Calculus Examples

$y = | x^{2} - 25 |$

Step 1

Write $y = | x^{2} - 25 |$ as a function.

$f (x) = | x^{2} - 25 |$

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x^{2} - 25$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $x^{2} - 25$ .

$\frac{d}{d u} [| u |] \frac{d}{d x} [x^{2} - 25]$

Step 2.1.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$\frac{u}{| u |} \frac{d}{d x} [x^{2} - 25]$

Step 2.1.3

Replace all occurrences of $u$ with $x^{2} - 25$ .

$\frac{x^{2} - 25}{| x^{2} - 25 |} \frac{d}{d x} [x^{2} - 25]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $x^{2} - 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]$ .

$\frac{x^{2} - 25}{| x^{2} - 25 |} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25])$

Step 2.2.2

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

$\frac{x^{2} - 25}{| x^{2} - 25 |} (2 x + \frac{d}{d x} [- 25])$

Step 2.2.3

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25$ with respect to $x$ is $0$ .

$\frac{x^{2} - 25}{| x^{2} - 25 |} (2 x + 0)$

Step 2.2.4

Combine fractions.

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Step 2.2.4.1

Add $2 x$ and $0$ .

$\frac{x^{2} - 25}{| x^{2} - 25 |} (2 x)$

Step 2.2.4.2

Combine $2$ and $\frac{x^{2} - 25}{| x^{2} - 25 |}$ .

$\frac{2 (x^{2} - 25)}{| x^{2} - 25 |} x$

Step 2.2.4.3

Combine $\frac{2 (x^{2} - 25)}{| x^{2} - 25 |}$ and $x$ .

$\frac{2 (x^{2} - 25) x}{| x^{2} - 25 |}$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$\frac{(2 x^{2} + 2 \cdot - 25) x}{| x^{2} - 25 |}$

Step 2.3.2

Apply the distributive property.

$\frac{2 x^{2} x + 2 \cdot - 25 x}{| x^{2} - 25 |}$

Step 2.3.3

Simplify each term.

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Step 2.3.3.1

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

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Step 2.3.3.1.1

Move $x$ .

$\frac{2 (x \cdot x^{2}) + 2 \cdot - 25 x}{| x^{2} - 25 |}$

Step 2.3.3.1.2

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

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Step 2.3.3.1.2.1

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

$\frac{2 (x^{1} x^{2}) + 2 \cdot - 25 x}{| x^{2} - 25 |}$

Step 2.3.3.1.2.2

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

$\frac{2 x^{1 + 2} + 2 \cdot - 25 x}{| x^{2} - 25 |}$

Step 2.3.3.1.3

Add $1$ and $2$ .

$\frac{2 x^{3} + 2 \cdot - 25 x}{| x^{2} - 25 |}$

Step 2.3.3.2

Multiply $2$ by $- 25$ .

$\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$

Step 3

Find the second derivative of the function.

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Step 3.1

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 $f (x) = 2 x^{3} - 50 x$ and $g (x) = | x^{2} - 25 |$ .

$f'' (x) = \frac{| x^{2} - 25 | \frac{d}{d x} (2 x^{3} - 50 x) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2

Differentiate.

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Step 3.2.1

By the Sum Rule, the derivative of $2 x^{3} - 50 x$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 50 x]$ .

$f'' (x) = \frac{| x^{2} - 25 | (\frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (- 50 x)) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = \frac{| x^{2} - 25 | (2 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 50 x)) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = \frac{| x^{2} - 25 | (2 (3 x^{2}) + \frac{d}{d x} (- 50 x)) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2.4

Multiply $3$ by $2$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} + \frac{d}{d x} (- 50 x)) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2.5

Since $- 50$ is constant with respect to $x$ , the derivative of $- 50 x$ with respect to $x$ is $- 50 \frac{d}{d x} [x]$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50 \frac{d}{d x} x) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2.6

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

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50 \cdot 1) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.2.7

Multiply $- 50$ by $1$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) \frac{d}{d x} | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x^{2} - 25$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $x^{2} - 25$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{d}{d u} (| u |) \frac{d}{d x} (x^{2} - 25))}{{| x^{2} - 25 |}^{2}}$

Step 3.3.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{u}{| u |} \cdot \frac{d}{d x} (x^{2} - 25))}{{| x^{2} - 25 |}^{2}}$

Step 3.3.3

Replace all occurrences of $u$ with $x^{2} - 25$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{x^{2} - 25}{| x^{2} - 25 |} \cdot \frac{d}{d x} (x^{2} - 25))}{{| x^{2} - 25 |}^{2}}$

Step 3.4

Differentiate.

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Step 3.4.1

By the Sum Rule, the derivative of $x^{2} - 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 25)))}{{| x^{2} - 25 |}^{2}}$

Step 3.4.2

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

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (2 x + \frac{d}{d x} (- 25)))}{{| x^{2} - 25 |}^{2}}$

Step 3.4.3

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25$ with respect to $x$ is $0$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (2 x + 0))}{{| x^{2} - 25 |}^{2}}$

Step 3.4.4

Combine fractions.

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Step 3.4.4.1

Add $2 x$ and $0$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (2 x))}{{| x^{2} - 25 |}^{2}}$

Step 3.4.4.2

Combine $2$ and $\frac{x^{2} - 25}{| x^{2} - 25 |}$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) (\frac{2 (x^{2} - 25)}{| x^{2} - 25 |} \cdot x)}{{| x^{2} - 25 |}^{2}}$

Step 3.4.4.3

Combine $\frac{2 (x^{2} - 25)}{| x^{2} - 25 |}$ and $x$ .

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) - (2 x^{3} - 50 x) \frac{2 (x^{2} - 25) x}{| x^{2} - 25 |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5

Simplify.

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Step 3.5.1

Apply the distributive property.

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) + (- (2 x^{3}) - (- 50 x)) (\frac{2 (x^{2} - 25) x}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.2

Apply the distributive property.

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) + (- (2 x^{3}) - (- 50 x)) (\frac{(2 x^{2} + 2 \cdot - 25) x}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.3

Apply the distributive property.

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2} - 50) + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{2} x + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4

Simplify the numerator.

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Step 3.5.4.1

Simplify each term.

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Step 3.5.4.1.1

Apply the distributive property.

$f'' (x) = \frac{| x^{2} - 25 | (6 x^{2}) + | x^{2} - 25 | \cdot - 50 + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{2} x + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.2

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} + | x^{2} - 25 | \cdot - 50 + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{2} x + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.3

Move $- 50$ to the left of $| x^{2} - 25 |$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 \cdot | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{2} x + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4

Simplify the numerator.

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Step 3.5.4.1.4.1

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

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Step 3.5.4.1.4.1.1

Move $x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 (x \cdot x^{2}) + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.1.2

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

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Step 3.5.4.1.4.1.2.1

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 (x \cdot x^{2}) + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.1.2.2

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{1 + 2} + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.1.3

Add $1$ and $2$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{3} + 2 \cdot (- 25 x)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.2

Multiply $2$ by $- 25$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x^{3} - 50 x}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.3

Rewrite $2 x^{3} - 50 x$ in a factored form.

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Step 3.5.4.1.4.3.1

Factor $2 x$ out of $2 x^{3} - 50 x$ .

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Step 3.5.4.1.4.3.1.1

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x^{2}) - 50 x}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.3.1.2

Factor $2 x$ out of $- 50 x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x^{2}) + 2 x (- 25)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.3.1.3

Factor $2 x$ out of $2 x (x^{2}) + 2 x (- 25)$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x^{2} - 25)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.3.2

Rewrite $25$ as $5^{2}$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x^{2} - 5^{2})}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.4.3.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5

Simplify the denominator.

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Step 3.5.4.1.5.1

Rewrite $25$ as $5^{2}$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x^{2} - 5^{2} |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| (x + 5) (x - 5) |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.3

Expand $(x + 5) (x - 5)$ using the FOIL Method.

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Step 3.5.4.1.5.3.1

Apply the distributive property.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x (x - 5) + 5 (x - 5) |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.3.2

Apply the distributive property.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x \cdot x + x \cdot - 5 + 5 (x - 5) |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.3.3

Apply the distributive property.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.4

Combine the opposite terms in $x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5$ .

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Step 3.5.4.1.5.4.1

Reorder the factors in the terms $x \cdot - 5$ and $5 x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x \cdot x - 5 x + 5 x + 5 \cdot - 5 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.4.2

Add $- 5 x$ and $5 x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x \cdot x + 0 + 5 \cdot - 5 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.4.3

Add $x \cdot x$ and $0$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x \cdot x + 5 \cdot - 5 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.5

Simplify each term.

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Step 3.5.4.1.5.5.1

Multiply $x$ by $x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x^{2} + 5 \cdot - 5 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.5.2

Multiply $5$ by $- 5$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x^{2} - 25 |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.6

Rewrite $25$ as $5^{2}$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| x^{2} - 5^{2} |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.5.7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- (2 x^{3}) - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| (x + 5) (x - 5) |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.6

Simplify each term.

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Step 3.5.4.1.6.1

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- 2 x^{3} - (- 50 x)) (\frac{2 x (x + 5) (x - 5)}{| (x + 5) (x - 5) |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.6.2

Multiply $- 50$ by $- 1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + (- 2 x^{3} + 50 x) (\frac{2 x (x + 5) (x - 5)}{| (x + 5) (x - 5) |})}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.7

Multiply $- 2 x^{3} + 50 x$ by $\frac{2 x (x + 5) (x - 5)}{| (x + 5) (x - 5) |}$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{(- 2 x^{3} + 50 x) (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8

Simplify the numerator.

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Step 3.5.4.1.8.1

Factor $2 x$ out of $- 2 x^{3} + 50 x$ .

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Step 3.5.4.1.8.1.1

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{(2 x (- x^{2}) + 50 x) \cdot (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.1.2

Factor $2 x$ out of $50 x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{(2 x (- x^{2}) + 2 x (25)) \cdot (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.1.3

Factor $2 x$ out of $2 x (- x^{2}) + 2 x (25)$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{2 x (- x^{2} + 25) \cdot (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.2

Rewrite $25$ as $5^{2}$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{2 x (- x^{2} + 5^{2}) \cdot (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.3

Reorder $- x^{2}$ and $5^{2}$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{2 x (5^{2} - x^{2}) \cdot (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{2 x (5 + x) (5 - x) \cdot (2 x (x + 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5

Combine exponents.

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Step 3.5.4.1.8.5.1

Multiply $2$ by $2$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x (5 + x) (5 - x) x (x + 5) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.2

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 (x \cdot x) (5 + x) (5 - x) (x + 5) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.3

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 (x \cdot x) (5 + x) (5 - x) (x + 5) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.4

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{1 + 1} (5 + x) (5 - x) (x + 5) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.5

Add $1$ and $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} (5 + x) (5 - x) (x + 5) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.6

Reorder terms.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} (x + 5) (5 - x) (x + 5) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.7

Raise $x + 5$ to the power of $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} ((x + 5) (x + 5)) (5 - x) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.8

Raise $x + 5$ to the power of $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} ((x + 5) (x + 5)) (5 - x) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.9

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{1 + 1} (5 - x) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.10

Add $1$ and $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} (5 - x) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.11

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} (- 1 \cdot - 5 - x) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.12

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} (- 1 \cdot - 5 - (x)) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.13

Factor $- 1$ out of $- 1 (- 5) - (x)$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} (- 1 (- 5 + x)) (x - 5)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.14

Reorder terms.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} \cdot (- 1 (x - 5) (x - 5))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.15

Raise $x - 5$ to the power of $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} \cdot (- 1 ((x - 5) (x - 5)))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.16

Raise $x - 5$ to the power of $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} \cdot (- 1 ((x - 5) (x - 5)))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.17

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

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} \cdot (- 1 {(x - 5)}^{1 + 1})}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.18

Add $1$ and $1$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{4 x^{2} {(x + 5)}^{2} \cdot (- 1 {(x - 5)}^{2})}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.8.5.19

Multiply $- 1$ by $4$ .

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | + \frac{- 4 x^{2} {(x + 5)}^{2} {(x - 5)}^{2}}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.1.9

Move the negative in front of the fraction.

$f'' (x) = \frac{6 | x^{2} - 25 | x^{2} - 50 | x^{2} - 25 | - \frac{4 x^{2} {(x + 5)}^{2} {(x - 5)}^{2}}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.2

To write $6 | x^{2} - 25 | x^{2}$ as a fraction with a common denominator, multiply by $\frac{| (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}$ .

$f'' (x) = \frac{\frac{6 | x^{2} - 25 | x^{2} | (x + 5) (x - 5) |}{| (x + 5) (x - 5) |} - \frac{4 x^{2} {(x + 5)}^{2} {(x - 5)}^{2}}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.3

Combine the numerators over the common denominator.

$f'' (x) = \frac{\frac{6 | x^{2} - 25 | x^{2} | (x + 5) (x - 5) | - 4 x^{2} {(x + 5)}^{2} {(x - 5)}^{2}}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4

Simplify each term.

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Step 3.5.4.4.1

Simplify the numerator.

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Step 3.5.4.4.1.1

Factor $2 x^{2}$ out of $6 | x^{2} - 25 | x^{2} | (x + 5) (x - 5) | - 4 x^{2} {(x + 5)}^{2} {(x - 5)}^{2}$ .

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Step 3.5.4.4.1.1.1

Factor $2 x^{2}$ out of $6 | x^{2} - 25 | x^{2} | (x + 5) (x - 5) |$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2} - 25 | | (x + 5) (x - 5) |) - 4 x^{2} {(x + 5)}^{2} {(x - 5)}^{2}}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.1.2

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

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2} - 25 | | (x + 5) (x - 5) |) + 2 x^{2} (- 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.1.3

Factor $2 x^{2}$ out of $2 x^{2} (3 | x^{2} - 25 | | (x + 5) (x - 5) |) + 2 x^{2} (- 2 {(x + 5)}^{2} {(x - 5)}^{2})$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2} - 25 | | (x + 5) (x - 5) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.2

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x + 5) (x - 5) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3

Simplify each term.

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Step 3.5.4.4.1.3.1

Expand $(x + 5) (x - 5)$ using the FOIL Method.

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Step 3.5.4.4.1.3.1.1

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x (x - 5) + 5 (x - 5)) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.1.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x \cdot x + x \cdot - 5 + 5 (x - 5)) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.1.3

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.2

Combine the opposite terms in $x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5$ .

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Step 3.5.4.4.1.3.2.1

Reorder the factors in the terms $x \cdot - 5$ and $5 x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x \cdot x - 5 x + 5 x + 5 \cdot - 5) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.2.2

Add $- 5 x$ and $5 x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x \cdot x + 0 + 5 \cdot - 5) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.2.3

Add $x \cdot x$ and $0$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x \cdot x + 5 \cdot - 5) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.3

Simplify each term.

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Step 3.5.4.4.1.3.3.1

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x^{2} + 5 \cdot - 5) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.3.2

Multiply $5$ by $- 5$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | (x^{2} - 25) (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.4

Expand $(x^{2} - 25) (x^{2} - 25)$ using the FOIL Method.

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Step 3.5.4.4.1.3.4.1

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2} (x^{2} - 25) - 25 (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.4.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2} x^{2} + x^{2} \cdot - 25 - 25 (x^{2} - 25) | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.4.3

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2} x^{2} + x^{2} \cdot - 25 - 25 x^{2} - 25 \cdot - 25 | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.5

Simplify and combine like terms.

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Step 3.5.4.4.1.3.5.1

Simplify each term.

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Step 3.5.4.4.1.3.5.1.1

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

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Step 3.5.4.4.1.3.5.1.1.1

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

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{2 + 2} + x^{2} \cdot - 25 - 25 x^{2} - 25 \cdot - 25 | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.5.1.1.2

Add $2$ and $2$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} + x^{2} \cdot - 25 - 25 x^{2} - 25 \cdot - 25 | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.5.1.2

Move $- 25$ to the left of $x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 25 \cdot x^{2} - 25 x^{2} - 25 \cdot - 25 | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.5.1.3

Multiply $- 25$ by $- 25$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 25 x^{2} - 25 x^{2} + 625 | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.5.2

Subtract $25 x^{2}$ from $- 25 x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 {(x + 5)}^{2} {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.6

Rewrite ${(x + 5)}^{2}$ as $(x + 5) (x + 5)$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 ((x + 5) (x + 5)) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.7

Expand $(x + 5) (x + 5)$ using the FOIL Method.

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Step 3.5.4.4.1.3.7.1

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x (x + 5) + 5 (x + 5)) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.7.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x \cdot x + x \cdot 5 + 5 (x + 5)) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.7.3

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x \cdot x + x \cdot 5 + 5 x + 5 \cdot 5) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.8

Simplify and combine like terms.

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Step 3.5.4.4.1.3.8.1

Simplify each term.

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Step 3.5.4.4.1.3.8.1.1

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x^{2} + x \cdot 5 + 5 x + 5 \cdot 5) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.8.1.2

Move $5$ to the left of $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x^{2} + 5 \cdot x + 5 x + 5 \cdot 5) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.8.1.3

Multiply $5$ by $5$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x^{2} + 5 x + 5 x + 25) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.8.2

Add $5 x$ and $5 x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x^{2} + 10 x + 25) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.9

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 2 (10 x) - 2 \cdot 25) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.10

Simplify.

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Step 3.5.4.4.1.3.10.1

Multiply $10$ by $- 2$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 2 \cdot 25) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.10.2

Multiply $- 2$ by $25$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) {(x - 5)}^{2})}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.11

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

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) ((x - 5) (x - 5)))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.12

Expand $(x - 5) (x - 5)$ using the FOIL Method.

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Step 3.5.4.4.1.3.12.1

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x (x - 5) - 5 (x - 5)))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.12.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x \cdot x + x \cdot - 5 - 5 (x - 5)))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.12.3

Apply the distributive property.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.13

Simplify and combine like terms.

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Step 3.5.4.4.1.3.13.1

Simplify each term.

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Step 3.5.4.4.1.3.13.1.1

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x^{2} + x \cdot - 5 - 5 x - 5 \cdot - 5))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.13.1.2

Move $- 5$ to the left of $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x^{2} - 5 \cdot x - 5 x - 5 \cdot - 5))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.13.1.3

Multiply $- 5$ by $- 5$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x^{2} - 5 x - 5 x + 25))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.13.2

Subtract $5 x$ from $- 5 x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | + (- 2 x^{2} - 20 x - 50) (x^{2} - 10 x + 25))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.14

Expand $(- 2 x^{2} - 20 x - 50) (x^{2} - 10 x + 25)$ by multiplying each term in the first expression by each term in the second expression.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{2} x^{2} - 2 x^{2} (- 10 x) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15

Simplify each term.

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Step 3.5.4.4.1.3.15.1

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

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Step 3.5.4.4.1.3.15.1.1

Move $x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 (x^{2} x^{2}) - 2 x^{2} (- 10 x) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.1.2

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

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{2 + 2} - 2 x^{2} (- 10 x) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.1.3

Add $2$ and $2$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 2 x^{2} (- 10 x) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.2

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 2 \cdot (- 10 x^{2} x) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.3

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

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Step 3.5.4.4.1.3.15.3.1

Move $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 2 \cdot (- 10 (x \cdot x^{2})) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.3.2

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

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Step 3.5.4.4.1.3.15.3.2.1

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

Step 3.5.4.4.1.3.15.3.2.2

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

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 2 \cdot (- 10 x^{1 + 2}) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.3.3

Add $1$ and $2$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 2 \cdot (- 10 x^{3}) - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.4

Multiply $- 2$ by $- 10$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 2 x^{2} \cdot 25 - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.5

Multiply $25$ by $- 2$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x \cdot x^{2} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.6

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

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Step 3.5.4.4.1.3.15.6.1

Move $x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 (x^{2} x) - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.6.2

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

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Step 3.5.4.4.1.3.15.6.2.1

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

Step 3.5.4.4.1.3.15.6.2.2

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

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{2 + 1} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.6.3

Add $2$ and $1$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} - 20 x (- 10 x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.7

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} - 20 \cdot (- 10 x \cdot x) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.8

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

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Step 3.5.4.4.1.3.15.8.1

Move $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} - 20 \cdot (- 10 (x \cdot x)) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.8.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} - 20 \cdot (- 10 x^{2}) - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.9

Multiply $- 20$ by $- 10$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} + 200 x^{2} - 20 x \cdot 25 - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.10

Multiply $25$ by $- 20$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} + 200 x^{2} - 500 x - 50 x^{2} - 50 (- 10 x) - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.11

Multiply $- 10$ by $- 50$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} + 200 x^{2} - 500 x - 50 x^{2} + 500 x - 50 \cdot 25)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.15.12

Multiply $- 50$ by $25$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} + 200 x^{2} - 500 x - 50 x^{2} + 500 x - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.16

Combine the opposite terms in $- 2 x^{4} + 20 x^{3} - 50 x^{2} - 20 x^{3} + 200 x^{2} - 500 x - 50 x^{2} + 500 x - 1250$ .

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Step 3.5.4.4.1.3.16.1

Subtract $20 x^{3}$ from $20 x^{3}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 50 x^{2} + 0 + 200 x^{2} - 500 x - 50 x^{2} + 500 x - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.16.2

Add $- 2 x^{4} - 50 x^{2}$ and $0$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 50 x^{2} + 200 x^{2} - 500 x - 50 x^{2} + 500 x - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.16.3

Add $- 500 x$ and $500 x$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 50 x^{2} + 200 x^{2} - 50 x^{2} + 0 - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.16.4

Add $- 2 x^{4} - 50 x^{2} + 200 x^{2} - 50 x^{2}$ and $0$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} - 50 x^{2} + 200 x^{2} - 50 x^{2} - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.17

Add $- 50 x^{2}$ and $200 x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 150 x^{2} - 50 x^{2} - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.3.18

Subtract $50 x^{2}$ from $150 x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (3 | x^{4} - 50 x^{2} + 625 | - 2 x^{4} + 100 x^{2} - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.4

Reorder terms.

$f'' (x) = \frac{\frac{2 x^{2} (- 2 x^{4} + 100 x^{2} + 3 | x^{4} - 50 x^{2} + 625 | - 1250)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5

Rewrite $- 2 x^{4} + 100 x^{2} + 3 | x^{4} - 50 x^{2} + 625 | - 1250$ in a factored form.

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Step 3.5.4.4.1.5.1

Regroup terms.

$f'' (x) = \frac{\frac{2 x^{2} (- 1250 - 2 x^{4} + 100 x^{2} + 3 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.2

Factor $- 1$ out of $- 2 x^{4} + 100 x^{2} + 3 | x^{4} - 50 x^{2} + 625 |$ .

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Step 3.5.4.4.1.5.2.1

Factor $- 1$ out of $- 2 x^{4}$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1250 - (2 x^{4}) + 100 x^{2} + 3 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.2.2

Factor $- 1$ out of $100 x^{2}$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1250 - (2 x^{4}) - (- 100 x^{2}) + 3 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.2.3

Factor $- 1$ out of $3 | x^{4} - 50 x^{2} + 625 |$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1250 - (2 x^{4}) - (- 100 x^{2}) - (- 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.2.4

Factor $- 1$ out of $- (2 x^{4}) - (- 100 x^{2})$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1250 - (2 x^{4} - 100 x^{2}) - (- 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.2.5

Factor $- 1$ out of $- (2 x^{4} - 100 x^{2}) - (- 3 | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1250 - (2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.3

Factor $- 1$ out of $- 1250 - (2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |)$ .

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Step 3.5.4.4.1.5.3.1

Rewrite $- 1250$ as $- 1 (1250)$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1 \cdot 1250 - (2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.3.2

Factor $- 1$ out of $- 1 (1250) - (2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 x^{2} (- 1 (1250 + 2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.3.3

Rewrite $- 1 (1250 + 2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |)$ as $- (1250 + 2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 x^{2} (- (1250 + 2 x^{4} - 100 x^{2} - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.5.4

Reorder terms.

$f'' (x) = \frac{\frac{2 x^{2} (- (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

$f'' (x) = \frac{\frac{2 x^{2} \cdot (- 1 (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.6

Combine exponents.

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Step 3.5.4.4.1.6.1

Factor out negative.

$f'' (x) = \frac{\frac{- (2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.1.6.2

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{\frac{- 2 (x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.4.2

Move the negative in front of the fraction.

$f'' (x) = \frac{- \frac{2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 |}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.5

To write $- 50 | x^{2} - 25 |$ as a fraction with a common denominator, multiply by $\frac{| (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}$ .

$f'' (x) = \frac{- \frac{2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} - 50 | x^{2} - 25 | \cdot \frac{| (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.6

Combine $- 50 | x^{2} - 25 |$ and $\frac{| (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}$ .

$f'' (x) = \frac{- \frac{2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} + \frac{- 50 | x^{2} - 25 | | (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.7

Combine the numerators over the common denominator.

$f'' (x) = \frac{\frac{- 2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |) - 50 | x^{2} - 25 | | (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8

Simplify the numerator.

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Step 3.5.4.8.1

Factor $2$ out of $- 2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |) - 50 | x^{2} - 25 | | (x + 5) (x - 5) |$ .

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Step 3.5.4.8.1.1

Factor $2$ out of $- 2 x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 (- x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)) - 50 | x^{2} - 25 | | (x + 5) (x - 5) |}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.1.2

Factor $2$ out of $- 50 | x^{2} - 25 | | (x + 5) (x - 5) |$ .

$f'' (x) = \frac{\frac{2 (- x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)) + 2 (- 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.1.3

Factor $2$ out of $2 (- x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |)) + 2 (- 25 | x^{2} - 25 | | (x + 5) (x - 5) |)$ .

$f'' (x) = \frac{\frac{2 (- x^{2} (2 x^{4} - 100 x^{2} + 1250 - 3 | x^{4} - 50 x^{2} + 625 |) - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- x^{2} (2 x^{4}) - x^{2} (- 100 x^{2}) - x^{2} \cdot 1250 - x^{2} (- 3 | x^{4} - 50 x^{2} + 625 |) - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.3

Simplify.

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Step 3.5.4.8.3.1

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 x^{2} x^{4}) - x^{2} (- 100 x^{2}) - x^{2} \cdot 1250 - x^{2} (- 3 | x^{4} - 50 x^{2} + 625 |) - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.3.2

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 x^{2} x^{4}) - 1 \cdot (- 100 x^{2} x^{2}) - x^{2} \cdot 1250 - x^{2} (- 3 | x^{4} - 50 x^{2} + 625 |) - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.3.3

Multiply $1250$ by $- 1$ .

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 x^{2} x^{4}) - 1 \cdot (- 100 x^{2} x^{2}) - 1250 x^{2} - x^{2} (- 3 | x^{4} - 50 x^{2} + 625 |) - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.3.4

Multiply $- 3$ by $- 1$ .

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 x^{2} x^{4}) - 1 \cdot (- 100 x^{2} x^{2}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4

Simplify each term.

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Step 3.5.4.8.4.1

Multiply $x^{2}$ by $x^{4}$ by adding the exponents.

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Step 3.5.4.8.4.1.1

Move $x^{4}$ .

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 (x^{4} x^{2})) - 1 \cdot (- 100 x^{2} x^{2}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.1.2

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

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 x^{4 + 2}) - 1 \cdot (- 100 x^{2} x^{2}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.1.3

Add $4$ and $2$ .

$f'' (x) = \frac{\frac{2 (- 1 \cdot (2 x^{6}) - 1 \cdot (- 100 x^{2} x^{2}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.2

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} - 1 \cdot (- 100 x^{2} x^{2}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.3

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

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Step 3.5.4.8.4.3.1

Move $x^{2}$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} - 1 \cdot (- 100 (x^{2} x^{2})) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.3.2

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

$f'' (x) = \frac{\frac{2 (- 2 x^{6} - 1 \cdot (- 100 x^{2 + 2}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.3.3

Add $2$ and $2$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} - 1 \cdot (- 100 x^{4}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.4.4

Multiply $- 1$ by $- 100$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | (x + 5) (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.5

Expand $(x + 5) (x - 5)$ using the FOIL Method.

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Step 3.5.4.8.5.1

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x (x - 5) + 5 (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.5.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x \cdot x + x \cdot - 5 + 5 (x - 5) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.5.3

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.6

Combine the opposite terms in $x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5$ .

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Step 3.5.4.8.6.1

Reorder the factors in the terms $x \cdot - 5$ and $5 x$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x \cdot x - 5 x + 5 x + 5 \cdot - 5 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.6.2

Add $- 5 x$ and $5 x$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x \cdot x + 0 + 5 \cdot - 5 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.6.3

Add $x \cdot x$ and $0$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x \cdot x + 5 \cdot - 5 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.7

Simplify each term.

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Step 3.5.4.8.7.1

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x^{2} + 5 \cdot - 5 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.7.2

Multiply $5$ by $- 5$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} - 25 | | x^{2} - 25 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.8

Multiply $- 25 | x^{2} - 25 | | x^{2} - 25 |$ .

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Step 3.5.4.8.8.1

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | (x^{2} - 25) (x^{2} - 25) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.8.2

Raise $x^{2} - 25$ to the power of $1$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | (x^{2} - 25) (x^{2} - 25) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.8.3

Raise $x^{2} - 25$ to the power of $1$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | (x^{2} - 25) (x^{2} - 25) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.8.4

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

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | {(x^{2} - 25)}^{1 + 1} |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.8.5

Add $1$ and $1$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | {(x^{2} - 25)}^{2} |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.9

Rewrite ${(x^{2} - 25)}^{2}$ as $(x^{2} - 25) (x^{2} - 25)$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | (x^{2} - 25) (x^{2} - 25) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.10

Expand $(x^{2} - 25) (x^{2} - 25)$ using the FOIL Method.

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Step 3.5.4.8.10.1

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} (x^{2} - 25) - 25 (x^{2} - 25) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.10.2

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} x^{2} + x^{2} \cdot - 25 - 25 (x^{2} - 25) |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.10.3

Apply the distributive property.

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2} x^{2} + x^{2} \cdot - 25 - 25 x^{2} - 25 \cdot - 25 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.11

Simplify and combine like terms.

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Step 3.5.4.8.11.1

Simplify each term.

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Step 3.5.4.8.11.1.1

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

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Step 3.5.4.8.11.1.1.1

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

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{2 + 2} + x^{2} \cdot - 25 - 25 x^{2} - 25 \cdot - 25 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.11.1.1.2

Add $2$ and $2$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} + x^{2} \cdot - 25 - 25 x^{2} - 25 \cdot - 25 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.11.1.2

Move $- 25$ to the left of $x^{2}$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 25 \cdot x^{2} - 25 x^{2} - 25 \cdot - 25 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.11.1.3

Multiply $- 25$ by $- 25$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 25 x^{2} - 25 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.8.11.2

Subtract $25 x^{2}$ from $- 25 x^{2}$ .

$f'' (x) = \frac{\frac{2 (- 2 x^{6} + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.9

Factor $- 1$ out of $- 2 x^{6}$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6}) + 100 x^{4} - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.10

Factor $- 1$ out of $100 x^{4}$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6}) - (- 100 x^{4}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.11

Factor $- 1$ out of $- (2 x^{6}) - (- 100 x^{4})$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4}) - 1250 x^{2} + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.12

Factor $- 1$ out of $- 1250 x^{2}$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4}) - (1250 x^{2}) + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.13

Factor $- 1$ out of $- (2 x^{6} - 100 x^{4}) - (1250 x^{2})$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4} + 1250 x^{2}) + 3 x^{2} | x^{4} - 50 x^{2} + 625 | - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.14

Factor $- 1$ out of $3 x^{2} | x^{4} - 50 x^{2} + 625 |$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4} + 1250 x^{2}) - (- 3 x^{2} | x^{4} - 50 x^{2} + 625 |) - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.15

Factor $- 1$ out of $- (2 x^{6} - 100 x^{4} + 1250 x^{2}) - (- 3 x^{2} | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 |) - 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.16

Factor $- 1$ out of $- 25 | x^{4} - 50 x^{2} + 625 |$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 |) - (25 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.17

Factor $- 1$ out of $- (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 |) - (25 | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 (- (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.18

Rewrite $- (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)$ as $- 1 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)$ .

$f'' (x) = \frac{\frac{2 (- 1 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |))}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.4.19

Move the negative in front of the fraction.

$f'' (x) = \frac{- \frac{2 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$

Step 3.5.5

Combine terms.

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Step 3.5.5.1

Rewrite $\frac{- \frac{2 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}}{{| x^{2} - 25 |}^{2}}$ as a product.

$f'' (x) = - \frac{2 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |} \cdot \frac{1}{{| x^{2} - 25 |}^{2}}$

Step 3.5.5.2

Multiply $\frac{1}{{| x^{2} - 25 |}^{2}}$ by $\frac{2 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)}{| (x + 5) (x - 5) |}$ .

$f'' (x) = - \frac{2 (2 x^{6} - 100 x^{4} + 1250 x^{2} - 3 x^{2} | x^{4} - 50 x^{2} + 625 | + 25 | x^{4} - 50 x^{2} + 625 |)}{{| x^{2} - 25 |}^{2} | (x + 5) (x - 5) |}$

Step 4

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

$\frac{2 x^{3} - 50 x}{| x^{2} - 25 |} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x^{2} - 25$ .

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Step 5.1.1.1

To apply the Chain Rule, set $u$ as $x^{2} - 25$ .

$f' (x) = \frac{d}{d u} (| u |) \frac{d}{d x} (x^{2} - 25)$

Step 5.1.1.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$f' (x) = \frac{u}{| u |} \cdot \frac{d}{d x} (x^{2} - 25)$

Step 5.1.1.3

Replace all occurrences of $u$ with $x^{2} - 25$ .

$f' (x) = \frac{x^{2} - 25}{| x^{2} - 25 |} \cdot \frac{d}{d x} (x^{2} - 25)$

Step 5.1.2

Differentiate.

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Step 5.1.2.1

By the Sum Rule, the derivative of $x^{2} - 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]$ .

$f' (x) = \frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 25))$

Step 5.1.2.2

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

$f' (x) = \frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (2 x + \frac{d}{d x} (- 25))$

Step 5.1.2.3

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25$ with respect to $x$ is $0$ .

$f' (x) = \frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (2 x + 0)$

Step 5.1.2.4

Combine fractions.

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Step 5.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = \frac{x^{2} - 25}{| x^{2} - 25 |} \cdot (2 x)$

Step 5.1.2.4.2

Combine $2$ and $\frac{x^{2} - 25}{| x^{2} - 25 |}$ .

$f' (x) = \frac{2 (x^{2} - 25)}{| x^{2} - 25 |} \cdot x$

Step 5.1.2.4.3

Combine $\frac{2 (x^{2} - 25)}{| x^{2} - 25 |}$ and $x$ .

$f' (x) = \frac{2 (x^{2} - 25) x}{| x^{2} - 25 |}$

Step 5.1.3

Simplify.

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Step 5.1.3.1

Apply the distributive property.

$f' (x) = \frac{(2 x^{2} + 2 \cdot - 25) x}{| x^{2} - 25 |}$

Step 5.1.3.2

Apply the distributive property.

$f' (x) = \frac{2 x^{2} x + 2 \cdot (- 25 x)}{| x^{2} - 25 |}$

Step 5.1.3.3

Simplify each term.

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Step 5.1.3.3.1

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

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Step 5.1.3.3.1.1

Move $x$ .

$f' (x) = \frac{2 (x \cdot x^{2}) + 2 \cdot (- 25 x)}{| x^{2} - 25 |}$

Step 5.1.3.3.1.2

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

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Step 5.1.3.3.1.2.1

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

$f' (x) = \frac{2 (x \cdot x^{2}) + 2 \cdot (- 25 x)}{| x^{2} - 25 |}$

Step 5.1.3.3.1.2.2

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

$f' (x) = \frac{2 x^{1 + 2} + 2 \cdot (- 25 x)}{| x^{2} - 25 |}$

Step 5.1.3.3.1.3

Add $1$ and $2$ .

$f' (x) = \frac{2 x^{3} + 2 \cdot (- 25 x)}{| x^{2} - 25 |}$

Step 5.1.3.3.2

Multiply $2$ by $- 25$ .

$f' (x) = \frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$ .

$\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$

Step 6

Set the first derivative equal to $0$ then solve the equation $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$\frac{2 x^{3} - 50 x}{| x^{2} - 25 |} = 0$

Step 6.2

Set the numerator equal to zero.

$2 x^{3} - 50 x = 0$

Step 6.3

Solve the equation for $x$ .

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Step 6.3.1

Factor the left side of the equation.

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Step 6.3.1.1

Factor $2 x$ out of $2 x^{3} - 50 x$ .

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Step 6.3.1.1.1

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

$2 x (x^{2}) - 50 x = 0$

Step 6.3.1.1.2

Factor $2 x$ out of $- 50 x$ .

$2 x (x^{2}) + 2 x (- 25) = 0$

Step 6.3.1.1.3

Factor $2 x$ out of $2 x (x^{2}) + 2 x (- 25)$ .

$2 x (x^{2} - 25) = 0$

Step 6.3.1.2

Rewrite $25$ as $5^{2}$ .

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

Step 6.3.1.3

Factor.

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Step 6.3.1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$2 x ((x + 5) (x - 5)) = 0$

Step 6.3.1.3.2

Remove unnecessary parentheses.

$2 x (x + 5) (x - 5) = 0$

Step 6.3.2

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

$x = 0$

$x + 5 = 0$

$x - 5 = 0$

Step 6.3.3

Set $x$ equal to $0$ .

$x = 0$

Step 6.3.4

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 6.3.4.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 6.3.4.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6.3.5

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 6.3.5.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 6.3.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 6.3.6

The final solution is all the values that make $2 x (x + 5) (x - 5) = 0$ true.

$x = 0, - 5, 5$

Step 6.4

Exclude the solutions that do not make $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |} = 0$ true.

$x = 0$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

Set the denominator in $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$ equal to $0$ to find where the expression is undefined.

$| x^{2} - 25 | = 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x^{2} - 25 = \pm 0$

Step 7.2.2

Plus or minus $0$ is $0$ .

$x^{2} - 25 = 0$

Step 7.2.3

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 7.2.4

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

$x = \pm \sqrt{25}$

Step 7.2.5

Simplify $\pm \sqrt{25}$ .

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Step 7.2.5.1

Rewrite $25$ as $5^{2}$ .

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

Step 7.2.5.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 7.2.6

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

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Step 7.2.6.1

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

$x = 5$

Step 7.2.6.2

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

$x = - 5$

Step 7.2.6.3

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

$x = 5, - 5$

Step 7.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 5, x = 5$

Step 8

Critical points to evaluate.

$x = 0, - 5, 5$

Step 9

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.

$- \frac{2 (2 {(0)}^{6} - 100 {(0)}^{4} + 1250 {(0)}^{2} - 3 {(0)}^{2} | {(0)}^{4} - 50 {(0)}^{2} + 625 | + 25 | {(0)}^{4} - 50 {(0)}^{2} + 625 |)}{{| {(0)}^{2} - 25 |}^{2} | ((0) + 5) ((0) - 5) |}$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify the numerator.

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Step 10.1.1

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

$- \frac{2 (2 \cdot 0 - 100 \cdot 0^{4} + 1250 \cdot 0^{2} - 3 \cdot 0^{2} | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.2

Multiply $2$ by $0$ .

$- \frac{2 (0 - 100 \cdot 0^{4} + 1250 \cdot 0^{2} - 3 \cdot 0^{2} | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.3

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

$- \frac{2 (0 - 100 \cdot 0 + 1250 \cdot 0^{2} - 3 \cdot 0^{2} | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.4

Multiply $- 100$ by $0$ .

$- \frac{2 (0 + 0 + 1250 \cdot 0^{2} - 3 \cdot 0^{2} | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.5

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

$- \frac{2 (0 + 0 + 1250 \cdot 0 - 3 \cdot 0^{2} | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.6

Multiply $1250$ by $0$ .

$- \frac{2 (0 + 0 + 0 - 3 \cdot 0^{2} | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.7

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

$- \frac{2 (0 + 0 + 0 - 3 \cdot 0 | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.8

Multiply $- 3$ by $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 0^{4} - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.9

Simplify each term.

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Step 10.1.9.1

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

$- \frac{2 (0 + 0 + 0 + 0 | 0 - 50 \cdot 0^{2} + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.9.2

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

$- \frac{2 (0 + 0 + 0 + 0 | 0 - 50 \cdot 0 + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.9.3

Multiply $- 50$ by $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 0 + 0 + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.10

Add $0$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 0 + 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.11

Add $0$ and $625$ .

$- \frac{2 (0 + 0 + 0 + 0 | 625 | + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.12

The absolute value is the distance between a number and zero. The distance between $0$ and $625$ is $625$ .

$- \frac{2 (0 + 0 + 0 + 0 \cdot 625 + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.13

Multiply $0$ by $625$ .

$- \frac{2 (0 + 0 + 0 + 0 + 25 | 0^{4} - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.14

Simplify each term.

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Step 10.1.14.1

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

$- \frac{2 (0 + 0 + 0 + 0 + 25 | 0 - 50 \cdot 0^{2} + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.14.2

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

$- \frac{2 (0 + 0 + 0 + 0 + 25 | 0 - 50 \cdot 0 + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.14.3

Multiply $- 50$ by $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 25 | 0 + 0 + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.15

Add $0$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 25 | 0 + 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.16

Add $0$ and $625$ .

$- \frac{2 (0 + 0 + 0 + 0 + 25 | 625 |)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.17

The absolute value is the distance between a number and zero. The distance between $0$ and $625$ is $625$ .

$- \frac{2 (0 + 0 + 0 + 0 + 25 \cdot 625)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.18

Multiply $25$ by $625$ .

$- \frac{2 (0 + 0 + 0 + 0 + 15625)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.19

Add $0$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 15625)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.20

Add $0$ and $0$ .

$- \frac{2 (0 + 0 + 15625)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.21

Add $0$ and $0$ .

$- \frac{2 (0 + 15625)}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.1.22

Add $0$ and $15625$ .

$- \frac{2 \cdot 15625}{{| 0^{2} - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.2

Simplify the denominator.

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Step 10.2.1

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

$- \frac{2 \cdot 15625}{{| 0 - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.2.2

Subtract $25$ from $0$ .

$- \frac{2 \cdot 15625}{{| - 25 |}^{2} | (0 + 5) (0 - 5) |}$

Step 10.2.3

Add $0$ and $5$ .

$- \frac{2 \cdot 15625}{{| - 25 |}^{2} | 5 (0 - 5) |}$

Step 10.2.4

Subtract $5$ from $0$ .

$- \frac{2 \cdot 15625}{{| - 25 |}^{2} | 5 \cdot - 5 |}$

Step 10.2.5

The absolute value is the distance between a number and zero. The distance between $- 25$ and $0$ is $25$ .

$- \frac{2 \cdot 15625}{25^{2} | 5 \cdot - 5 |}$

Step 10.2.6

Raise $25$ to the power of $2$ .

$- \frac{2 \cdot 15625}{625 | 5 \cdot - 5 |}$

Step 10.2.7

Multiply $5$ by $- 5$ .

$- \frac{2 \cdot 15625}{625 | - 25 |}$

Step 10.2.8

The absolute value is the distance between a number and zero. The distance between $- 25$ and $0$ is $25$ .

$- \frac{2 \cdot 15625}{625 \cdot 25}$

Step 10.3

Simplify the expression.

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Step 10.3.1

Multiply $2$ by $15625$ .

$- \frac{31250}{625 \cdot 25}$

Step 10.3.2

Multiply $625$ by $25$ .

$- \frac{31250}{15625}$

Step 10.3.3

Divide $31250$ by $15625$ .

$- 1 \cdot 2$

Step 10.3.4

Multiply $- 1$ by $2$ .

$- 2$

Step 11

$x = 0$ 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 12

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

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Step 12.1

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

$f (0) = | {(0)}^{2} - 25 |$

Step 12.2

Simplify the result.

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Step 12.2.1

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

$f (0) = | 0 - 25 |$

Step 12.2.2

Subtract $25$ from $0$ .

$f (0) = | - 25 |$

Step 12.2.3

The absolute value is the distance between a number and zero. The distance between $- 25$ and $0$ is $25$ .

$f (0) = 25$

Step 12.2.4

The final answer is $25$ .

$y = 25$

Step 13

Evaluate the second derivative at $x = - 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{{| {(- 5)}^{2} - 25 |}^{2} | ((- 5) + 5) ((- 5) - 5) |}$

Step 14

Evaluate the second derivative.

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Step 14.1

Raise $- 5$ to the power of $2$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{{| 25 - 25 |}^{2} | ((- 5) + 5) ((- 5) - 5) |}$

Step 14.2

Subtract $25$ from $25$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{{| 0 |}^{2} | ((- 5) + 5) ((- 5) - 5) |}$

Step 14.3

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0^{2} | ((- 5) + 5) ((- 5) - 5) |}$

Step 14.4

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

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0 | ((- 5) + 5) ((- 5) - 5) |}$

Step 14.5

Add $- 5$ and $5$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0 | 0 ((- 5) - 5) |}$

Step 14.6

Subtract $5$ from $- 5$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0 | 0 \cdot - 10 |}$

Step 14.7

Multiply $0$ by $- 10$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0 | 0 |}$

Step 14.8

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0 \cdot 0}$

Step 14.9

Multiply $0$ by $0$ .

$- \frac{2 (2 {(- 5)}^{6} - 100 {(- 5)}^{4} + 1250 {(- 5)}^{2} - 3 {(- 5)}^{2} | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 | + 25 | {(- 5)}^{4} - 50 {(- 5)}^{2} + 625 |)}{0}$

Step 14.10

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 15

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 15.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 5) \cup (- 5, 0) \cup (0, 5) \cup (5, \infty)$

Step 15.2

Substitute any number, such as $- 8$ , from the interval $(- \infty, - 5)$ in the first derivative $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$ to check if the result is negative or positive.

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Step 15.2.1

Replace the variable $x$ with $- 8$ in the expression.

$f' (- 8) = \frac{2 {(- 8)}^{3} - 50 \cdot - 8}{| {(- 8)}^{2} - 25 |}$

Step 15.2.2

Simplify the result.

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Step 15.2.2.1

Simplify the numerator.

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Step 15.2.2.1.1

Raise $- 8$ to the power of $3$ .

$f' (- 8) = \frac{2 \cdot - 512 - 50 \cdot - 8}{| {(- 8)}^{2} - 25 |}$

Step 15.2.2.1.2

Multiply $2$ by $- 512$ .

$f' (- 8) = \frac{- 1024 - 50 \cdot - 8}{| {(- 8)}^{2} - 25 |}$

Step 15.2.2.1.3

Multiply $- 50$ by $- 8$ .

$f' (- 8) = \frac{- 1024 + 400}{| {(- 8)}^{2} - 25 |}$

Step 15.2.2.1.4

Add $- 1024$ and $400$ .

$f' (- 8) = \frac{- 624}{| {(- 8)}^{2} - 25 |}$

Step 15.2.2.2

Simplify the denominator.

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Step 15.2.2.2.1

Raise $- 8$ to the power of $2$ .

$f' (- 8) = \frac{- 624}{| 64 - 25 |}$

Step 15.2.2.2.2

Subtract $25$ from $64$ .

$f' (- 8) = \frac{- 624}{| 39 |}$

Step 15.2.2.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $39$ is $39$ .

$f' (- 8) = \frac{- 624}{39}$

Step 15.2.2.3

Divide $- 624$ by $39$ .

$f' (- 8) = - 16$

Step 15.2.2.4

The final answer is $- 16$ .

$- 16$

Step 15.3

Substitute any number, such as $- 2$ , from the interval $(- 5, 0)$ in the first derivative $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$ to check if the result is negative or positive.

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Step 15.3.1

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

$f' (- 2) = \frac{2 {(- 2)}^{3} - 50 \cdot - 2}{| {(- 2)}^{2} - 25 |}$

Step 15.3.2

Simplify the result.

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Step 15.3.2.1

Simplify the numerator.

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Step 15.3.2.1.1

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

$f' (- 2) = \frac{2 \cdot - 8 - 50 \cdot - 2}{| {(- 2)}^{2} - 25 |}$

Step 15.3.2.1.2

Multiply $2$ by $- 8$ .

$f' (- 2) = \frac{- 16 - 50 \cdot - 2}{| {(- 2)}^{2} - 25 |}$

Step 15.3.2.1.3

Multiply $- 50$ by $- 2$ .

$f' (- 2) = \frac{- 16 + 100}{| {(- 2)}^{2} - 25 |}$

Step 15.3.2.1.4

Add $- 16$ and $100$ .

$f' (- 2) = \frac{84}{| {(- 2)}^{2} - 25 |}$

Step 15.3.2.2

Simplify the denominator.

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Step 15.3.2.2.1

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

$f' (- 2) = \frac{84}{| 4 - 25 |}$

Step 15.3.2.2.2

Subtract $25$ from $4$ .

$f' (- 2) = \frac{84}{| - 21 |}$

Step 15.3.2.2.3

The absolute value is the distance between a number and zero. The distance between $- 21$ and $0$ is $21$ .

$f' (- 2) = \frac{84}{21}$

Step 15.3.2.3

Divide $84$ by $21$ .

$f' (- 2) = 4$

Step 15.3.2.4

The final answer is $4$ .

$4$

Step 15.4

Substitute any number, such as $2$ , from the interval $(0, 5)$ in the first derivative $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$ to check if the result is negative or positive.

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Step 15.4.1

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

$f' (2) = \frac{2 {(2)}^{3} - 50 \cdot 2}{| {(2)}^{2} - 25 |}$

Step 15.4.2

Simplify the result.

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Step 15.4.2.1

Simplify the numerator.

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Step 15.4.2.1.1

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

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Step 15.4.2.1.1.1

Multiply $2$ by $2^{3}$ .

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Step 15.4.2.1.1.1.1

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

$f' (2) = \frac{2 \cdot 2^{3} - 50 \cdot 2}{| 2^{2} - 25 |}$

Step 15.4.2.1.1.1.2

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

$f' (2) = \frac{2^{1 + 3} - 50 \cdot 2}{| 2^{2} - 25 |}$

Step 15.4.2.1.1.2

Add $1$ and $3$ .

$f' (2) = \frac{2^{4} - 50 \cdot 2}{| 2^{2} - 25 |}$

Step 15.4.2.1.2

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

$f' (2) = \frac{16 - 50 \cdot 2}{| 2^{2} - 25 |}$

Step 15.4.2.1.3

Multiply $- 50$ by $2$ .

$f' (2) = \frac{16 - 100}{| 2^{2} - 25 |}$

Step 15.4.2.1.4

Subtract $100$ from $16$ .

$f' (2) = \frac{- 84}{| 2^{2} - 25 |}$

Step 15.4.2.2

Simplify the denominator.

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Step 15.4.2.2.1

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

$f' (2) = \frac{- 84}{| 4 - 25 |}$

Step 15.4.2.2.2

Subtract $25$ from $4$ .

$f' (2) = \frac{- 84}{| - 21 |}$

Step 15.4.2.2.3

The absolute value is the distance between a number and zero. The distance between $- 21$ and $0$ is $21$ .

$f' (2) = \frac{- 84}{21}$

Step 15.4.2.3

Divide $- 84$ by $21$ .

$f' (2) = - 4$

Step 15.4.2.4

The final answer is $- 4$ .

$- 4$

Step 15.5

Substitute any number, such as $8$ , from the interval $(5, \infty)$ in the first derivative $\frac{2 x^{3} - 50 x}{| x^{2} - 25 |}$ to check if the result is negative or positive.

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Step 15.5.1

Replace the variable $x$ with $8$ in the expression.

$f' (8) = \frac{2 {(8)}^{3} - 50 \cdot 8}{| {(8)}^{2} - 25 |}$

Step 15.5.2

Simplify the result.

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Step 15.5.2.1

Simplify the numerator.

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Step 15.5.2.1.1

Raise $8$ to the power of $3$ .

$f' (8) = \frac{2 \cdot 512 - 50 \cdot 8}{| 8^{2} - 25 |}$

Step 15.5.2.1.2

Multiply $2$ by $512$ .

$f' (8) = \frac{1024 - 50 \cdot 8}{| 8^{2} - 25 |}$

Step 15.5.2.1.3

Multiply $- 50$ by $8$ .

$f' (8) = \frac{1024 - 400}{| 8^{2} - 25 |}$

Step 15.5.2.1.4

Subtract $400$ from $1024$ .

$f' (8) = \frac{624}{| 8^{2} - 25 |}$

Step 15.5.2.2

Simplify the denominator.

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Step 15.5.2.2.1

Raise $8$ to the power of $2$ .

$f' (8) = \frac{624}{| 64 - 25 |}$

Step 15.5.2.2.2

Subtract $25$ from $64$ .

$f' (8) = \frac{624}{| 39 |}$

Step 15.5.2.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $39$ is $39$ .

$f' (8) = \frac{624}{39}$

Step 15.5.2.3

Divide $624$ by $39$ .

$f' (8) = 16$

Step 15.5.2.4

The final answer is $16$ .

$16$

Step 15.6

Since the first derivative changed signs from negative to positive around $x = - 5$ , then $x = - 5$ is a local minimum.

$x = - 5$ is a local minimum

Step 15.7

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 15.8

Since the first derivative changed signs from negative to positive around $x = 5$ , then $x = 5$ is a local minimum.

$x = 5$ is a local minimum

Step 15.9

These are the local extrema for $f (x) = | x^{2} - 25 |$ .

$x = - 5$ is a local minimum

$x = 0$ is a local maximum

$x = 5$ is a local minimum

$x = - 5$ is a local minimum

$x = 0$ is a local maximum

$x = 5$ is a local minimum

Step 16