Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.1

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

$8 \frac{d}{d x} [\frac{1}{x^{2} + 1}]$

Step 1.1.2

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

$8 \frac{d}{d x} [{(x^{2} + 1)}^{- 1}]$

Step 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 $g (x) = x^{2} + 1$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

$8 (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x^{2} + 1])$

Step 1.2.2

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

$8 (- u^{- 2} \frac{d}{d x} [x^{2} + 1])$

Step 1.2.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

$8 (- {(x^{2} + 1)}^{- 2} \frac{d}{d x} [x^{2} + 1])$

Step 1.3

Differentiate.

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Step 1.3.1

Multiply $- 1$ by $8$ .

$- 8 ({(x^{2} + 1)}^{- 2} \frac{d}{d x} [x^{2} + 1])$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$- 8 {(x^{2} + 1)}^{- 2} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1])$

Step 1.3.3

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

$- 8 {(x^{2} + 1)}^{- 2} (2 x + \frac{d}{d x} [1])$

Step 1.3.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$- 8 {(x^{2} + 1)}^{- 2} (2 x + 0)$

Step 1.3.5

Simplify the expression.

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Step 1.3.5.1

Add $2 x$ and $0$ .

$- 8 {(x^{2} + 1)}^{- 2} (2 x)$

Step 1.3.5.2

Multiply $2$ by $- 8$ .

$- 16 {(x^{2} + 1)}^{- 2} x$

Step 1.4

Simplify.

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Step 1.4.1

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

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

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Combine $- 16$ and $\frac{1}{{(x^{2} + 1)}^{2}}$ .

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

Step 1.4.2.2

Move the negative in front of the fraction.

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

Step 1.4.2.3

Combine $x$ and $\frac{16}{{(x^{2} + 1)}^{2}}$ .

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

Step 1.4.2.4

Move $16$ to the left of $x$ .

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

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- \frac{16 x}{{(x^{2} + 1)}^{2}}$ with respect to $x$ is $- 16 \frac{d}{d x} [\frac{x}{{(x^{2} + 1)}^{2}}]$ .

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

Step 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 $g (x) = {(x^{2} + 1)}^{2}$ .

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

Multiply the exponents in ${({(x^{2} + 1)}^{2})}^{2}$ .

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Step 2.3.1.1

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

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

Step 2.3.1.2

Multiply $2$ by $2$ .

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{4}}$

Step 2.3.2

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

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} \cdot 1 - x \frac{d}{d x} {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{4}}$

Step 2.3.3

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

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} - x \frac{d}{d x} {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{4}}$

Step 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 $g (x) = x^{2} + 1$ .

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Step 2.4.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} - x (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x^{2} + 1))}{{(x^{2} + 1)}^{4}}$

Step 2.4.2

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

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} - x (2 u \frac{d}{d x} (x^{2} + 1))}{{(x^{2} + 1)}^{4}}$

Step 2.4.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} - x (2 (x^{2} + 1) \frac{d}{d x} (x^{2} + 1))}{{(x^{2} + 1)}^{4}}$

Step 2.5

Simplify with factoring out.

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Step 2.5.1

Multiply $2$ by $- 1$ .

$f'' (x) = - 16 \frac{{(x^{2} + 1)}^{2} - 2 x ((x^{2} + 1) \frac{d}{d x} (x^{2} + 1))}{{(x^{2} + 1)}^{4}}$

Step 2.5.2

Factor $x^{2} + 1$ out of ${(x^{2} + 1)}^{2} - 2 x ((x^{2} + 1) \frac{d}{d x} [x^{2} + 1])$ .

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Step 2.5.2.1

Factor $x^{2} + 1$ out of ${(x^{2} + 1)}^{2}$ .

$f'' (x) = - 16 \frac{(x^{2} + 1) (x^{2} + 1) - 2 x ((x^{2} + 1) \frac{d}{d x} (x^{2} + 1))}{{(x^{2} + 1)}^{4}}$

Step 2.5.2.2

Factor $x^{2} + 1$ out of $- 2 x ((x^{2} + 1) \frac{d}{d x} [x^{2} + 1])$ .

$f'' (x) = - 16 \frac{(x^{2} + 1) (x^{2} + 1) + (x^{2} + 1) (- 2 x (\frac{d}{d x} (x^{2} + 1)))}{{(x^{2} + 1)}^{4}}$

Step 2.5.2.3

Factor $x^{2} + 1$ out of $(x^{2} + 1) (x^{2} + 1) + (x^{2} + 1) (- 2 x (\frac{d}{d x} [x^{2} + 1]))$ .

$f'' (x) = - 16 \frac{(x^{2} + 1) (x^{2} + 1 - 2 x (\frac{d}{d x} (x^{2} + 1)))}{{(x^{2} + 1)}^{4}}$

Step 2.6

Cancel the common factors.

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Step 2.6.1

Factor $x^{2} + 1$ out of ${(x^{2} + 1)}^{4}$ .

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

Step 2.6.2

Cancel the common factor.

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

Step 2.6.3

Rewrite the expression.

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

Step 2.7

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 2.8

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

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

Step 2.9

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = - 16 \frac{x^{2} + 1 - 2 x (2 x + 0)}{{(x^{2} + 1)}^{3}}$

Step 2.10

Simplify the expression.

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Step 2.10.1

Add $2 x$ and $0$ .

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

Step 2.10.2

Multiply $2$ by $- 2$ .

$f'' (x) = - 16 \frac{x^{2} + 1 - 4 x \cdot x}{{(x^{2} + 1)}^{3}}$

Step 2.11

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

$f'' (x) = - 16 \frac{x^{2} + 1 - 4 (x \cdot x)}{{(x^{2} + 1)}^{3}}$

Step 2.12

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

$f'' (x) = - 16 \frac{x^{2} + 1 - 4 (x \cdot x)}{{(x^{2} + 1)}^{3}}$

Step 2.13

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

$f'' (x) = - 16 \frac{x^{2} + 1 - 4 x^{1 + 1}}{{(x^{2} + 1)}^{3}}$

Step 2.14

Add $1$ and $1$ .

$f'' (x) = - 16 \frac{x^{2} + 1 - 4 x^{2}}{{(x^{2} + 1)}^{3}}$

Step 2.15

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

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

Step 2.16

Combine $- 16$ and $\frac{- 3 x^{2} + 1}{{(x^{2} + 1)}^{3}}$ .

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

Step 2.17

Move the negative in front of the fraction.

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

Step 2.18

Simplify.

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Step 2.18.1

Apply the distributive property.

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

Step 2.18.2

Simplify each term.

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Step 2.18.2.1

Multiply $- 3$ by $16$ .

$f'' (x) = - \frac{- 48 x^{2} + 16 \cdot 1}{{(x^{2} + 1)}^{3}}$

Step 2.18.2.2

Multiply $16$ by $1$ .

$f'' (x) = - \frac{- 48 x^{2} + 16}{{(x^{2} + 1)}^{3}}$

Step 2.18.3

Factor $16$ out of $- 48 x^{2} + 16$ .

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Step 2.18.3.1

Factor $16$ out of $- 48 x^{2}$ .

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

Step 2.18.3.2

Factor $16$ out of $16$ .

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

Step 2.18.3.3

Factor $16$ out of $16 (- 3 x^{2}) + 16 (1)$ .

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

Step 2.18.4

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

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

Step 2.18.5

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

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

Step 2.18.6

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

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

Step 2.18.7

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

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

Step 2.18.8

Move the negative in front of the fraction.

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

Step 2.18.9

Multiply $- 1$ by $- 1$ .

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

Step 2.18.10

Multiply $\frac{16 (3 x^{2} - 1)}{{(x^{2} + 1)}^{3}}$ by $1$ .

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- \frac{16 x}{{(x^{2} + 1)}^{2}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate using the Constant Multiple Rule.

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Step 4.1.1.1

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

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

Step 4.1.1.2

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

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

Step 4.1.2

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 $g (x) = x^{2} + 1$ .

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Step 4.1.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

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

Step 4.1.3

Differentiate.

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Step 4.1.3.1

Multiply $- 1$ by $8$ .

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

Step 4.1.3.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 4.1.3.3

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

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

Step 4.1.3.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f' (x) = - 8 {(x^{2} + 1)}^{- 2} (2 x + 0)$

Step 4.1.3.5

Simplify the expression.

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Step 4.1.3.5.1

Add $2 x$ and $0$ .

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

Step 4.1.3.5.2

Multiply $2$ by $- 8$ .

$f' (x) = - 16 {(x^{2} + 1)}^{- 2} x$

Step 4.1.4

Simplify.

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Step 4.1.4.1

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

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

Step 4.1.4.2

Combine terms.

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Step 4.1.4.2.1

Combine $- 16$ and $\frac{1}{{(x^{2} + 1)}^{2}}$ .

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

Step 4.1.4.2.2

Move the negative in front of the fraction.

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

Step 4.1.4.2.3

Combine $x$ and $\frac{16}{{(x^{2} + 1)}^{2}}$ .

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

Step 4.1.4.2.4

Move $16$ to the left of $x$ .

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

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{16 x}{{(x^{2} + 1)}^{2}}$ .

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

Step 5

Set the first derivative equal to $0$ then solve the equation $- \frac{16 x}{{(x^{2} + 1)}^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- \frac{16 x}{{(x^{2} + 1)}^{2}} = 0$

Step 5.2

Set the numerator equal to zero.

$16 x = 0$

Step 5.3

Divide each term in $16 x = 0$ by $16$ and simplify.

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Step 5.3.1

Divide each term in $16 x = 0$ by $16$ .

$\frac{16 x}{16} = \frac{0}{16}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $16$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{16 x}{16} = \frac{0}{16}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{16}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $0$ by $16$ .

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 0$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{16 (3 {(0)}^{2} - 1)}{{({(0)}^{2} + 1)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

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

$\frac{16 (3 \cdot 0 - 1)}{{(0^{2} + 1)}^{3}}$

Step 9.1.2

Multiply $3$ by $0$ .

$\frac{16 (0 - 1)}{{(0^{2} + 1)}^{3}}$

Step 9.1.3

Subtract $1$ from $0$ .

$\frac{16 \cdot - 1}{{(0^{2} + 1)}^{3}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

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

$\frac{16 \cdot - 1}{{(0 + 1)}^{3}}$

Step 9.2.2

Add $0$ and $1$ .

$\frac{16 \cdot - 1}{1^{3}}$

Step 9.2.3

One to any power is one.

$\frac{16 \cdot - 1}{1}$

Step 9.3

Simplify the expression.

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Step 9.3.1

Multiply $16$ by $- 1$ .

$\frac{- 16}{1}$

Step 9.3.2

Divide $- 16$ by $1$ .

$- 16$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

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

$f (0) = \frac{8}{{(0)}^{2} + 1}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify the denominator.

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Step 11.2.1.1

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

$f (0) = \frac{8}{0 + 1}$

Step 11.2.1.2

Add $0$ and $1$ .

$f (0) = \frac{8}{1}$

Step 11.2.2

Divide $8$ by $1$ .

$f (0) = 8$

Step 11.2.3

The final answer is $8$ .

$y = 8$

Step 12

These are the local extrema for $f (x) = \frac{8}{x^{2} + 1}$ .

$(0, 8)$ is a local maxima

Step 13