Calculus Examples

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

Find the first derivative of the function.

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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 $f (x) = x - 1$ and $g (x) = x^{2} + 3 x + 5$ .

$\frac{(x^{2} + 3 x + 5) \frac{d}{d x} [x - 1] - (x - 1) \frac{d}{d x} [x^{2} + 3 x + 5]}{{(x^{2} + 3 x + 5)}^{2}}$

Differentiate.

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

$\frac{(x^{2} + 3 x + 5) (\frac{d}{d x} [x] + \frac{d}{d x} [- 1]) - (x - 1) \frac{d}{d x} [x^{2} + 3 x + 5]}{{(x^{2} + 3 x + 5)}^{2}}$

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

$\frac{(x^{2} + 3 x + 5) (1 + \frac{d}{d x} [- 1]) - (x - 1) \frac{d}{d x} [x^{2} + 3 x + 5]}{{(x^{2} + 3 x + 5)}^{2}}$

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

$\frac{(x^{2} + 3 x + 5) (1 + 0) - (x - 1) \frac{d}{d x} [x^{2} + 3 x + 5]}{{(x^{2} + 3 x + 5)}^{2}}$

Simplify the expression.

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Add $1$ and $0$ .

$\frac{(x^{2} + 3 x + 5) \cdot 1 - (x - 1) \frac{d}{d x} [x^{2} + 3 x + 5]}{{(x^{2} + 3 x + 5)}^{2}}$

Multiply $x^{2} + 3 x + 5$ by $1$ .

$\frac{x^{2} + 3 x + 5 - (x - 1) \frac{d}{d x} [x^{2} + 3 x + 5]}{{(x^{2} + 3 x + 5)}^{2}}$

By the Sum Rule, the derivative of $x^{2} + 3 x + 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [5]$ .

$\frac{x^{2} + 3 x + 5 - (x - 1) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [5])}{{(x^{2} + 3 x + 5)}^{2}}$

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

$\frac{x^{2} + 3 x + 5 - (x - 1) (2 x + \frac{d}{d x} [3 x] + \frac{d}{d x} [5])}{{(x^{2} + 3 x + 5)}^{2}}$

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$\frac{x^{2} + 3 x + 5 - (x - 1) (2 x + 3 \frac{d}{d x} [x] + \frac{d}{d x} [5])}{{(x^{2} + 3 x + 5)}^{2}}$

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

$\frac{x^{2} + 3 x + 5 - (x - 1) (2 x + 3 \cdot 1 + \frac{d}{d x} [5])}{{(x^{2} + 3 x + 5)}^{2}}$

Multiply $3$ by $1$ .

$\frac{x^{2} + 3 x + 5 - (x - 1) (2 x + 3 + \frac{d}{d x} [5])}{{(x^{2} + 3 x + 5)}^{2}}$

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

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

Add $2 x + 3$ and $0$ .

$\frac{x^{2} + 3 x + 5 - (x - 1) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{2}}$

Simplify.

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Apply the distributive property.

$\frac{x^{2} + 3 x + 5 + (- x - - 1) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{2}}$

Simplify the numerator.

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Simplify each term.

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Multiply $- 1$ by $- 1$ .

$\frac{x^{2} + 3 x + 5 + (- x + 1) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{2}}$

Expand $(- x + 1) (2 x + 3)$ using the FOIL Method.

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Apply the distributive property.

$\frac{x^{2} + 3 x + 5 - x (2 x + 3) + 1 (2 x + 3)}{{(x^{2} + 3 x + 5)}^{2}}$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

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

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $2$ .

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

Multiply $3$ by $- 1$ .

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

Multiply $2 x$ by $1$ .

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

Multiply $3$ by $1$ .

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

Add $- 3 x$ and $2 x$ .

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

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

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

Subtract $x$ from $3 x$ .

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

Add $5$ and $3$ .

$\frac{- x^{2} + 2 x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 8 = - 8$ and whose sum is $b = 2$ .

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Factor $2$ out of $2 x$ .

$\frac{- x^{2} + 2 (x) + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Rewrite $2$ as $- 2$ plus $4$

$\frac{- x^{2} + (- 2 + 4) x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Apply the distributive property.

$\frac{- x^{2} - 2 x + 4 x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$\frac{(- x^{2} - 2 x) + 4 x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x - 2$ .

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

Factor $- 1$ out of $- x$ .

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

Rewrite $- 2$ as $- 1 (2)$ .

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

Factor $- 1$ out of $- (x) - 1 (2)$ .

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

Rewrite $- (x + 2)$ as $- 1 (x + 2)$ .

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

Move the negative in front of the fraction.

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

Find the second derivative of the function.

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{(x + 2) (x - 4)}{{(x^{2} + 3 x + 5)}^{2}}$ .

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

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

Multiply the exponents in ${({(x^{2} + 3 x + 5)}^{2})}^{2}$ .

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

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

Multiply $2$ by $2$ .

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

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 2$ and $g (x) = x - 4$ .

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

Differentiate.

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By the Sum Rule, the derivative of $x - 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4]$ .

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

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

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

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

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

Simplify the expression.

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Add $1$ and $0$ .

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

Multiply $x + 2$ by $1$ .

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

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

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

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

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

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Simplify by adding terms.

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Add $1$ and $0$ .

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

Multiply $x - 4$ by $1$ .

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

Add $x$ and $x$ .

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

Subtract $4$ from $2$ .

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

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 $g (x) = x^{2} + 3 x + 5$ .

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To apply the Chain Rule, set $u$ as $x^{2} + 3 x + 5$ .

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

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

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

Replace all occurrences of $u$ with $x^{2} + 3 x + 5$ .

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

Simplify with factoring out.

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Multiply $2$ by $- 1$ .

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

Factor $x^{2} + 3 x + 5$ out of ${(x^{2} + 3 x + 5)}^{2} (2 x - 2) - 2 (x + 2) (x - 4) ((x^{2} + 3 x + 5) \frac{d}{d x} [x^{2} + 3 x + 5])$ .

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

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

Factor $x^{2} + 3 x + 5$ out of $- 2 (x + 2) (x - 4) ((x^{2} + 3 x + 5) \frac{d}{d x} [x^{2} + 3 x + 5])$ .

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

Factor $x^{2} + 3 x + 5$ out of $(x^{2} + 3 x + 5) ((x^{2} + 3 x + 5) (2 x - 2)) + (x^{2} + 3 x + 5) (- 2 (x + 2) (x - 4) (\frac{d}{d x} [x^{2} + 3 x + 5]))$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

Rewrite the expression.

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

By the Sum Rule, the derivative of $x^{2} + 3 x + 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [5]$ .

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

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

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

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

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

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

Multiply $3$ by $1$ .

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

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

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

Add $2 x + 3$ and $0$ .

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

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

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

Simplify the expression.

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

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

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

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

Simplify.

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Apply the distributive property.

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

Simplify the numerator.

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Simplify each term.

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Expand $(x^{2} + 3 x + 5) (2 x - 2)$ by multiplying each term in the first expression by each term in the second expression.

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

Simplify each term.

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Rewrite using the commutative property of multiplication.

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

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

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Move $x$ .

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

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

Rewrite using the commutative property of multiplication.

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

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $3$ by $2$ .

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

Multiply $- 2$ by $3$ .

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

Multiply $2$ by $5$ .

$h'' (x) = - \frac{2 x^{3} - 2 x^{2} + 6 x^{2} - 6 x + 10 x + 5 \cdot - 2 + (- 2 x - 2 \cdot 2) (x - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $5$ by $- 2$ .

$h'' (x) = - \frac{2 x^{3} - 2 x^{2} + 6 x^{2} - 6 x + 10 x - 10 + (- 2 x - 2 \cdot 2) (x - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Add $- 2 x^{2}$ and $6 x^{2}$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} - 6 x + 10 x - 10 + (- 2 x - 2 \cdot 2) (x - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Add $- 6 x$ and $10 x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x - 2 \cdot 2) (x - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $- 2$ by $2$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x - 4) (x - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Expand $(- 2 x - 4) (x - 4)$ using the FOIL Method.

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Apply the distributive property.

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x (x - 4) - 4 (x - 4)) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Apply the distributive property.

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x \cdot x - 2 x \cdot - 4 - 4 (x - 4)) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Apply the distributive property.

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x \cdot x - 2 x \cdot - 4 - 4 x - 4 \cdot - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Simplify and combine like terms.

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Simplify each term.

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Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 (x \cdot x) - 2 x \cdot - 4 - 4 x - 4 \cdot - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $x$ by $x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x^{2} - 2 x \cdot - 4 - 4 x - 4 \cdot - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $- 4$ by $- 2$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x^{2} + 8 x - 4 x - 4 \cdot - 4) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $- 4$ by $- 4$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x^{2} + 8 x - 4 x + 16) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Subtract $4 x$ from $8 x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 + (- 2 x^{2} + 4 x + 16) (2 x + 3)}{{(x^{2} + 3 x + 5)}^{3}}$

Expand $(- 2 x^{2} + 4 x + 16) (2 x + 3)$ by multiplying each term in the first expression by each term in the second expression.

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 2 x^{2} (2 x) - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Simplify each term.

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Rewrite using the commutative property of multiplication.

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 2 \cdot (2 x^{2} x) - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

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

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Move $x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 2 \cdot (2 (x \cdot x^{2})) - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

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

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

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 2 \cdot (2 (x \cdot x^{2})) - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

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

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 2 \cdot (2 x^{1 + 2}) - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Add $1$ and $2$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 2 \cdot (2 x^{3}) - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $- 2$ by $2$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 2 x^{2} \cdot 3 + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $3$ by $- 2$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 4 x (2 x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Rewrite using the commutative property of multiplication.

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 4 \cdot (2 x \cdot x) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 4 \cdot (2 (x \cdot x)) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $x$ by $x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 4 \cdot (2 x^{2}) + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $4$ by $2$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 8 x^{2} + 4 x \cdot 3 + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $3$ by $4$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 8 x^{2} + 12 x + 16 (2 x) + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $2$ by $16$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 8 x^{2} + 12 x + 32 x + 16 \cdot 3}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $16$ by $3$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} - 6 x^{2} + 8 x^{2} + 12 x + 32 x + 48}{{(x^{2} + 3 x + 5)}^{3}}$

Add $- 6 x^{2}$ and $8 x^{2}$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} + 2 x^{2} + 12 x + 32 x + 48}{{(x^{2} + 3 x + 5)}^{3}}$

Add $12 x$ and $32 x$ .

$h'' (x) = - \frac{2 x^{3} + 4 x^{2} + 4 x - 10 - 4 x^{3} + 2 x^{2} + 44 x + 48}{{(x^{2} + 3 x + 5)}^{3}}$

Subtract $4 x^{3}$ from $2 x^{3}$ .

$h'' (x) = - \frac{- 2 x^{3} + 4 x^{2} + 4 x - 10 + 2 x^{2} + 44 x + 48}{{(x^{2} + 3 x + 5)}^{3}}$

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

$h'' (x) = - \frac{- 2 x^{3} + 6 x^{2} + 4 x - 10 + 44 x + 48}{{(x^{2} + 3 x + 5)}^{3}}$

Add $4 x$ and $44 x$ .

$h'' (x) = - \frac{- 2 x^{3} + 6 x^{2} + 48 x - 10 + 48}{{(x^{2} + 3 x + 5)}^{3}}$

Add $- 10$ and $48$ .

$h'' (x) = - \frac{- 2 x^{3} + 6 x^{2} + 48 x + 38}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $2$ out of $- 2 x^{3} + 6 x^{2} + 48 x + 38$ .

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

$h'' (x) = - \frac{2 (- x^{3}) + 6 x^{2} + 48 x + 38}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $2$ out of $6 x^{2}$ .

$h'' (x) = - \frac{2 (- x^{3}) + 2 (3 x^{2}) + 48 x + 38}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $2$ out of $48 x$ .

$h'' (x) = - \frac{2 (- x^{3}) + 2 (3 x^{2}) + 2 (24 x) + 38}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $2$ out of $38$ .

$h'' (x) = - \frac{2 (- x^{3}) + 2 (3 x^{2}) + 2 (24 x) + 2 (19)}{{(x^{2} + 3 x + 5)}^{3}}$

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

$h'' (x) = - \frac{2 (- x^{3} + 3 x^{2}) + 2 (24 x) + 2 (19)}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $2$ out of $2 (- x^{3} + 3 x^{2}) + 2 (24 x)$ .

$h'' (x) = - \frac{2 (- x^{3} + 3 x^{2} + 24 x) + 2 (19)}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $2$ out of $2 (- x^{3} + 3 x^{2} + 24 x) + 2 (19)$ .

$h'' (x) = - \frac{2 (- x^{3} + 3 x^{2} + 24 x + 19)}{{(x^{2} + 3 x + 5)}^{3}}$

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

$h'' (x) = - \frac{2 (- (x^{3}) + 3 x^{2} + 24 x + 19)}{{(x^{2} + 3 x + 5)}^{3}}$

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

$h'' (x) = - \frac{2 (- (x^{3}) - (- 3 x^{2}) + 24 x + 19)}{{(x^{2} + 3 x + 5)}^{3}}$

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

$h'' (x) = - \frac{2 (- (x^{3} - 3 x^{2}) + 24 x + 19)}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $- 1$ out of $24 x$ .

$h'' (x) = - \frac{2 (- (x^{3} - 3 x^{2}) - (- 24 x) + 19)}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $- 1$ out of $- (x^{3} - 3 x^{2}) - (- 24 x)$ .

$h'' (x) = - \frac{2 (- (x^{3} - 3 x^{2} - 24 x) + 19)}{{(x^{2} + 3 x + 5)}^{3}}$

Rewrite $19$ as $- 1 (- 19)$ .

$h'' (x) = - \frac{2 (- (x^{3} - 3 x^{2} - 24 x) - 1 \cdot - 19)}{{(x^{2} + 3 x + 5)}^{3}}$

Factor $- 1$ out of $- (x^{3} - 3 x^{2} - 24 x) - 1 (- 19)$ .

$h'' (x) = - \frac{2 (- (x^{3} - 3 x^{2} - 24 x - 19))}{{(x^{2} + 3 x + 5)}^{3}}$

Rewrite $- (x^{3} - 3 x^{2} - 24 x - 19)$ as $- 1 (x^{3} - 3 x^{2} - 24 x - 19)$ .

$h'' (x) = - \frac{2 (- 1 (x^{3} - 3 x^{2} - 24 x - 19))}{{(x^{2} + 3 x + 5)}^{3}}$

Move the negative in front of the fraction.

$h'' (x) = \frac{2 (x^{3} - 3 x^{2} - 24 x - 19)}{{(x^{2} + 3 x + 5)}^{3}}$

Multiply $- 1$ by $- 1$ .

$h'' (x) = 1 (\frac{2 (x^{3} - 3 x^{2} - 24 x - 19)}{{(x^{2} + 3 x + 5)}^{3}})$

Multiply $\frac{2 (x^{3} - 3 x^{2} - 24 x - 19)}{{(x^{2} + 3 x + 5)}^{3}}$ by $1$ .

$h'' (x) = \frac{2 (x^{3} - 3 x^{2} - 24 x - 19)}{{(x^{2} + 3 x + 5)}^{3}}$

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

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

Find the first derivative.

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Find the first derivative.

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

Differentiate.

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

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

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

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

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

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

Simplify the expression.

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Add $1$ and $0$ .

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

Multiply $x^{2} + 3 x + 5$ by $1$ .

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

By the Sum Rule, the derivative of $x^{2} + 3 x + 5$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x] + \frac{d}{d x} [5]$ .

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

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

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

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

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

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

Multiply $3$ by $1$ .

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

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

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

Add $2 x + 3$ and $0$ .

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

Simplify.

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Apply the distributive property.

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

Simplify the numerator.

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Simplify each term.

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Multiply $- 1$ by $- 1$ .

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

Expand $(- x + 1) (2 x + 3)$ using the FOIL Method.

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

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

Multiply $x$ by $x$ by adding the exponents.

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Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $2$ .

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

Multiply $3$ by $- 1$ .

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

Multiply $2 x$ by $1$ .

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

Multiply $3$ by $1$ .

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

Add $- 3 x$ and $2 x$ .

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

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

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

Subtract $x$ from $3 x$ .

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

Add $5$ and $3$ .

$f' (x) = \frac{- x^{2} + 2 x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 8 = - 8$ and whose sum is $b = 2$ .

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Factor $2$ out of $2 x$ .

$f' (x) = \frac{- x^{2} + 2 (x) + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Rewrite $2$ as $- 2$ plus $4$

$f' (x) = \frac{- x^{2} + (- 2 + 4) x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Apply the distributive property.

$f' (x) = \frac{- x^{2} - 2 x + 4 x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$f' (x) = \frac{(- x^{2} - 2 x) + 4 x + 8}{{(x^{2} + 3 x + 5)}^{2}}$

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- x - 2$ .

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

Factor $- 1$ out of $- x$ .

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

Rewrite $- 2$ as $- 1 (2)$ .

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

Factor $- 1$ out of $- (x) - 1 (2)$ .

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

Rewrite $- (x + 2)$ as $- 1 (x + 2)$ .

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

Move the negative in front of the fraction.

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

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

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

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

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Set the first derivative equal to $0$ .

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

Set the numerator equal to zero.

$(x + 2) (x - 4) = 0$

Solve the equation for $x$ .

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

$x + 2 = 0$

$x - 4 = 0$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

$x = - 2, 4$

Find the values where the derivative is undefined.

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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.

Critical points to evaluate.

$x = - 2, 4$

Evaluate the second derivative at $x = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{2 ({(- 2)}^{3} - 3 {(- 2)}^{2} - 24 \cdot - 2 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Evaluate the second derivative.

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Simplify the numerator.

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

$\frac{2 (- 8 - 3 {(- 2)}^{2} - 24 \cdot - 2 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

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

$\frac{2 (- 8 - 3 \cdot 4 - 24 \cdot - 2 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Multiply $- 3$ by $4$ .

$\frac{2 (- 8 - 12 - 24 \cdot - 2 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Multiply $- 24$ by $- 2$ .

$\frac{2 (- 8 - 12 + 48 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Subtract $12$ from $- 8$ .

$\frac{2 (- 20 + 48 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Add $- 20$ and $48$ .

$\frac{2 (28 - 19)}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Subtract $19$ from $28$ .

$\frac{2 \cdot 9}{{({(- 2)}^{2} + 3 (- 2) + 5)}^{3}}$

Simplify the denominator.

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

$\frac{2 \cdot 9}{{(4 + 3 (- 2) + 5)}^{3}}$

Multiply $3$ by $- 2$ .

$\frac{2 \cdot 9}{{(4 - 6 + 5)}^{3}}$

Subtract $6$ from $4$ .

$\frac{2 \cdot 9}{{(- 2 + 5)}^{3}}$

Add $- 2$ and $5$ .

$\frac{2 \cdot 9}{3^{3}}$

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

$\frac{2 \cdot 9}{27}$

Reduce the expression by cancelling the common factors.

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Multiply $2$ by $9$ .

$\frac{18}{27}$

Cancel the common factor of $18$ and $27$ .

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Factor $9$ out of $18$ .

$\frac{9 (2)}{27}$

Cancel the common factors.

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Factor $9$ out of $27$ .

$\frac{9 \cdot 2}{9 \cdot 3}$

Cancel the common factor.

$\frac{9 \cdot 2}{9 \cdot 3}$

Rewrite the expression.

$\frac{2}{3}$

$x = - 2$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 2$ is a local minimum

Find the y-value when $x = - 2$ .

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

$h (- 2) = \frac{(- 2) - 1}{{(- 2)}^{2} + 3 (- 2) + 5}$

Simplify the result.

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Subtract $1$ from $- 2$ .

$h (- 2) = \frac{- 3}{{(- 2)}^{2} + 3 (- 2) + 5}$

Simplify the denominator.

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

$h (- 2) = \frac{- 3}{4 + 3 (- 2) + 5}$

Multiply $3$ by $- 2$ .

$h (- 2) = \frac{- 3}{4 - 6 + 5}$

Subtract $6$ from $4$ .

$h (- 2) = \frac{- 3}{- 2 + 5}$

Add $- 2$ and $5$ .

$h (- 2) = \frac{- 3}{3}$

Divide $- 3$ by $3$ .

$h (- 2) = - 1$

The final answer is $- 1$ .

$y = - 1$

Evaluate the second derivative at $x = 4$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{2 ({(4)}^{3} - 3 {(4)}^{2} - 24 \cdot 4 - 19)}{{({(4)}^{2} + 3 (4) + 5)}^{3}}$

Evaluate the second derivative.

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Simplify the numerator.

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Raise $4$ to the power of $3$ .

$\frac{2 (64 - 3 \cdot 4^{2} - 24 \cdot 4 - 19)}{{(4^{2} + 3 (4) + 5)}^{3}}$

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

$\frac{2 (64 - 3 \cdot 16 - 24 \cdot 4 - 19)}{{(4^{2} + 3 (4) + 5)}^{3}}$

Multiply $- 3$ by $16$ .

$\frac{2 (64 - 48 - 24 \cdot 4 - 19)}{{(4^{2} + 3 (4) + 5)}^{3}}$

Multiply $- 24$ by $4$ .

$\frac{2 (64 - 48 - 96 - 19)}{{(4^{2} + 3 (4) + 5)}^{3}}$

Subtract $48$ from $64$ .

$\frac{2 (16 - 96 - 19)}{{(4^{2} + 3 (4) + 5)}^{3}}$

Subtract $96$ from $16$ .

$\frac{2 (- 80 - 19)}{{(4^{2} + 3 (4) + 5)}^{3}}$

Subtract $19$ from $- 80$ .

$\frac{2 \cdot - 99}{{(4^{2} + 3 (4) + 5)}^{3}}$

Simplify the denominator.

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

$\frac{2 \cdot - 99}{{(16 + 3 (4) + 5)}^{3}}$

Multiply $3$ by $4$ .

$\frac{2 \cdot - 99}{{(16 + 12 + 5)}^{3}}$

Add $16$ and $12$ .

$\frac{2 \cdot - 99}{{(28 + 5)}^{3}}$

Add $28$ and $5$ .

$\frac{2 \cdot - 99}{33^{3}}$

Raise $33$ to the power of $3$ .

$\frac{2 \cdot - 99}{35937}$

Reduce the expression by cancelling the common factors.

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Multiply $2$ by $- 99$ .

$\frac{- 198}{35937}$

Cancel the common factor of $- 198$ and $35937$ .

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Factor $99$ out of $- 198$ .

$\frac{99 (- 2)}{35937}$

Cancel the common factors.

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Factor $99$ out of $35937$ .

$\frac{99 \cdot - 2}{99 \cdot 363}$

Cancel the common factor.

$\frac{99 \cdot - 2}{99 \cdot 363}$

Rewrite the expression.

$\frac{- 2}{363}$

Move the negative in front of the fraction.

$- \frac{2}{363}$

$x = 4$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 4$ is a local maximum

Find the y-value when $x = 4$ .

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

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

Simplify the result.

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Subtract $1$ from $4$ .

$h (4) = \frac{3}{4^{2} + 3 (4) + 5}$

Simplify the denominator.

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

$h (4) = \frac{3}{16 + 3 (4) + 5}$

Multiply $3$ by $4$ .

$h (4) = \frac{3}{16 + 12 + 5}$

Add $16$ and $12$ .

$h (4) = \frac{3}{28 + 5}$

Add $28$ and $5$ .

$h (4) = \frac{3}{33}$

Cancel the common factor of $3$ and $33$ .

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Factor $3$ out of $3$ .

$h (4) = \frac{3 (1)}{33}$

Cancel the common factors.

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Factor $3$ out of $33$ .

$h (4) = \frac{3 \cdot 1}{3 \cdot 11}$

Cancel the common factor.

$h (4) = \frac{3 \cdot 1}{3 \cdot 11}$

Rewrite the expression.

$h (4) = \frac{1}{11}$

The final answer is $\frac{1}{11}$ .

$y = \frac{1}{11}$

These are the local extrema for $h (x) = \frac{x - 1}{x^{2} + 3 x + 5}$ .

$(- 2, - 1)$ is a local minima

$(4, \frac{1}{11})$ is a local maxima