Find the Inflection Points f(x)=(4x)/(x^2-4)

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

Step 1

Find the second derivative.

Find the first derivative.

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

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

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

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

Differentiate.

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

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

Multiply $x^{2} - 4$ by $1$ .

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

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

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

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

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

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

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

Simplify the expression.

Add $2 x$ and $0$ .

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

Multiply $2$ by $- 1$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

Simplify.

Apply the distributive property.

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

Simplify each term.

Multiply $- 1$ by $4$ .

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

Multiply $4$ by $- 4$ .

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

Find the second derivative.

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

Differentiate.

Multiply the exponents in ${({(x^{2} - 4)}^{2})}^{2}$ .

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

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

Multiply $2$ by $2$ .

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

By the Sum Rule, the derivative of $- 4 x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [- 4 x^{2}] + \frac{d}{d x} [- 16]$ .

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

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

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

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

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

Multiply $2$ by $- 4$ .

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

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x + 0) - (- 4 x^{2} - 16) \frac{d}{d x} {(x^{2} - 4)}^{2}}{{(x^{2} - 4)}^{4}}$

Add $- 8 x$ and $0$ .

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - (- 4 x^{2} - 16) \frac{d}{d x} {(x^{2} - 4)}^{2}}{{(x^{2} - 4)}^{4}}$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = x^{2} - 4$ .

To apply the Chain Rule, set $u$ as $x^{2} - 4$ .

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - (- 4 x^{2} - 16) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x^{2} - 4))}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - (- 4 x^{2} - 16) (2 u \frac{d}{d x} (x^{2} - 4))}{{(x^{2} - 4)}^{4}}$

Replace all occurrences of $u$ with $x^{2} - 4$ .

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - (- 4 x^{2} - 16) (2 (x^{2} - 4) \frac{d}{d x} (x^{2} - 4))}{{(x^{2} - 4)}^{4}}$

Differentiate.

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - 2 (- 4 x^{2} - 16) ((x^{2} - 4) \frac{d}{d x} (x^{2} - 4))}{{(x^{2} - 4)}^{4}}$

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

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

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

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

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

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - 2 (- 4 x^{2} - 16) ((x^{2} - 4) (2 x + 0))}{{(x^{2} - 4)}^{4}}$

Simplify the expression.

Add $2 x$ and $0$ .

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

Move $2$ to the left of $x^{2} - 4$ .

$f'' (x) = \frac{{(x^{2} - 4)}^{2} (- 8 x) - 2 (- 4 x^{2} - 16) (2 \cdot ((x^{2} - 4) x))}{{(x^{2} - 4)}^{4}}$

Multiply $2$ by $- 2$ .

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

Simplify.

Apply the distributive property.

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

Apply the distributive property.

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

Simplify the numerator.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

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

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

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

Add $2$ and $2$ .

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

Move $- 4$ to the left of $x^{2}$ .

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

Multiply $- 4$ by $- 4$ .

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

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

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

Apply the distributive property.

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

Simplify.

Multiply $- 8$ by $- 8$ .

$f'' (x) = \frac{(- 8 x^{4} + 64 x^{2} - 8 \cdot 16) x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Multiply $- 8$ by $16$ .

$f'' (x) = \frac{(- 8 x^{4} + 64 x^{2} - 128) x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Apply the distributive property.

$f'' (x) = \frac{- 8 x^{4} x + 64 x^{2} x - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Simplify.

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

Move $x$ .

$f'' (x) = \frac{- 8 (x \cdot x^{4}) + 64 x^{2} x - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

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

$f'' (x) = \frac{- 8 (x \cdot x^{4}) + 64 x^{2} x - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{- 8 x^{1 + 4} + 64 x^{2} x - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Add $1$ and $4$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{2} x - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

Move $x$ .

$f'' (x) = \frac{- 8 x^{5} + 64 (x \cdot x^{2}) - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

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

$f'' (x) = \frac{- 8 x^{5} + 64 (x \cdot x^{2}) - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{- 8 x^{5} + 64 x^{1 + 2} - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Add $1$ and $2$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + (- 4 (- 4 x^{2}) - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Simplify each term.

Multiply $- 4$ by $- 4$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + (16 x^{2} - 4 \cdot - 16) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

Multiply $- 4$ by $- 16$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + (16 x^{2} + 64) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

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

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

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + (16 x^{2} + 64) (x^{2} x - 4 x)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + (16 x^{2} + 64) (x^{2 + 1} - 4 x)}{{(x^{2} - 4)}^{4}}$

Add $2$ and $1$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + (16 x^{2} + 64) (x^{3} - 4 x)}{{(x^{2} - 4)}^{4}}$

Expand $(16 x^{2} + 64) (x^{3} - 4 x)$ using the FOIL Method.

Apply the distributive property.

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{2} (x^{3} - 4 x) + 64 (x^{3} - 4 x)}{{(x^{2} - 4)}^{4}}$

Apply the distributive property.

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{2} x^{3} + 16 x^{2} (- 4 x) + 64 (x^{3} - 4 x)}{{(x^{2} - 4)}^{4}}$

Apply the distributive property.

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{2} x^{3} + 16 x^{2} (- 4 x) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

Simplify and combine like terms.

Simplify each term.

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

Move $x^{3}$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 (x^{3} x^{2}) + 16 x^{2} (- 4 x) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{3 + 2} + 16 x^{2} (- 4 x) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

Add $3$ and $2$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 16 x^{2} (- 4 x) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 16 \cdot (- 4 x^{2} x) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

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

Move $x$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 16 \cdot (- 4 (x \cdot x^{2})) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

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

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

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 16 \cdot (- 4 (x \cdot x^{2})) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 16 \cdot (- 4 x^{1 + 2}) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

Add $1$ and $2$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 16 \cdot (- 4 x^{3}) + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

Multiply $16$ by $- 4$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} - 64 x^{3} + 64 x^{3} + 64 (- 4 x)}{{(x^{2} - 4)}^{4}}$

Multiply $- 4$ by $64$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} - 64 x^{3} + 64 x^{3} - 256 x}{{(x^{2} - 4)}^{4}}$

Add $- 64 x^{3}$ and $64 x^{3}$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} + 0 - 256 x}{{(x^{2} - 4)}^{4}}$

Add $16 x^{5}$ and $0$ .

$f'' (x) = \frac{- 8 x^{5} + 64 x^{3} - 128 x + 16 x^{5} - 256 x}{{(x^{2} - 4)}^{4}}$

Add $- 8 x^{5}$ and $16 x^{5}$ .

$f'' (x) = \frac{8 x^{5} + 64 x^{3} - 128 x - 256 x}{{(x^{2} - 4)}^{4}}$

Subtract $256 x$ from $- 128 x$ .

$f'' (x) = \frac{8 x^{5} + 64 x^{3} - 384 x}{{(x^{2} - 4)}^{4}}$

Simplify the numerator.

Factor $8 x$ out of $8 x^{5} + 64 x^{3} - 384 x$ .

Factor $8 x$ out of $8 x^{5}$ .

$f'' (x) = \frac{8 x (x^{4}) + 64 x^{3} - 384 x}{{(x^{2} - 4)}^{4}}$

Factor $8 x$ out of $64 x^{3}$ .

$f'' (x) = \frac{8 x (x^{4}) + 8 x (8 x^{2}) - 384 x}{{(x^{2} - 4)}^{4}}$

Factor $8 x$ out of $- 384 x$ .

$f'' (x) = \frac{8 x (x^{4}) + 8 x (8 x^{2}) + 8 x (- 48)}{{(x^{2} - 4)}^{4}}$

Factor $8 x$ out of $8 x (x^{4}) + 8 x (8 x^{2})$ .

$f'' (x) = \frac{8 x (x^{4} + 8 x^{2}) + 8 x (- 48)}{{(x^{2} - 4)}^{4}}$

Factor $8 x$ out of $8 x (x^{4} + 8 x^{2}) + 8 x (- 48)$ .

$f'' (x) = \frac{8 x (x^{4} + 8 x^{2} - 48)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{8 x ({(x^{2})}^{2} + 8 x^{2} - 48)}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{8 x (u^{2} + 8 u - 48)}{{(x^{2} - 4)}^{4}}$

Factor $u^{2} + 8 u - 48$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 48$ and whose sum is $8$ .

$- 4, 12$

Write the factored form using these integers.

$f'' (x) = \frac{8 x ((u - 4) (u + 12))}{{(x^{2} - 4)}^{4}}$

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

$f'' (x) = \frac{8 x ((x^{2} - 4) (x^{2} + 12))}{{(x^{2} - 4)}^{4}}$

Rewrite $4$ as $2^{2}$ .

$f'' (x) = \frac{8 x ((x^{2} - 2^{2}) (x^{2} + 12))}{{(x^{2} - 4)}^{4}}$

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 $b = 2$ .

$f'' (x) = \frac{8 x (x + 2) (x - 2) (x^{2} + 12)}{{(x^{2} - 4)}^{4}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$f'' (x) = \frac{8 x (x + 2) (x - 2) (x^{2} + 12)}{{(x^{2} - 2^{2})}^{4}}$

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 $b = 2$ .

$f'' (x) = \frac{8 x (x + 2) (x - 2) (x^{2} + 12)}{{((x + 2) (x - 2))}^{4}}$

Apply the product rule to $(x + 2) (x - 2)$ .

$f'' (x) = \frac{8 x (x + 2) (x - 2) (x^{2} + 12)}{{(x + 2)}^{4} {(x - 2)}^{4}}$

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

Factor $x + 2$ out of $8 x (x + 2) (x - 2) (x^{2} + 12)$ .

$f'' (x) = \frac{(x + 2) ((8 x (x - 2)) (x^{2} + 12))}{{(x + 2)}^{4} {(x - 2)}^{4}}$

Cancel the common factors.

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

$f'' (x) = \frac{(x + 2) ((8 x (x - 2)) (x^{2} + 12))}{(x + 2) ({(x + 2)}^{3} {(x - 2)}^{4})}$

Cancel the common factor.

$f'' (x) = \frac{(x + 2) ((8 x (x - 2)) (x^{2} + 12))}{(x + 2) ({(x + 2)}^{3} {(x - 2)}^{4})}$

Rewrite the expression.

$f'' (x) = \frac{(8 x (x - 2)) (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{4}}$

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

Factor $x - 2$ out of $(8 x (x - 2)) (x^{2} + 12)$ .

$f'' (x) = \frac{(x - 2) ((8 x) (x^{2} + 12))}{{(x + 2)}^{3} {(x - 2)}^{4}}$

Cancel the common factors.

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

$f'' (x) = \frac{(x - 2) ((8 x) (x^{2} + 12))}{(x - 2) ({(x + 2)}^{3} {(x - 2)}^{3})}$

Cancel the common factor.

$f'' (x) = \frac{(x - 2) ((8 x) (x^{2} + 12))}{(x - 2) ({(x + 2)}^{3} {(x - 2)}^{3})}$

Rewrite the expression.

$f'' (x) = \frac{(8 x) (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{3}}$

$f'' (x) = \frac{8 x (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{3}}$

The second derivative of $f (x)$ with respect to $x$ is $\frac{8 x (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{3}}$ .

$\frac{8 x (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{3}}$

Step 2

Set the second derivative equal to $0$ then solve the equation $\frac{8 x (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{3}} = 0$ .

Set the second derivative equal to $0$ .

$\frac{8 x (x^{2} + 12)}{{(x + 2)}^{3} {(x - 2)}^{3}} = 0$

Set the numerator equal to zero.

$8 x (x^{2} + 12) = 0$

Solve the equation for $x$ .

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

$x = 0$

$x^{2} + 12 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} + 12$ equal to $0$ and solve for $x$ .

Set $x^{2} + 12$ equal to $0$ .

$x^{2} + 12 = 0$

Solve $x^{2} + 12 = 0$ for $x$ .

Subtract $12$ from both sides of the equation.

$x^{2} = - 12$

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

$x = \pm \sqrt{- 12}$

Simplify $\pm \sqrt{- 12}$ .

Rewrite $- 12$ as $- 1 (12)$ .

$x = \pm \sqrt{- 1 (12)}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{12}$

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

$x = \pm i \cdot \sqrt{12}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \pm i \cdot \sqrt{4 (3)}$

Rewrite $4$ as $2^{2}$ .

$x = \pm i \cdot \sqrt{2^{2} \cdot 3}$

Pull terms out from under the radical.

$x = \pm i \cdot (2 \sqrt{3})$

Move $2$ to the left of $i$ .

$x = \pm 2 i \sqrt{3}$

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

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

$x = 2 i \sqrt{3}$

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

$x = - 2 i \sqrt{3}$

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

$x = 2 i \sqrt{3}, - 2 i \sqrt{3}$

The final solution is all the values that make $8 x (x^{2} + 12) = 0$ true.

$x = 0, 2 i \sqrt{3}, - 2 i \sqrt{3}$

Step 3

Find the points where the second derivative is $0$ .

Substitute $0$ in $f (x) = \frac{4 x}{x^{2} - 4}$ to find the value of $y$ .

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

$f (0) = \frac{4 (0)}{{(0)}^{2} - 4}$

Simplify the result.

Multiply $4$ by $0$ .

$f (0) = \frac{0}{0^{2} - 4}$

Simplify the denominator.

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

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

Subtract $4$ from $0$ .

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

Divide $0$ by $- 4$ .

$f (0) = 0$

The final answer is $0$ .

$0$

The point found by substituting $0$ in $f (x) = \frac{4 x}{x^{2} - 4}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Step 5

Substitute a value from the interval $(- \infty, 0)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = \frac{8 (- 0.1) ({(- 0.1)}^{2} + 12)}{{((- 0.1) + 2)}^{3} {((- 0.1) - 2)}^{3}}$

Simplify the result.

Simplify the numerator.

Multiply $8$ by $- 0.1$ .

$f'' (- 0.1) = \frac{- 0.8 ({(- 0.1)}^{2} + 12)}{{(- 0.1 + 2)}^{3} {(- 0.1 - 2)}^{3}}$

Multiply $- 0.8$ by ${(- 0.1)}^{2} + 12$ .

$f'' (- 0.1) = \frac{- 9.608}{{(- 0.1 + 2)}^{3} {(- 0.1 - 2)}^{3}}$

Simplify the denominator.

Add $- 0.1$ and $2$ .

$f'' (- 0.1) = \frac{- 9.608}{{1.9}^{3} {(- 0.1 - 2)}^{3}}$

Subtract $2$ from $- 0.1$ .

$f'' (- 0.1) = \frac{- 9.608}{{1.9}^{3} {(- 2.1)}^{3}}$

Raise $1.9$ to the power of $3$ .

$f'' (- 0.1) = \frac{- 9.608}{6.859 {(- 2.1)}^{3}}$

Raise $- 2.1$ to the power of $3$ .

$f'' (- 0.1) = \frac{- 9.608}{6.859 \cdot - 9.261}$

Simplify the expression.

Multiply $6.859$ by $- 9.261$ .

$f'' (- 0.1) = \frac{- 9.608}{- 63.521199}$

Divide $- 9.608$ by $- 63.521199$ .

$f'' (- 0.1) = 0.15125659$

The final answer is $0.15125659$ .

$0.15125659$

At $- 0.1$ , the second derivative is $0.15125659$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(0, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = \frac{8 (0.1) ({(0.1)}^{2} + 12)}{{((0.1) + 2)}^{3} {((0.1) - 2)}^{3}}$

Simplify the result.

Simplify the numerator.

Multiply $8$ by $0.1$ .

$f'' (0.1) = \frac{0.8 ({0.1}^{2} + 12)}{{(0.1 + 2)}^{3} {(0.1 - 2)}^{3}}$

Multiply $0.8$ by ${0.1}^{2} + 12$ .

$f'' (0.1) = \frac{9.608}{{(0.1 + 2)}^{3} {(0.1 - 2)}^{3}}$

Simplify the denominator.

Add $0.1$ and $2$ .

$f'' (0.1) = \frac{9.608}{{2.1}^{3} {(0.1 - 2)}^{3}}$

Subtract $2$ from $0.1$ .

$f'' (0.1) = \frac{9.608}{{2.1}^{3} {(- 1.9)}^{3}}$

Raise $2.1$ to the power of $3$ .

$f'' (0.1) = \frac{9.608}{9.261 {(- 1.9)}^{3}}$

Raise $- 1.9$ to the power of $3$ .

$f'' (0.1) = \frac{9.608}{9.261 \cdot - 6.859}$

Simplify the expression.

Multiply $9.261$ by $- 6.859$ .

$f'' (0.1) = \frac{9.608}{- 63.521199}$

Divide $9.608$ by $- 63.521199$ .

$f'' (0.1) = - 0.15125659$

The final answer is $- 0.15125659$ .

$- 0.15125659$

At $0.1$ , the second derivative is $- 0.15125659$ . Since this is negative, the second derivative is decreasing on the interval $(0, \infty)$

Decreasing on $(0, \infty)$ since $f'' (x) < 0$

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(0, 0)$ .

$(0, 0)$

Step 8

Find the Inflection Points f(x)=(4x)/(x^2-4)

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