Calculus Examples

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

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

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

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

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

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

Step 1.1.1.2.5

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

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

Step 1.1.1.2.6

Simplify the expression.

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Step 1.1.1.2.6.1

Add $2 x$ and $0$ .

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

Step 1.1.1.2.6.2

Multiply $2$ by $- 1$ .

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

Step 1.1.1.3

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

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

Step 1.1.1.4

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

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

Step 1.1.1.5

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

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

Step 1.1.1.6

Add $1$ and $1$ .

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

Step 1.1.1.7

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

Step 1.1.2.2

Differentiate.

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Step 1.1.2.2.1

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

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Step 1.1.2.2.1.1

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

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

Step 1.1.2.2.1.2

Multiply $2$ by $2$ .

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

Step 1.1.2.2.2

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

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

Step 1.1.2.2.3

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

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

Step 1.1.2.2.4

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

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

Step 1.1.2.2.5

Multiply $2$ by $- 1$ .

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

Step 1.1.2.2.6

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

$f'' (x) = \frac{{(x^{2} + 1)}^{2} (- 2 x + 0) - (- x^{2} + 1) \frac{d}{d x} {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{4}}$

Step 1.1.2.2.7

Add $- 2 x$ and $0$ .

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

Step 1.1.2.3

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

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Step 1.1.2.3.1

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

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

Step 1.1.2.3.2

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

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

Step 1.1.2.3.3

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

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

Step 1.1.2.4

Differentiate.

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Step 1.1.2.4.1

Multiply $2$ by $- 1$ .

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

Step 1.1.2.4.2

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

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

Step 1.1.2.4.3

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

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

Step 1.1.2.4.4

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

$f'' (x) = \frac{{(x^{2} + 1)}^{2} (- 2 x) - 2 (- x^{2} + 1) ((x^{2} + 1) (2 x + 0))}{{(x^{2} + 1)}^{4}}$

Step 1.1.2.4.5

Simplify the expression.

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Step 1.1.2.4.5.1

Add $2 x$ and $0$ .

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

Step 1.1.2.4.5.2

Move $2$ to the left of $x^{2} + 1$ .

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

Step 1.1.2.4.5.3

Multiply $2$ by $- 2$ .

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

Step 1.1.2.5

Simplify.

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Step 1.1.2.5.1

Apply the distributive property.

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

Step 1.1.2.5.2

Apply the distributive property.

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

Step 1.1.2.5.3

Simplify the numerator.

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Step 1.1.2.5.3.1

Simplify each term.

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Step 1.1.2.5.3.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.2.5.3.1.2

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

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

Step 1.1.2.5.3.1.3

Expand $(x^{2} + 1) (x^{2} + 1)$ using the FOIL Method.

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Step 1.1.2.5.3.1.3.1

Apply the distributive property.

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

Step 1.1.2.5.3.1.3.2

Apply the distributive property.

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

Step 1.1.2.5.3.1.3.3

Apply the distributive property.

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

Step 1.1.2.5.3.1.4

Simplify and combine like terms.

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Step 1.1.2.5.3.1.4.1

Simplify each term.

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Step 1.1.2.5.3.1.4.1.1

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 1.1.2.5.3.1.4.1.1.1

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

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

Step 1.1.2.5.3.1.4.1.1.2

Add $2$ and $2$ .

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

Step 1.1.2.5.3.1.4.1.2

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

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

Step 1.1.2.5.3.1.4.1.3

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

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

Step 1.1.2.5.3.1.4.1.4

Multiply $1$ by $1$ .

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

Step 1.1.2.5.3.1.4.2

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

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

Step 1.1.2.5.3.1.5

Apply the distributive property.

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

Step 1.1.2.5.3.1.6

Simplify.

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Step 1.1.2.5.3.1.6.1

Multiply $2$ by $- 2$ .

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

Step 1.1.2.5.3.1.6.2

Multiply $- 2$ by $1$ .

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

Step 1.1.2.5.3.1.7

Apply the distributive property.

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

Step 1.1.2.5.3.1.8

Simplify.

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Step 1.1.2.5.3.1.8.1

Multiply $x^{4}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.8.1.1

Move $x$ .

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

Step 1.1.2.5.3.1.8.1.2

Multiply $x$ by $x^{4}$ .

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Step 1.1.2.5.3.1.8.1.2.1

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

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

Step 1.1.2.5.3.1.8.1.2.2

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

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

Step 1.1.2.5.3.1.8.1.3

Add $1$ and $4$ .

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

Step 1.1.2.5.3.1.8.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.8.2.1

Move $x$ .

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

Step 1.1.2.5.3.1.8.2.2

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

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Step 1.1.2.5.3.1.8.2.2.1

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

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

Step 1.1.2.5.3.1.8.2.2.2

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

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

Step 1.1.2.5.3.1.8.2.3

Add $1$ and $2$ .

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

Step 1.1.2.5.3.1.9

Simplify each term.

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Step 1.1.2.5.3.1.9.1

Multiply $- 1$ by $- 4$ .

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

Step 1.1.2.5.3.1.9.2

Multiply $- 4$ by $1$ .

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

Step 1.1.2.5.3.1.10

Simplify each term.

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Step 1.1.2.5.3.1.10.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.10.1.1

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

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Step 1.1.2.5.3.1.10.1.1.1

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

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

Step 1.1.2.5.3.1.10.1.1.2

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

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

Step 1.1.2.5.3.1.10.1.2

Add $2$ and $1$ .

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

Step 1.1.2.5.3.1.10.2

Multiply $x$ by $1$ .

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

Step 1.1.2.5.3.1.11

Expand $(4 x^{2} - 4) (x^{3} + x)$ using the FOIL Method.

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Step 1.1.2.5.3.1.11.1

Apply the distributive property.

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

Step 1.1.2.5.3.1.11.2

Apply the distributive property.

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

Step 1.1.2.5.3.1.11.3

Apply the distributive property.

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

Step 1.1.2.5.3.1.12

Simplify and combine like terms.

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Step 1.1.2.5.3.1.12.1

Simplify each term.

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Step 1.1.2.5.3.1.12.1.1

Multiply $x^{2}$ by $x^{3}$ by adding the exponents.

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Step 1.1.2.5.3.1.12.1.1.1

Move $x^{3}$ .

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

Step 1.1.2.5.3.1.12.1.1.2

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

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

Step 1.1.2.5.3.1.12.1.1.3

Add $3$ and $2$ .

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

Step 1.1.2.5.3.1.12.1.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.5.3.1.12.1.2.1

Move $x$ .

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

Step 1.1.2.5.3.1.12.1.2.2

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

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Step 1.1.2.5.3.1.12.1.2.2.1

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

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

Step 1.1.2.5.3.1.12.1.2.2.2

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

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

Step 1.1.2.5.3.1.12.1.2.3

Add $1$ and $2$ .

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

Step 1.1.2.5.3.1.12.2

Subtract $4 x^{3}$ from $4 x^{3}$ .

$f'' (x) = \frac{- 2 x^{5} - 4 x^{3} - 2 x + 4 x^{5} + 0 - 4 x}{{(x^{2} + 1)}^{4}}$

Step 1.1.2.5.3.1.12.3

Add $4 x^{5}$ and $0$ .

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

Step 1.1.2.5.3.2

Add $- 2 x^{5}$ and $4 x^{5}$ .

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

Step 1.1.2.5.3.3

Subtract $4 x$ from $- 2 x$ .

$f'' (x) = \frac{2 x^{5} - 4 x^{3} - 6 x}{{(x^{2} + 1)}^{4}}$

Step 1.1.2.5.4

Simplify the numerator.

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Step 1.1.2.5.4.1

Factor $2 x$ out of $2 x^{5} - 4 x^{3} - 6 x$ .

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Step 1.1.2.5.4.1.1

Factor $2 x$ out of $2 x^{5}$ .

$f'' (x) = \frac{2 x (x^{4}) - 4 x^{3} - 6 x}{{(x^{2} + 1)}^{4}}$

Step 1.1.2.5.4.1.2

Factor $2 x$ out of $- 4 x^{3}$ .

$f'' (x) = \frac{2 x (x^{4}) + 2 x (- 2 x^{2}) - 6 x}{{(x^{2} + 1)}^{4}}$

Step 1.1.2.5.4.1.3

Factor $2 x$ out of $- 6 x$ .

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

Step 1.1.2.5.4.1.4

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

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

Step 1.1.2.5.4.1.5

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

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

Step 1.1.2.5.4.2

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

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

Step 1.1.2.5.4.3

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

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

Step 1.1.2.5.4.4

Factor $u^{2} - 2 u - 3$ using the AC method.

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Step 1.1.2.5.4.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 1.1.2.5.4.4.2

Write the factored form using these integers.

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

Step 1.1.2.5.4.5

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

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

Step 1.1.2.5.5

Cancel the common factor of $x^{2} + 1$ and ${(x^{2} + 1)}^{4}$ .

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Step 1.1.2.5.5.1

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

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

Step 1.1.2.5.5.2

Cancel the common factors.

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Step 1.1.2.5.5.2.1

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

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

Step 1.1.2.5.5.2.2

Cancel the common factor.

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

Step 1.1.2.5.5.2.3

Rewrite the expression.

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2 x (x^{2} - 3)}{{(x^{2} + 1)}^{3}}$ .

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

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{2 x (x^{2} - 3)}{{(x^{2} + 1)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

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

Step 1.2.2

Set the numerator equal to zero.

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

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

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

$x = 0$

$x^{2} - 3 = 0$

Step 1.2.3.2

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.3.3

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

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Step 1.2.3.3.1

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Step 1.2.3.3.2

Solve $x^{2} - 3 = 0$ for $x$ .

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Step 1.2.3.3.2.1

Add $3$ to both sides of the equation.

$x^{2} = 3$

Step 1.2.3.3.2.2

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

$x = \pm \sqrt{3}$

Step 1.2.3.3.2.3

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

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Step 1.2.3.3.2.3.1

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

$x = \sqrt{3}$

Step 1.2.3.3.2.3.2

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

$x = - \sqrt{3}$

Step 1.2.3.3.2.3.3

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

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

Step 1.2.3.4

The final solution is all the values that make $2 x (x^{2} - 3) = 0$ true.

$x = 0, \sqrt{3}, - \sqrt{3}$

Step 2

Find the domain of $f (x) = \frac{x}{x^{2} + 1}$ .

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Step 2.1

Set the denominator in $\frac{x}{x^{2} + 1}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 1 = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.2.2

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

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

Step 2.2.3

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

$x = \pm i$

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.

$x = i$

Step 2.2.4.2

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

$x = - i$

Step 2.2.4.3

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

$x = i, - i$

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.

$(- \infty, - \sqrt{3}) \cup (- \sqrt{3}, 0) \cup (0, \sqrt{3}) \cup (\sqrt{3}, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - \sqrt{3})$ 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.

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

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply $2$ by $- 4$ .

$f'' (- 4) = \frac{- 8 ({(- 4)}^{2} - 3)}{{({(- 4)}^{2} + 1)}^{3}}$

Step 4.2.2

Simplify the denominator.

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Step 4.2.2.1

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

$f'' (- 4) = \frac{- 8 ({(- 4)}^{2} - 3)}{{(16 + 1)}^{3}}$

Step 4.2.2.2

Add $16$ and $1$ .

$f'' (- 4) = \frac{- 8 ({(- 4)}^{2} - 3)}{17^{3}}$

Step 4.2.2.3

Raise $17$ to the power of $3$ .

$f'' (- 4) = \frac{- 8 ({(- 4)}^{2} - 3)}{4913}$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

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

$f'' (- 4) = \frac{- 8 (16 - 3)}{4913}$

Step 4.2.3.2

Subtract $3$ from $16$ .

$f'' (- 4) = \frac{- 8 \cdot 13}{4913}$

Step 4.2.4

Simplify the expression.

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Step 4.2.4.1

Multiply $- 8$ by $13$ .

$f'' (- 4) = \frac{- 104}{4913}$

Step 4.2.4.2

Move the negative in front of the fraction.

$f'' (- 4) = - \frac{104}{4913}$

Step 4.2.5

The final answer is $- \frac{104}{4913}$ .

$- \frac{104}{4913}$

Step 4.3

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

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

Step 5

Substitute any number from the interval $(- \sqrt{3}, 0)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

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

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $2$ by $- 1$ .

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

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

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

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

Step 5.2.2.2

Add $1$ and $1$ .

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

Step 5.2.2.3

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

$f'' (- 1) = \frac{- 2 ({(- 1)}^{2} - 3)}{8}$

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

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

$f'' (- 1) = \frac{- 2 (1 - 3)}{8}$

Step 5.2.3.2

Subtract $3$ from $1$ .

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

Step 5.2.4

Reduce the expression by cancelling the common factors.

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Step 5.2.4.1

Multiply $- 2$ by $- 2$ .

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

Step 5.2.4.2

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

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Step 5.2.4.2.1

Factor $4$ out of $4$ .

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

Step 5.2.4.2.2

Cancel the common factors.

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Step 5.2.4.2.2.1

Factor $4$ out of $8$ .

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

Step 5.2.4.2.2.2

Cancel the common factor.

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

Step 5.2.4.2.2.3

Rewrite the expression.

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

Step 5.2.5

The final answer is $\frac{1}{2}$ .

$\frac{1}{2}$

Step 5.3

The graph is concave up on the interval $(- \sqrt{3}, 0)$ because $f'' (- 1)$ is positive.

Concave up on $(- \sqrt{3}, 0)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(0, \sqrt{3})$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $1$ .

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

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

One to any power is one.

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

Step 6.2.2.2

Add $1$ and $1$ .

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

Step 6.2.2.3

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

$f'' (1) = \frac{2 (1^{2} - 3)}{8}$

Step 6.2.3

Simplify the numerator.

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Step 6.2.3.1

One to any power is one.

$f'' (1) = \frac{2 (1 - 3)}{8}$

Step 6.2.3.2

Subtract $3$ from $1$ .

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

Step 6.2.4

Reduce the expression by cancelling the common factors.

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Step 6.2.4.1

Multiply $2$ by $- 2$ .

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

Step 6.2.4.2

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

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Step 6.2.4.2.1

Factor $4$ out of $- 4$ .

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

Step 6.2.4.2.2

Cancel the common factors.

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Step 6.2.4.2.2.1

Factor $4$ out of $8$ .

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

Step 6.2.4.2.2.2

Cancel the common factor.

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

Step 6.2.4.2.2.3

Rewrite the expression.

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

Step 6.2.4.3

Move the negative in front of the fraction.

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

Step 6.2.5

The final answer is $- \frac{1}{2}$ .

$- \frac{1}{2}$

Step 6.3

The graph is concave down on the interval $(0, \sqrt{3})$ because $f'' (1)$ is negative.

Concave down on $(0, \sqrt{3})$ since $f'' (x)$ is negative

Step 7

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

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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $4$ .

$f'' (4) = \frac{8 (4^{2} - 3)}{{(4^{2} + 1)}^{3}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

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

$f'' (4) = \frac{8 (4^{2} - 3)}{{(16 + 1)}^{3}}$

Step 7.2.2.2

Add $16$ and $1$ .

$f'' (4) = \frac{8 (4^{2} - 3)}{17^{3}}$

Step 7.2.2.3

Raise $17$ to the power of $3$ .

$f'' (4) = \frac{8 (4^{2} - 3)}{4913}$

Step 7.2.3

Simplify the numerator.

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Step 7.2.3.1

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

$f'' (4) = \frac{8 (16 - 3)}{4913}$

Step 7.2.3.2

Subtract $3$ from $16$ .

$f'' (4) = \frac{8 \cdot 13}{4913}$

Step 7.2.4

Multiply $8$ by $13$ .

$f'' (4) = \frac{104}{4913}$

Step 7.2.5

The final answer is $\frac{104}{4913}$ .

$\frac{104}{4913}$

Step 7.3

The graph is concave up on the interval $(\sqrt{3}, \infty)$ because $f'' (4)$ is positive.

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

Step 8

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

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

Concave up on $(- \sqrt{3}, 0)$ since $f'' (x)$ is positive

Concave down on $(0, \sqrt{3})$ since $f'' (x)$ is negative

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

Step 9