1.1.2.17.1

Apply the distributive property.

$f'' (x) = \frac{4 (- 3 x^{2}) + 4 \cdot 5}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.2

Simplify each term.

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Step 1.1.2.17.2.1

Multiply $- 3$ by $4$ .

$f'' (x) = \frac{- 12 x^{2} + 4 \cdot 5}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.2.2

Multiply $4$ by $5$ .

$f'' (x) = \frac{- 12 x^{2} + 20}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.3

Factor $4$ out of $- 12 x^{2} + 20$ .

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Step 1.1.2.17.3.1

Factor $4$ out of $- 12 x^{2}$ .

$f'' (x) = \frac{4 (- 3 x^{2}) + 20}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.3.2

Factor $4$ out of $20$ .

$f'' (x) = \frac{4 (- 3 x^{2}) + 4 (5)}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.3.3

Factor $4$ out of $4 (- 3 x^{2}) + 4 (5)$ .

$f'' (x) = \frac{4 (- 3 x^{2} + 5)}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.4

Factor $- 1$ out of $- 3 x^{2}$ .

$f'' (x) = \frac{4 (- (3 x^{2}) + 5)}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.5

Rewrite $5$ as $- 1 (- 5)$ .

$f'' (x) = \frac{4 (- (3 x^{2}) - 1 \cdot - 5)}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.6

Factor $- 1$ out of $- (3 x^{2}) - 1 (- 5)$ .

$f'' (x) = \frac{4 (- (3 x^{2} - 5))}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.7

Rewrite $- (3 x^{2} - 5)$ as $- 1 (3 x^{2} - 5)$ .

$f'' (x) = \frac{4 (- 1 (3 x^{2} - 5))}{{(x^{2} + 5)}^{3}}$

Step 1.1.2.17.8

Move the negative in front of the fraction.

$f'' (x) = - \frac{4 (3 x^{2} - 5)}{{(x^{2} + 5)}^{3}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{4 (3 x^{2} - 5)}{{(x^{2} + 5)}^{3}}$ .

$- \frac{4 (3 x^{2} - 5)}{{(x^{2} + 5)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{4 (3 x^{2} - 5)}{{(x^{2} + 5)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{4 (3 x^{2} - 5)}{{(x^{2} + 5)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$4 (3 x^{2} - 5) = 0$

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

Divide each term in $4 (3 x^{2} - 5) = 0$ by $4$ and simplify.

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Step 1.2.3.1.1

Divide each term in $4 (3 x^{2} - 5) = 0$ by $4$ .

$\frac{4 (3 x^{2} - 5)}{4} = \frac{0}{4}$

Step 1.2.3.1.2

Simplify the left side.

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Step 1.2.3.1.2.1

Cancel the common factor of $4$ .

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Step 1.2.3.1.2.1.1

Cancel the common factor.

$\frac{4 (3 x^{2} - 5)}{4} = \frac{0}{4}$

Step 1.2.3.1.2.1.2

Divide $3 x^{2} - 5$ by $1$ .

$3 x^{2} - 5 = \frac{0}{4}$

Step 1.2.3.1.3

Simplify the right side.

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Step 1.2.3.1.3.1

Divide $0$ by $4$ .

$3 x^{2} - 5 = 0$

Step 1.2.3.2

Add $5$ to both sides of the equation.

$3 x^{2} = 5$

Step 1.2.3.3

Divide each term in $3 x^{2} = 5$ by $3$ and simplify.

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Step 1.2.3.3.1

Divide each term in $3 x^{2} = 5$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{5}{3}$

Step 1.2.3.3.2

Simplify the left side.

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Step 1.2.3.3.2.1

Cancel the common factor of $3$ .

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Step 1.2.3.3.2.1.1

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{5}{3}$

Step 1.2.3.3.2.1.2

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

$x^{2} = \frac{5}{3}$

Step 1.2.3.4

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

$x = \pm \sqrt{\frac{5}{3}}$

Step 1.2.3.5

Simplify $\pm \sqrt{\frac{5}{3}}$ .

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Step 1.2.3.5.1

Rewrite $\sqrt{\frac{5}{3}}$ as $\frac{\sqrt{5}}{\sqrt{3}}$ .

$x = \pm \frac{\sqrt{5}}{\sqrt{3}}$

Step 1.2.3.5.2

Multiply $\frac{\sqrt{5}}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$x = \pm \frac{\sqrt{5}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

Step 1.2.3.5.3

Combine and simplify the denominator.

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Step 1.2.3.5.3.1

Multiply $\frac{\sqrt{5}}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$x = \pm \frac{\sqrt{5} \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Step 1.2.3.5.3.2

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm \frac{\sqrt{5} \sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Step 1.2.3.5.3.3

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm \frac{\sqrt{5} \sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Step 1.2.3.5.3.4

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

$x = \pm \frac{\sqrt{5} \sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Step 1.2.3.5.3.5

Add $1$ and $1$ .

$x = \pm \frac{\sqrt{5} \sqrt{3}}{{\sqrt{3}}^{2}}$

Step 1.2.3.5.3.6

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 1.2.3.5.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$x = \pm \frac{\sqrt{5} \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Step 1.2.3.5.3.6.2

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

$x = \pm \frac{\sqrt{5} \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Step 1.2.3.5.3.6.3

Combine $\frac{1}{2}$ and $2$ .

$x = \pm \frac{\sqrt{5} \sqrt{3}}{3^{\frac{2}{2}}}$

Step 1.2.3.5.3.6.4

Cancel the common factor of $2$ .

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Step 1.2.3.5.3.6.4.1

Cancel the common factor.

$x = \pm \frac{\sqrt{5} \sqrt{3}}{3^{\frac{2}{2}}}$

Step 1.2.3.5.3.6.4.2

Rewrite the expression.

$x = \pm \frac{\sqrt{5} \sqrt{3}}{3^{1}}$

Step 1.2.3.5.3.6.5

Evaluate the exponent.

$x = \pm \frac{\sqrt{5} \sqrt{3}}{3}$

Step 1.2.3.5.4

Simplify the numerator.

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Step 1.2.3.5.4.1

Combine using the product rule for radicals.

$x = \pm \frac{\sqrt{5 \cdot 3}}{3}$

Step 1.2.3.5.4.2

Multiply $5$ by $3$ .

$x = \pm \frac{\sqrt{15}}{3}$

Step 1.2.3.6

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

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Step 1.2.3.6.1

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

$x = \frac{\sqrt{15}}{3}$

Step 1.2.3.6.2

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

$x = - \frac{\sqrt{15}}{3}$

Step 1.2.3.6.3

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

$x = \frac{\sqrt{15}}{3}, - \frac{\sqrt{15}}{3}$

Step 2

Find the domain of $f (x) = \frac{- 2}{x^{2} + 5}$ .

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Step 2.1

Set the denominator in $\frac{- 2}{x^{2} + 5}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 5 = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $5$ from both sides of the equation.

$x^{2} = - 5$

Step 2.2.2

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

$x = \pm \sqrt{- 5}$

Step 2.2.3

Simplify $\pm \sqrt{- 5}$ .

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Step 2.2.3.1

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Step 2.2.3.2

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{5}$

Step 2.2.3.3

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

$x = \pm i \sqrt{5}$

Step 2.2.4

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

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Step 2.2.4.1

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

$x = i \sqrt{5}$

Step 2.2.4.2

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

$x = - i \sqrt{5}$

Step 2.2.4.3

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

$x = i \sqrt{5}, - i \sqrt{5}$

Step 2.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - \frac{\sqrt{15}}{3}) \cup (- \frac{\sqrt{15}}{3}, \frac{\sqrt{15}}{3}) \cup (\frac{\sqrt{15}}{3}, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - \frac{\sqrt{15}}{3})$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 4$ in the expression.

$f'' (- 4) = - \frac{4 (3 {(- 4)}^{2} - 5)}{{({(- 4)}^{2} + 5)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the numerator.

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Step 4.2.1.1

Raise $- 4$ to the power of $2$ .

$f'' (- 4) = - \frac{4 (3 \cdot 16 - 5)}{{({(- 4)}^{2} + 5)}^{3}}$

Step 4.2.1.2

Multiply $3$ by $16$ .

$f'' (- 4) = - \frac{4 (48 - 5)}{{({(- 4)}^{2} + 5)}^{3}}$

Step 4.2.1.3

Subtract $5$ from $48$ .

$f'' (- 4) = - \frac{4 \cdot 43}{{({(- 4)}^{2} + 5)}^{3}}$

Step 4.2.2

Simplify the denominator.

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Step 4.2.2.1

Raise $- 4$ to the power of $2$ .

$f'' (- 4) = - \frac{4 \cdot 43}{{(16 + 5)}^{3}}$

Step 4.2.2.2

Add $16$ and $5$ .

$f'' (- 4) = - \frac{4 \cdot 43}{21^{3}}$

Step 4.2.2.3

Raise $21$ to the power of $3$ .

$f'' (- 4) = - \frac{4 \cdot 43}{9261}$

Step 4.2.3

Multiply $4$ by $43$ .

$f'' (- 4) = - \frac{172}{9261}$

Step 4.2.4

The final answer is $- \frac{172}{9261}$ .

$- \frac{172}{9261}$

Step 4.3

The graph is concave down on the interval $(- \infty, - \frac{\sqrt{15}}{3})$ because $f'' (- 4)$ is negative.

Concave down on $(- \infty, - \frac{\sqrt{15}}{3})$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- \frac{\sqrt{15}}{3}, \frac{\sqrt{15}}{3})$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = - \frac{4 (3 {(0)}^{2} - 5)}{{({(0)}^{2} + 5)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

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

$f'' (0) = - \frac{4 (3 \cdot 0 - 5)}{{(0^{2} + 5)}^{3}}$

Step 5.2.1.2

Multiply $3$ by $0$ .

$f'' (0) = - \frac{4 (0 - 5)}{{(0^{2} + 5)}^{3}}$

Step 5.2.1.3

Subtract $5$ from $0$ .

$f'' (0) = - \frac{4 \cdot - 5}{{(0^{2} + 5)}^{3}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

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

$f'' (0) = - \frac{4 \cdot - 5}{{(0 + 5)}^{3}}$

Step 5.2.2.2

Add $0$ and $5$ .

$f'' (0) = - \frac{4 \cdot - 5}{5^{3}}$

Step 5.2.2.3

Raise $5$ to the power of $3$ .

$f'' (0) = - \frac{4 \cdot - 5}{125}$

Step 5.2.3

Reduce the expression by cancelling the common factors.

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Step 5.2.3.1

Multiply $4$ by $- 5$ .

$f'' (0) = - \frac{- 20}{125}$

Step 5.2.3.2

Cancel the common factor of $- 20$ and $125$ .

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Step 5.2.3.2.1

Factor $5$ out of $- 20$ .

$f'' (0) = - \frac{5 (- 4)}{125}$

Step 5.2.3.2.2

Cancel the common factors.

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Step 5.2.3.2.2.1

Factor $5$ out of $125$ .

$f'' (0) = - \frac{5 \cdot - 4}{5 \cdot 25}$

Step 5.2.3.2.2.2

Cancel the common factor.

$f'' (0) = - \frac{5 \cdot - 4}{5 \cdot 25}$

Step 5.2.3.2.2.3

Rewrite the expression.

$f'' (0) = - \frac{- 4}{25}$

Step 5.2.3.3

Move the negative in front of the fraction.

$f'' (0) = \frac{4}{25}$

Step 5.2.4

The final answer is $\frac{4}{25}$ .

$\frac{4}{25}$

Step 5.3

The graph is concave up on the interval $(- \frac{\sqrt{15}}{3}, \frac{\sqrt{15}}{3})$ because $f'' (0)$ is positive.

Concave up on $(- \frac{\sqrt{15}}{3}, \frac{\sqrt{15}}{3})$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(\frac{\sqrt{15}}{3}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $4$ in the expression.

$f'' (4) = - \frac{4 (3 {(4)}^{2} - 5)}{{({(4)}^{2} + 5)}^{3}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Raise $4$ to the power of $2$ .

$f'' (4) = - \frac{4 (3 \cdot 16 - 5)}{{(4^{2} + 5)}^{3}}$

Step 6.2.1.2

Multiply $3$ by $16$ .

$f'' (4) = - \frac{4 (48 - 5)}{{(4^{2} + 5)}^{3}}$

Step 6.2.1.3

Subtract $5$ from $48$ .

$f'' (4) = - \frac{4 \cdot 43}{{(4^{2} + 5)}^{3}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $4$ to the power of $2$ .

$f'' (4) = - \frac{4 \cdot 43}{{(16 + 5)}^{3}}$

Step 6.2.2.2

Add $16$ and $5$ .

$f'' (4) = - \frac{4 \cdot 43}{21^{3}}$

Step 6.2.2.3

Raise $21$ to the power of $3$ .

$f'' (4) = - \frac{4 \cdot 43}{9261}$

Step 6.2.3

Multiply $4$ by $43$ .

$f'' (4) = - \frac{172}{9261}$

Step 6.2.4

The final answer is $- \frac{172}{9261}$ .

$- \frac{172}{9261}$

Step 6.3

The graph is concave down on the interval $(\frac{\sqrt{15}}{3}, \infty)$ because $f'' (4)$ is negative.

Concave down on $(\frac{\sqrt{15}}{3}, \infty)$ since $f'' (x)$ is negative

Step 7

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(- \infty, - \frac{\sqrt{15}}{3})$ since $f'' (x)$ is negative

Concave up on $(- \frac{\sqrt{15}}{3}, \frac{\sqrt{15}}{3})$ since $f'' (x)$ is positive

Concave down on $(\frac{\sqrt{15}}{3}, \infty)$ since $f'' (x)$ is negative

Step 8