Calculus Examples

$\frac{x^{2}}{\sqrt{x + 1}}$

Step 1

Write $\frac{x^{2}}{\sqrt{x + 1}}$ as a function.

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

Step 2

Find the $x$ values where the second derivative is equal to $0$ .

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

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

Step 2.1.1.2

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

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

Step 2.1.1.3

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

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Step 2.1.1.3.1

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

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

Step 2.1.1.3.2

Cancel the common factor of $2$ .

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Step 2.1.1.3.2.1

Cancel the common factor.

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

Step 2.1.1.3.2.2

Rewrite the expression.

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

Step 2.1.1.4

Simplify.

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

Step 2.1.1.5

Differentiate using the Power Rule.

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Step 2.1.1.5.1

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

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

Step 2.1.1.5.2

Move $2$ to the left of ${(x + 1)}^{\frac{1}{2}}$ .

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

Step 2.1.1.6

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

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Step 2.1.1.6.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.1.1.6.2

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

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

Step 2.1.1.6.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 2.1.1.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.1.1.8

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 2.1.1.9

Combine the numerators over the common denominator.

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

Step 2.1.1.10

Simplify the numerator.

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Step 2.1.1.10.1

Multiply $- 1$ by $2$ .

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

Step 2.1.1.10.2

Subtract $2$ from $1$ .

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

Step 2.1.1.11

Combine fractions.

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Step 2.1.1.11.1

Move the negative in front of the fraction.

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

Step 2.1.1.11.2

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

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

Step 2.1.1.11.3

Move ${(x + 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.1.1.11.4

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

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

Step 2.1.1.12

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

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

Step 2.1.1.13

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

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

Step 2.1.1.14

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

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

Step 2.1.1.15

Simplify the expression.

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Step 2.1.1.15.1

Add $1$ and $0$ .

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

Step 2.1.1.15.2

Multiply $- 1$ by $1$ .

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

Step 2.1.1.16

Combine $2 {(x + 1)}^{\frac{1}{2}} x$ and $- \frac{x^{2}}{2 {(x + 1)}^{\frac{1}{2}}}$ using a common denominator.

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Step 2.1.1.16.1

Move ${(x + 1)}^{\frac{1}{2}}$ .

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

Step 2.1.1.16.2

To write $2 x {(x + 1)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(x + 1)}^{\frac{1}{2}}}{2 {(x + 1)}^{\frac{1}{2}}}$ .

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

Step 2.1.1.16.3

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

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

Step 2.1.1.16.4

Combine the numerators over the common denominator.

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

Step 2.1.1.17

Multiply $2$ by $2$ .

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

Step 2.1.1.18

Multiply ${(x + 1)}^{\frac{1}{2}}$ by ${(x + 1)}^{\frac{1}{2}}$ by adding the exponents.

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Step 2.1.1.18.1

Move ${(x + 1)}^{\frac{1}{2}}$ .

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

Step 2.1.1.18.2

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

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

Step 2.1.1.18.3

Combine the numerators over the common denominator.

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

Step 2.1.1.18.4

Add $1$ and $1$ .

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

Step 2.1.1.18.5

Divide $2$ by $2$ .

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

Step 2.1.1.19

Simplify $4 x {(x + 1)}^{1}$ .

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

Step 2.1.1.20

Rewrite $\frac{\frac{4 x (x + 1) - x^{2}}{2 {(x + 1)}^{\frac{1}{2}}}}{x + 1}$ as a product.

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

Step 2.1.1.21

Multiply $\frac{4 x (x + 1) - x^{2}}{2 {(x + 1)}^{\frac{1}{2}}}$ by $\frac{1}{x + 1}$ .

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

Step 2.1.1.22

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

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

Step 2.1.1.23

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

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

Step 2.1.1.24

Write $1$ as a fraction with a common denominator.

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

Step 2.1.1.25

Combine the numerators over the common denominator.

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

Step 2.1.1.26

Add $2$ and $1$ .

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

Step 2.1.1.27

Simplify.

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Step 2.1.1.27.1

Apply the distributive property.

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

Step 2.1.1.27.2

Simplify the numerator.

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Step 2.1.1.27.2.1

Simplify each term.

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Step 2.1.1.27.2.1.1

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

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Step 2.1.1.27.2.1.1.1

Move $x$ .

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

Step 2.1.1.27.2.1.1.2

Multiply $x$ by $x$ .

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

Step 2.1.1.27.2.1.2

Multiply $4$ by $1$ .

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

Step 2.1.1.27.2.2

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

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

Step 2.1.1.27.3

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

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Step 2.1.1.27.3.1

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

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

Step 2.1.1.27.3.2

Factor $x$ out of $4 x$ .

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

Step 2.1.1.27.3.3

Factor $x$ out of $x (3 x) + x \cdot 4$ .

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

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x (3 x + 4)}{2 {(x + 1)}^{\frac{3}{2}}}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\frac{x (3 x + 4)}{{(x + 1)}^{\frac{3}{2}}}]$ .

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

Step 2.1.2.2

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

Step 2.1.2.3

Multiply the exponents in ${({(x + 1)}^{\frac{3}{2}})}^{2}$ .

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Step 2.1.2.3.1

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

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

Step 2.1.2.3.2

Cancel the common factor of $2$ .

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Step 2.1.2.3.2.1

Cancel the common factor.

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

Step 2.1.2.3.2.2

Rewrite the expression.

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

Step 2.1.2.4

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

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

Step 2.1.2.5

Differentiate.

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Step 2.1.2.5.1

By the Sum Rule, the derivative of $3 x + 4$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [4]$ .

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

Step 2.1.2.5.2

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

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

Step 2.1.2.5.3

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

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

Step 2.1.2.5.4

Multiply $3$ by $1$ .

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

Step 2.1.2.5.5

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

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

Step 2.1.2.5.6

Simplify the expression.

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Step 2.1.2.5.6.1

Add $3$ and $0$ .

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

Step 2.1.2.5.6.2

Move $3$ to the left of $x$ .

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

Step 2.1.2.5.7

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

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

Step 2.1.2.5.8

Simplify by adding terms.

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Step 2.1.2.5.8.1

Multiply $3 x + 4$ by $1$ .

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

Step 2.1.2.5.8.2

Add $3 x$ and $3 x$ .

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

Step 2.1.2.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{3}{2}}$ and $g (x) = x + 1$ .

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Step 2.1.2.6.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.1.2.6.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{3}{2}$ .

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

Step 2.1.2.6.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 2.1.2.7

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.1.2.8

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 2.1.2.9

Combine the numerators over the common denominator.

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

Step 2.1.2.10

Simplify the numerator.

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Step 2.1.2.10.1

Multiply $- 1$ by $2$ .

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

Step 2.1.2.10.2

Subtract $2$ from $3$ .

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

Step 2.1.2.11

Combine fractions.

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Step 2.1.2.11.1

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

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

Step 2.1.2.11.2

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

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

Step 2.1.2.12

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

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

Step 2.1.2.13

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

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

Step 2.1.2.14

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

$f'' (x) = \frac{1}{2} \cdot \frac{{(x + 1)}^{\frac{3}{2}} (6 x + 4) - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot ((3 x + 4) (1 + 0))}{{(x + 1)}^{3}}$

Step 2.1.2.15

Combine fractions.

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Step 2.1.2.15.1

Add $1$ and $0$ .

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

Step 2.1.2.15.2

Multiply $- 1$ by $1$ .

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

Step 2.1.2.15.3

Multiply $\frac{1}{2}$ by $\frac{{(x + 1)}^{\frac{3}{2}} (6 x + 4) - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} (3 x + 4)}{{(x + 1)}^{3}}$ .

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

Step 2.1.2.16

Simplify.

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Step 2.1.2.16.1

Simplify the numerator.

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Step 2.1.2.16.1.1

Apply the distributive property.

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

Step 2.1.2.16.1.2

Rewrite using the commutative property of multiplication.

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

Step 2.1.2.16.1.3

Move $4$ to the left of ${(x + 1)}^{\frac{3}{2}}$ .

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

Step 2.1.2.16.1.4

Apply the distributive property.

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

Step 2.1.2.16.1.5

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

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Step 2.1.2.16.1.5.1

Multiply $3$ by $- 1$ .

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

Step 2.1.2.16.1.5.2

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

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

Step 2.1.2.16.1.5.3

Multiply $3$ by $- 3$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x)}{2} \cdot x - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.5.4

Combine $\frac{- 9 ({(x + 1)}^{\frac{1}{2}} x)}{2}$ and $x$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x) x}{2} - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.5.5

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

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} (x \cdot x))}{2} - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.5.6

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

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} (x \cdot x))}{2} - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.5.7

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

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{1 + 1})}{2} - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.5.8

Add $1$ and $1$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{2})}{2} - \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.6

Cancel the common factor of $2$ .

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Step 2.1.2.16.1.6.1

Move the leading negative in $- \frac{3 {(x + 1)}^{\frac{1}{2}} x}{2}$ into the numerator.

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{2})}{2} + \frac{- 3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot 4}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.6.2

Factor $2$ out of $4$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{2})}{2} + \frac{- 3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot (2 (2))}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.6.3

Cancel the common factor.

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{2})}{2} + \frac{- 3 {(x + 1)}^{\frac{1}{2}} x}{2} \cdot (2 \cdot 2)}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.6.4

Rewrite the expression.

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{2})}{2} - 3 {(x + 1)}^{\frac{1}{2}} x \cdot 2}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.7

Multiply $2$ by $- 3$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 ({(x + 1)}^{\frac{1}{2}} x^{2})}{2} - 6 {(x + 1)}^{\frac{1}{2}} x}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.8

Move the negative in front of the fraction.

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} - \frac{9 {(x + 1)}^{\frac{1}{2}} x^{2}}{2} - 6 {(x + 1)}^{\frac{1}{2}} x}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.9

To write $- 6 {(x + 1)}^{\frac{1}{2}} x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} - \frac{9 {(x + 1)}^{\frac{1}{2}} x^{2}}{2} - 6 {(x + 1)}^{\frac{1}{2}} x \cdot \frac{2}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.10

Combine $- 6 {(x + 1)}^{\frac{1}{2}} x$ and $\frac{2}{2}$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} - \frac{9 {(x + 1)}^{\frac{1}{2}} x^{2}}{2} + \frac{- 6 {(x + 1)}^{\frac{1}{2}} x \cdot 2}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.11

Combine the numerators over the common denominator.

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 {(x + 1)}^{\frac{1}{2}} x^{2} - 6 {(x + 1)}^{\frac{1}{2}} x \cdot 2}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.12

Simplify the numerator.

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Step 2.1.2.16.1.12.1

Factor $3 {(x + 1)}^{\frac{1}{2}} x$ out of $- 9 {(x + 1)}^{\frac{1}{2}} x^{2} - 6 {(x + 1)}^{\frac{1}{2}} x \cdot 2$ .

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Step 2.1.2.16.1.12.1.1

Reorder the expression.

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Step 2.1.2.16.1.12.1.1.1

Move ${(x + 1)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 x^{2} {(x + 1)}^{\frac{1}{2}} - 6 {(x + 1)}^{\frac{1}{2}} x \cdot 2}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.12.1.1.2

Move ${(x + 1)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 x^{2} {(x + 1)}^{\frac{1}{2}} - 6 x \cdot (2 {(x + 1)}^{\frac{1}{2}})}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.12.1.1.3

Move $x$ .

$f'' (x) = \frac{6 {(x + 1)}^{\frac{3}{2}} x + 4 {(x + 1)}^{\frac{3}{2}} + \frac{- 9 x^{2} {(x + 1)}^{\frac{1}{2}} - 6 \cdot (2 x {(x + 1)}^{\frac{1}{2}})}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.12.1.2

Factor $3 {(x + 1)}^{\frac{1}{2}} x$ out of $- 9 x^{2} {(x + 1)}^{\frac{1}{2}}$ .

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

Step 2.1.2.16.1.12.1.3

Factor $3 {(x + 1)}^{\frac{1}{2}} x$ out of $- 6 \cdot 2 x {(x + 1)}^{\frac{1}{2}}$ .

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

Step 2.1.2.16.1.12.1.4

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

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

Step 2.1.2.16.1.12.2

Multiply $- 2$ by $2$ .

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

Step 2.1.2.16.1.13

To write $6 {(x + 1)}^{\frac{3}{2}} x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.1.2.16.1.14

Combine $6 {(x + 1)}^{\frac{3}{2}} x$ and $\frac{2}{2}$ .

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

Step 2.1.2.16.1.15

Combine the numerators over the common denominator.

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

Step 2.1.2.16.1.16

To write $4 {(x + 1)}^{\frac{3}{2}}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.1.2.16.1.17

Combine $4 {(x + 1)}^{\frac{3}{2}}$ and $\frac{2}{2}$ .

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

Step 2.1.2.16.1.18

Combine the numerators over the common denominator.

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

Step 2.1.2.16.1.19

Reorder terms.

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

Step 2.1.2.16.1.20

Rewrite $\frac{2 \cdot 4 {(x + 1)}^{\frac{3}{2}} + 2 \cdot 6 x {(x + 1)}^{\frac{3}{2}} + 3 x {(x + 1)}^{\frac{1}{2}} (- 3 x - 4)}{2}$ in a factored form.

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Step 2.1.2.16.1.20.1

Factor ${(x + 1)}^{\frac{1}{2}}$ out of $2 \cdot 4 {(x + 1)}^{\frac{3}{2}} + 2 \cdot 6 x {(x + 1)}^{\frac{3}{2}} + 3 x {(x + 1)}^{\frac{1}{2}} (- 3 x - 4)$ .

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Step 2.1.2.16.1.20.1.1

Factor ${(x + 1)}^{\frac{1}{2}}$ out of $2 \cdot 4 {(x + 1)}^{\frac{3}{2}}$ .

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

Step 2.1.2.16.1.20.1.2

Factor ${(x + 1)}^{\frac{1}{2}}$ out of $2 \cdot 6 x {(x + 1)}^{\frac{3}{2}}$ .

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

Step 2.1.2.16.1.20.1.3

Factor ${(x + 1)}^{\frac{1}{2}}$ out of $3 x {(x + 1)}^{\frac{1}{2}} (- 3 x - 4)$ .

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

Step 2.1.2.16.1.20.1.4

Factor ${(x + 1)}^{\frac{1}{2}}$ out of ${(x + 1)}^{\frac{1}{2}} (2 \cdot 4 {(x + 1)}^{\frac{2}{2}}) + {(x + 1)}^{\frac{1}{2}} (2 \cdot 6 x {(x + 1)}^{\frac{2}{2}})$ .

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

Step 2.1.2.16.1.20.1.5

Factor ${(x + 1)}^{\frac{1}{2}}$ out of ${(x + 1)}^{\frac{1}{2}} (2 \cdot 4 {(x + 1)}^{\frac{2}{2}} + 2 \cdot 6 x {(x + 1)}^{\frac{2}{2}}) + {(x + 1)}^{\frac{1}{2}} (3 x (- 3 x - 4))$ .

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

Step 2.1.2.16.1.20.2

Multiply $2$ by $4$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 {(x + 1)}^{\frac{2}{2}} + 2 \cdot (6 x {(x + 1)}^{\frac{2}{2}}) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.3

Divide $2$ by $2$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 (x + 1) + 2 \cdot (6 x {(x + 1)}^{\frac{2}{2}}) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.4

Simplify.

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 (x + 1) + 2 \cdot (6 x {(x + 1)}^{\frac{2}{2}}) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.5

Apply the distributive property.

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 \cdot 1 + 2 \cdot (6 x {(x + 1)}^{\frac{2}{2}}) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.6

Multiply $8$ by $1$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 2 \cdot (6 x {(x + 1)}^{\frac{2}{2}}) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.7

Multiply $2$ by $6$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x {(x + 1)}^{\frac{2}{2}} + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.8

Divide $2$ by $2$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x (x + 1) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.9

Simplify.

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x (x + 1) + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.10

Apply the distributive property.

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x \cdot x + 12 x \cdot 1 + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.11

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

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Step 2.1.2.16.1.20.11.1

Move $x$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 (x \cdot x) + 12 x \cdot 1 + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.11.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x \cdot 1 + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.12

Multiply $12$ by $1$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x + 3 x (- 3 x - 4))}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.13

Apply the distributive property.

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x + 3 x (- 3 x) + 3 x \cdot - 4)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.14

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x + 3 \cdot (- 3 x \cdot x) + 3 x \cdot - 4)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.15

Multiply $- 4$ by $3$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x + 3 \cdot (- 3 x \cdot x) - 12 x)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.16

Simplify each term.

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Step 2.1.2.16.1.20.16.1

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

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Step 2.1.2.16.1.20.16.1.1

Move $x$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x + 3 \cdot (- 3 (x \cdot x)) - 12 x)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.16.1.2

Multiply $x$ by $x$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x + 3 \cdot (- 3 x^{2}) - 12 x)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.16.2

Multiply $3$ by $- 3$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 x + 8 + 12 x^{2} + 12 x - 9 x^{2} - 12 x)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.17

Add $8 x$ and $12 x$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (20 x + 8 + 12 x^{2} - 9 x^{2} - 12 x)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.18

Subtract $12 x$ from $20 x$ .

$f'' (x) = \frac{\frac{{(x + 1)}^{\frac{1}{2}} (8 + 12 x^{2} - 9 x^{2} + 8 x)}{2}}{2 {(x + 1)}^{3}}$

Step 2.1.2.16.1.20.19

Subtract $9 x^{2}$ from $12 x^{2}$ .

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

Step 2.1.2.16.1.20.20

Reorder terms.

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

Step 2.1.2.16.2

Combine terms.

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Step 2.1.2.16.2.1

Rewrite $\frac{\frac{{(x + 1)}^{\frac{1}{2}} (3 x^{2} + 8 x + 8)}{2}}{2 {(x + 1)}^{3}}$ as a product.

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

Step 2.1.2.16.2.2

Multiply $\frac{{(x + 1)}^{\frac{1}{2}} (3 x^{2} + 8 x + 8)}{2}$ by $\frac{1}{2 {(x + 1)}^{3}}$ .

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

Step 2.1.2.16.2.3

Multiply $2$ by $2$ .

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

Step 2.1.2.16.2.4

Move ${(x + 1)}^{\frac{1}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

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

Step 2.1.2.16.2.5

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

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Step 2.1.2.16.2.5.1

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

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

Step 2.1.2.16.2.5.2

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

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

Step 2.1.2.16.2.5.3

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.1.2.16.2.5.4

Combine $3$ and $\frac{2}{2}$ .

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

Step 2.1.2.16.2.5.5

Combine the numerators over the common denominator.

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

Step 2.1.2.16.2.5.6

Simplify the numerator.

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Step 2.1.2.16.2.5.6.1

Multiply $3$ by $2$ .

$f'' (x) = \frac{3 x^{2} + 8 x + 8}{4 {(x + 1)}^{\frac{- 1 + 6}{2}}}$

Step 2.1.2.16.2.5.6.2

Add $- 1$ and $6$ .

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

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{3 x^{2} + 8 x + 8}{4 {(x + 1)}^{\frac{5}{2}}}$ .

$\frac{3 x^{2} + 8 x + 8}{4 {(x + 1)}^{\frac{5}{2}}}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $\frac{3 x^{2} + 8 x + 8}{4 {(x + 1)}^{\frac{5}{2}}} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$\frac{3 x^{2} + 8 x + 8}{4 {(x + 1)}^{\frac{5}{2}}} = 0$

Step 2.2.2

Set the numerator equal to zero.

$3 x^{2} + 8 x + 8 = 0$

Step 2.2.3

Solve the equation for $x$ .

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Step 2.2.3.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.2.3.2

Substitute the values $a = 3$ , $b = 8$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (3 \cdot 8)}}{2 \cdot 3}$

Step 2.2.3.3

Simplify.

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Step 2.2.3.3.1

Simplify the numerator.

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Step 2.2.3.3.1.1

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

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 3 \cdot 8}}{2 \cdot 3}$

Step 2.2.3.3.1.2

Multiply $- 4 \cdot 3 \cdot 8$ .

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Step 2.2.3.3.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{- 8 \pm \sqrt{64 - 12 \cdot 8}}{2 \cdot 3}$

Step 2.2.3.3.1.2.2

Multiply $- 12$ by $8$ .

$x = \frac{- 8 \pm \sqrt{64 - 96}}{2 \cdot 3}$

Step 2.2.3.3.1.3

Subtract $96$ from $64$ .

$x = \frac{- 8 \pm \sqrt{- 32}}{2 \cdot 3}$

Step 2.2.3.3.1.4

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{- 8 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 3}$

Step 2.2.3.3.1.5

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{- 8 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 3}$

Step 2.2.3.3.1.6

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

$x = \frac{- 8 \pm i \cdot \sqrt{32}}{2 \cdot 3}$

Step 2.2.3.3.1.7

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 2.2.3.3.1.7.1

Factor $16$ out of $32$ .

$x = \frac{- 8 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 3}$

Step 2.2.3.3.1.7.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 8 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 3}$

Step 2.2.3.3.1.8

Pull terms out from under the radical.

$x = \frac{- 8 \pm i \cdot (4 \sqrt{2})}{2 \cdot 3}$

Step 2.2.3.3.1.9

Move $4$ to the left of $i$ .

$x = \frac{- 8 \pm 4 i \sqrt{2}}{2 \cdot 3}$

Step 2.2.3.3.2

Multiply $2$ by $3$ .

$x = \frac{- 8 \pm 4 i \sqrt{2}}{6}$

Step 2.2.3.3.3

Simplify $\frac{- 8 \pm 4 i \sqrt{2}}{6}$ .

$x = \frac{- 4 \pm 2 i \sqrt{2}}{3}$

Step 2.2.3.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.2.3.4.1

Simplify the numerator.

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Step 2.2.3.4.1.1

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

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 3 \cdot 8}}{2 \cdot 3}$

Step 2.2.3.4.1.2

Multiply $- 4 \cdot 3 \cdot 8$ .

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Step 2.2.3.4.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{- 8 \pm \sqrt{64 - 12 \cdot 8}}{2 \cdot 3}$

Step 2.2.3.4.1.2.2

Multiply $- 12$ by $8$ .

$x = \frac{- 8 \pm \sqrt{64 - 96}}{2 \cdot 3}$

Step 2.2.3.4.1.3

Subtract $96$ from $64$ .

$x = \frac{- 8 \pm \sqrt{- 32}}{2 \cdot 3}$

Step 2.2.3.4.1.4

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{- 8 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 3}$

Step 2.2.3.4.1.5

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{- 8 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 3}$

Step 2.2.3.4.1.6

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

$x = \frac{- 8 \pm i \cdot \sqrt{32}}{2 \cdot 3}$

Step 2.2.3.4.1.7

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 2.2.3.4.1.7.1

Factor $16$ out of $32$ .

$x = \frac{- 8 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 3}$

Step 2.2.3.4.1.7.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 8 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 3}$

Step 2.2.3.4.1.8

Pull terms out from under the radical.

$x = \frac{- 8 \pm i \cdot (4 \sqrt{2})}{2 \cdot 3}$

Step 2.2.3.4.1.9

Move $4$ to the left of $i$ .

$x = \frac{- 8 \pm 4 i \sqrt{2}}{2 \cdot 3}$

Step 2.2.3.4.2

Multiply $2$ by $3$ .

$x = \frac{- 8 \pm 4 i \sqrt{2}}{6}$

Step 2.2.3.4.3

Simplify $\frac{- 8 \pm 4 i \sqrt{2}}{6}$ .

$x = \frac{- 4 \pm 2 i \sqrt{2}}{3}$

Step 2.2.3.4.4

Change the $\pm$ to $+$ .

$x = \frac{- 4 + 2 i \sqrt{2}}{3}$

Step 2.2.3.4.5

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{- 1 \cdot 4 + 2 i \sqrt{2}}{3}$

Step 2.2.3.4.6

Factor $- 1$ out of $2 i \sqrt{2}$ .

$x = \frac{- 1 \cdot 4 - (- 2 i \sqrt{2})}{3}$

Step 2.2.3.4.7

Factor $- 1$ out of $- 1 (4) - (- 2 i \sqrt{2})$ .

$x = \frac{- 1 (4 - 2 i \sqrt{2})}{3}$

Step 2.2.3.4.8

Move the negative in front of the fraction.

$x = - \frac{4 - 2 i \sqrt{2}}{3}$

Step 2.2.3.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.2.3.5.1

Simplify the numerator.

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Step 2.2.3.5.1.1

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

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 3 \cdot 8}}{2 \cdot 3}$

Step 2.2.3.5.1.2

Multiply $- 4 \cdot 3 \cdot 8$ .

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Step 2.2.3.5.1.2.1

Multiply $- 4$ by $3$ .

$x = \frac{- 8 \pm \sqrt{64 - 12 \cdot 8}}{2 \cdot 3}$

Step 2.2.3.5.1.2.2

Multiply $- 12$ by $8$ .

$x = \frac{- 8 \pm \sqrt{64 - 96}}{2 \cdot 3}$

Step 2.2.3.5.1.3

Subtract $96$ from $64$ .

$x = \frac{- 8 \pm \sqrt{- 32}}{2 \cdot 3}$

Step 2.2.3.5.1.4

Rewrite $- 32$ as $- 1 (32)$ .

$x = \frac{- 8 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 3}$

Step 2.2.3.5.1.5

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$x = \frac{- 8 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 3}$

Step 2.2.3.5.1.6

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

$x = \frac{- 8 \pm i \cdot \sqrt{32}}{2 \cdot 3}$

Step 2.2.3.5.1.7

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 2.2.3.5.1.7.1

Factor $16$ out of $32$ .

$x = \frac{- 8 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 3}$

Step 2.2.3.5.1.7.2

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 8 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 3}$

Step 2.2.3.5.1.8

Pull terms out from under the radical.

$x = \frac{- 8 \pm i \cdot (4 \sqrt{2})}{2 \cdot 3}$

Step 2.2.3.5.1.9

Move $4$ to the left of $i$ .

$x = \frac{- 8 \pm 4 i \sqrt{2}}{2 \cdot 3}$

Step 2.2.3.5.2

Multiply $2$ by $3$ .

$x = \frac{- 8 \pm 4 i \sqrt{2}}{6}$

Step 2.2.3.5.3

Simplify $\frac{- 8 \pm 4 i \sqrt{2}}{6}$ .

$x = \frac{- 4 \pm 2 i \sqrt{2}}{3}$

Step 2.2.3.5.4

Change the $\pm$ to $-$ .

$x = \frac{- 4 - 2 i \sqrt{2}}{3}$

Step 2.2.3.5.5

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{- 1 \cdot 4 - 2 i \sqrt{2}}{3}$

Step 2.2.3.5.6

Factor $- 1$ out of $- 2 i \sqrt{2}$ .

$x = \frac{- 1 \cdot 4 - (2 i \sqrt{2})}{3}$

Step 2.2.3.5.7

Factor $- 1$ out of $- 1 (4) - (2 i \sqrt{2})$ .

$x = \frac{- 1 (4 + 2 i \sqrt{2})}{3}$

Step 2.2.3.5.8

Move the negative in front of the fraction.

$x = - \frac{4 + 2 i \sqrt{2}}{3}$

Step 2.2.3.6

The final answer is the combination of both solutions.

$x = - \frac{4 - 2 i \sqrt{2}}{3}, - \frac{4 + 2 i \sqrt{2}}{3}$

Step 3

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

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Step 3.1

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Step 3.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 3.3

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

$\sqrt{x + 1} = 0$

Step 3.4

Solve for $x$ .

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Step 3.4.1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x + 1}}^{2} = 0^{2}$

Step 3.4.2

Simplify each side of the equation.

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Step 3.4.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

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

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Simplify ${({(x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 3.4.2.2.1.1

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

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Step 3.4.2.2.1.1.1

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

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

Step 3.4.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.4.2.2.1.1.2.1

Cancel the common factor.

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

Step 3.4.2.2.1.1.2.2

Rewrite the expression.

${(x + 1)}^{1} = 0^{2}$

Step 3.4.2.2.1.2

Simplify.

$x + 1 = 0^{2}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

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

$x + 1 = 0$

Step 3.4.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Step 4

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- 1, \infty)$

Step 5

Substitute any number from the interval $(- 1, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = \frac{3 {(0)}^{2} + 8 (0) + 8}{4 {((0) + 1)}^{\frac{5}{2}}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

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

$f'' (0) = \frac{3 \cdot 0 + 8 (0) + 8}{4 {(0 + 1)}^{\frac{5}{2}}}$

Step 5.2.1.2

Multiply $3$ by $0$ .

$f'' (0) = \frac{0 + 8 (0) + 8}{4 {(0 + 1)}^{\frac{5}{2}}}$

Step 5.2.1.3

Multiply $8$ by $0$ .

$f'' (0) = \frac{0 + 0 + 8}{4 {(0 + 1)}^{\frac{5}{2}}}$

Step 5.2.1.4

Add $0$ and $0$ .

$f'' (0) = \frac{0 + 8}{4 {(0 + 1)}^{\frac{5}{2}}}$

Step 5.2.1.5

Add $0$ and $8$ .

$f'' (0) = \frac{8}{4 {(0 + 1)}^{\frac{5}{2}}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Add $0$ and $1$ .

$f'' (0) = \frac{8}{4 \cdot 1^{\frac{5}{2}}}$

Step 5.2.2.2

One to any power is one.

$f'' (0) = \frac{8}{4 \cdot 1}$

Step 5.2.3

Simplify the expression.

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Step 5.2.3.1

Multiply $4$ by $1$ .

$f'' (0) = \frac{8}{4}$

Step 5.2.3.2

Divide $8$ by $4$ .

$f'' (0) = 2$

Step 5.2.4

The final answer is $2$ .

$2$

Step 5.3

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

Concave up on $(- 1, \infty)$ since $f'' (x)$ is positive

Step 6