Calculus Examples

$f (x) = x \sqrt{x + 9}$

Step 1

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

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 9}$ as ${(x + 9)}^{\frac{1}{2}}$ .

$f' (x) = \frac{d}{d x} (x {(x + 9)}^{\frac{1}{2}})$

Step 1.1.1.2

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 $g (x) = {(x + 9)}^{\frac{1}{2}}$ .

$f' (x) = x \frac{d}{d x} ({(x + 9)}^{\frac{1}{2}}) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x + 9$ .

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Step 1.1.1.3.1

To apply the Chain Rule, set $u$ as $x + 9$ .

$f' (x) = x (\frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (x + 9)) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.3.2

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

$f' (x) = x (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (x + 9)) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.3.3

Replace all occurrences of $u$ with $x + 9$ .

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

Step 1.1.1.4

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

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

Step 1.1.1.5

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

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

Step 1.1.1.6

Combine the numerators over the common denominator.

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

Step 1.1.1.7

Simplify the numerator.

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Step 1.1.1.7.1

Multiply $- 1$ by $2$ .

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

Step 1.1.1.7.2

Subtract $2$ from $1$ .

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

Step 1.1.1.8

Combine fractions.

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Step 1.1.1.8.1

Move the negative in front of the fraction.

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

Step 1.1.1.8.2

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

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

Step 1.1.1.8.3

Move ${(x + 9)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = x (\frac{1}{2 {(x + 9)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x + 9)) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.8.4

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

$f' (x) = \frac{x}{2 {(x + 9)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x + 9) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.9

By the Sum Rule, the derivative of $x + 9$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [9]$ .

$f' (x) = \frac{x}{2 {(x + 9)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (9)) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.10

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

$f' (x) = \frac{x}{2 {(x + 9)}^{\frac{1}{2}}} \cdot (1 + \frac{d}{d x} (9)) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.11

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f' (x) = \frac{x}{2 {(x + 9)}^{\frac{1}{2}}} \cdot (1 + 0) + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.12

Simplify the expression.

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Step 1.1.1.12.1

Add $1$ and $0$ .

$f' (x) = \frac{x}{2 {(x + 9)}^{\frac{1}{2}}} \cdot 1 + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.12.2

Multiply $\frac{x}{2 {(x + 9)}^{\frac{1}{2}}}$ by $1$ .

$f' (x) = \frac{x}{2 {(x + 9)}^{\frac{1}{2}}} + {(x + 9)}^{\frac{1}{2}} \frac{d}{d x} (x)$

Step 1.1.1.13

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

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

Step 1.1.1.14

Multiply ${(x + 9)}^{\frac{1}{2}}$ by $1$ .

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

Step 1.1.1.15

To write ${(x + 9)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(x + 9)}^{\frac{1}{2}}}{2 {(x + 9)}^{\frac{1}{2}}}$ .

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

Step 1.1.1.16

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

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

Step 1.1.1.17

Combine the numerators over the common denominator.

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

Step 1.1.1.18

Multiply ${(x + 9)}^{\frac{1}{2}}$ by ${(x + 9)}^{\frac{1}{2}}$ by adding the exponents.

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Step 1.1.1.18.1

Move ${(x + 9)}^{\frac{1}{2}}$ .

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

Step 1.1.1.18.2

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

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

Step 1.1.1.18.3

Combine the numerators over the common denominator.

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

Step 1.1.1.18.4

Add $1$ and $1$ .

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

Step 1.1.1.18.5

Divide $2$ by $2$ .

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

Step 1.1.1.19

Simplify ${(x + 9)}^{1} \cdot 2$ .

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

Step 1.1.1.20

Move $2$ to the left of $x + 9$ .

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

Step 1.1.1.21

Simplify.

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Step 1.1.1.21.1

Apply the distributive property.

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

Step 1.1.1.21.2

Simplify the numerator.

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Step 1.1.1.21.2.1

Multiply $2$ by $9$ .

$f' (x) = \frac{x + 2 x + 18}{2 {(x + 9)}^{\frac{1}{2}}}$

Step 1.1.1.21.2.2

Add $x$ and $2 x$ .

$f' (x) = \frac{3 x + 18}{2 {(x + 9)}^{\frac{1}{2}}}$

Step 1.1.1.21.3

Factor $3$ out of $3 x + 18$ .

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Step 1.1.1.21.3.1

Factor $3$ out of $3 x$ .

$f' (x) = \frac{3 (x) + 18}{2 {(x + 9)}^{\frac{1}{2}}}$

Step 1.1.1.21.3.2

Factor $3$ out of $18$ .

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

Step 1.1.1.21.3.3

Factor $3$ out of $3 x + 3 \cdot 6$ .

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Since $\frac{3}{2}$ is constant with respect to $x$ , the derivative of $\frac{3 (x + 6)}{2 {(x + 9)}^{\frac{1}{2}}}$ with respect to $x$ is $\frac{3}{2} \frac{d}{d x} [\frac{x + 6}{{(x + 9)}^{\frac{1}{2}}}]$ .

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

Step 1.1.2.2

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 $f (x) = x + 6$ and $g (x) = {(x + 9)}^{\frac{1}{2}}$ .

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

Step 1.1.2.3

Multiply the exponents in ${({(x + 9)}^{\frac{1}{2}})}^{2}$ .

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Step 1.1.2.3.1

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

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

Step 1.1.2.3.2

Cancel the common factor of $2$ .

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Step 1.1.2.3.2.1

Cancel the common factor.

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

Step 1.1.2.3.2.2

Rewrite the expression.

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

Step 1.1.2.4

Simplify.

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

Step 1.1.2.5

Differentiate.

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Step 1.1.2.5.1

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

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

Step 1.1.2.5.2

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

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

Step 1.1.2.5.3

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$f'' (x) = \frac{3}{2} \cdot \frac{{(x + 9)}^{\frac{1}{2}} (1 + 0) - (x + 6) \frac{d}{d x} {(x + 9)}^{\frac{1}{2}}}{x + 9}$

Step 1.1.2.5.4

Simplify the expression.

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Step 1.1.2.5.4.1

Add $1$ and $0$ .

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

Step 1.1.2.5.4.2

Multiply ${(x + 9)}^{\frac{1}{2}}$ by $1$ .

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

Step 1.1.2.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x + 9$ .

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Step 1.1.2.6.1

To apply the Chain Rule, set $u_{1}$ as $x + 9$ .

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

Step 1.1.2.6.2

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

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

Step 1.1.2.6.3

Replace all occurrences of $u_{1}$ with $x + 9$ .

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

Step 1.1.2.7

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

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

Step 1.1.2.8

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

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

Step 1.1.2.9

Combine the numerators over the common denominator.

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

Step 1.1.2.10

Simplify the numerator.

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Step 1.1.2.10.1

Multiply $- 1$ by $2$ .

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

Step 1.1.2.10.2

Subtract $2$ from $1$ .

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

Step 1.1.2.11

Combine fractions.

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Step 1.1.2.11.1

Move the negative in front of the fraction.

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

Step 1.1.2.11.2

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

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

Step 1.1.2.11.3

Move ${(x + 9)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.1.2.12

By the Sum Rule, the derivative of $x + 9$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [9]$ .

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

Step 1.1.2.13

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

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

Step 1.1.2.14

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f'' (x) = \frac{3}{2} \cdot \frac{{(x + 9)}^{\frac{1}{2}} - (x + 6) (\frac{1}{2 {(x + 9)}^{\frac{1}{2}}} \cdot (1 + 0))}{x + 9}$

Step 1.1.2.15

Combine fractions.

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Step 1.1.2.15.1

Add $1$ and $0$ .

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

Step 1.1.2.15.2

Multiply $\frac{1}{2 {(x + 9)}^{\frac{1}{2}}}$ by $1$ .

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

Step 1.1.2.15.3

Multiply $\frac{3}{2}$ by $\frac{{(x + 9)}^{\frac{1}{2}} - (x + 6) \frac{1}{2 {(x + 9)}^{\frac{1}{2}}}}{x + 9}$ .

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

Step 1.1.2.16

Simplify.

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Step 1.1.2.16.1

Apply the distributive property.

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

Step 1.1.2.16.2

Apply the distributive property.

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

Step 1.1.2.16.3

Apply the distributive property.

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

Step 1.1.2.16.4

Simplify the numerator.

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Step 1.1.2.16.4.1

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

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Step 1.1.2.16.4.1.1

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

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

Step 1.1.2.16.4.1.2

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

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

Step 1.1.2.16.4.1.3

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

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

Step 1.1.2.16.4.2

Let $u_{2} = {(x + 9)}^{\frac{1}{2}}$ . Substitute $u_{2}$ for all occurrences of ${(x + 9)}^{\frac{1}{2}}$ .

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Step 1.1.2.16.4.2.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.2.16.4.2.2

Multiply $u_{2}$ by $u_{2}$ by adding the exponents.

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Step 1.1.2.16.4.2.2.1

Move $u_{2}$ .

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

Step 1.1.2.16.4.2.2.2

Multiply $u_{2}$ by $u_{2}$ .

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

Step 1.1.2.16.4.3

Replace all occurrences of $u_{2}$ with ${(x + 9)}^{\frac{1}{2}}$ .

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

Step 1.1.2.16.4.4

Simplify.

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Step 1.1.2.16.4.4.1

Simplify each term.

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Step 1.1.2.16.4.4.1.1

Multiply the exponents in ${({(x + 9)}^{\frac{1}{2}})}^{2}$ .

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Step 1.1.2.16.4.4.1.1.1

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

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

Step 1.1.2.16.4.4.1.1.2

Cancel the common factor of $2$ .

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Step 1.1.2.16.4.4.1.1.2.1

Cancel the common factor.

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

Step 1.1.2.16.4.4.1.1.2.2

Rewrite the expression.

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

Step 1.1.2.16.4.4.1.2

Simplify.

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

Step 1.1.2.16.4.4.1.3

Apply the distributive property.

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

Step 1.1.2.16.4.4.1.4

Multiply $2$ by $9$ .

$f'' (x) = \frac{3 (\frac{2 x + 18 - x - 6}{2 {(x + 9)}^{\frac{1}{2}}})}{2 x + 2 \cdot 9}$

Step 1.1.2.16.4.4.2

Subtract $x$ from $2 x$ .

$f'' (x) = \frac{3 (\frac{x + 18 - 6}{2 {(x + 9)}^{\frac{1}{2}}})}{2 x + 2 \cdot 9}$

Step 1.1.2.16.4.4.3

Subtract $6$ from $18$ .

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

Step 1.1.2.16.5

Combine terms.

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Step 1.1.2.16.5.1

Combine $3$ and $\frac{x + 12}{2 {(x + 9)}^{\frac{1}{2}}}$ .

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

Step 1.1.2.16.5.2

Multiply $2$ by $9$ .

$f'' (x) = \frac{\frac{3 (x + 12)}{2 {(x + 9)}^{\frac{1}{2}}}}{2 x + 18}$

Step 1.1.2.16.5.3

Rewrite $\frac{\frac{3 (x + 12)}{2 {(x + 9)}^{\frac{1}{2}}}}{2 x + 18}$ as a product.

$f'' (x) = \frac{3 (x + 12)}{2 {(x + 9)}^{\frac{1}{2}}} \cdot \frac{1}{2 x + 18}$

Step 1.1.2.16.5.4

Multiply $\frac{3 (x + 12)}{2 {(x + 9)}^{\frac{1}{2}}}$ by $\frac{1}{2 x + 18}$ .

$f'' (x) = \frac{3 (x + 12)}{2 {(x + 9)}^{\frac{1}{2}} (2 x + 18)}$

Step 1.1.2.16.6

Simplify the denominator.

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Step 1.1.2.16.6.1

Factor $2$ out of $2 x + 18$ .

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Step 1.1.2.16.6.1.1

Factor $2$ out of $2 x$ .

$f'' (x) = \frac{3 (x + 12)}{2 {(x + 9)}^{\frac{1}{2}} (2 (x) + 18)}$

Step 1.1.2.16.6.1.2

Factor $2$ out of $18$ .

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

Step 1.1.2.16.6.1.3

Factor $2$ out of $2 x + 2 \cdot 9$ .

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

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

Step 1.1.2.16.6.2

Combine exponents.

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Step 1.1.2.16.6.2.1

Multiply $2$ by $2$ .

$f'' (x) = \frac{3 (x + 12)}{4 {(x + 9)}^{\frac{1}{2}} (x + 9)}$

Step 1.1.2.16.6.2.2

Raise $x + 9$ to the power of $1$ .

$f'' (x) = \frac{3 (x + 12)}{4 ((x + 9) {(x + 9)}^{\frac{1}{2}})}$

Step 1.1.2.16.6.2.3

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

$f'' (x) = \frac{3 (x + 12)}{4 {(x + 9)}^{1 + \frac{1}{2}}}$

Step 1.1.2.16.6.2.4

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

$f'' (x) = \frac{3 (x + 12)}{4 {(x + 9)}^{\frac{2}{2} + \frac{1}{2}}}$

Step 1.1.2.16.6.2.5

Combine the numerators over the common denominator.

$f'' (x) = \frac{3 (x + 12)}{4 {(x + 9)}^{\frac{2 + 1}{2}}}$

Step 1.1.2.16.6.2.6

Add $2$ and $1$ .

$f'' (x) = \frac{3 (x + 12)}{4 {(x + 9)}^{\frac{3}{2}}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{3 (x + 12)}{4 {(x + 9)}^{\frac{3}{2}}}$ .

$\frac{3 (x + 12)}{4 {(x + 9)}^{\frac{3}{2}}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{3 (x + 12)}{4 {(x + 9)}^{\frac{3}{2}}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{3 (x + 12)}{4 {(x + 9)}^{\frac{3}{2}}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$3 (x + 12) = 0$

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

Divide each term in $3 (x + 12) = 0$ by $3$ and simplify.

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Step 1.2.3.1.1

Divide each term in $3 (x + 12) = 0$ by $3$ .

$\frac{3 (x + 12)}{3} = \frac{0}{3}$

Step 1.2.3.1.2

Simplify the left side.

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Step 1.2.3.1.2.1

Cancel the common factor of $3$ .

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Step 1.2.3.1.2.1.1

Cancel the common factor.

$\frac{3 (x + 12)}{3} = \frac{0}{3}$

Step 1.2.3.1.2.1.2

Divide $x + 12$ by $1$ .

$x + 12 = \frac{0}{3}$

Step 1.2.3.1.3

Simplify the right side.

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Step 1.2.3.1.3.1

Divide $0$ by $3$ .

$x + 12 = 0$

Step 1.2.3.2

Subtract $12$ from both sides of the equation.

$x = - 12$

Step 1.2.4

Exclude the solutions that do not make $\frac{3 (x + 12)}{4 {(x + 9)}^{\frac{3}{2}}} = 0$ true.

$No solution$

Step 2

Find the domain of $f (x) = x \sqrt{x + 9}$ .

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Step 2.1

Set the radicand in $\sqrt{x + 9}$ greater than or equal to $0$ to find where the expression is defined.

$x + 9 \geq 0$

Step 2.2

Subtract $9$ from both sides of the inequality.

$x \geq - 9$

Step 2.3

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

Interval Notation:

$[- 9, \infty)$

Set-Builder Notation:

${x | x \geq - 9}$

Interval Notation:

$[- 9, \infty)$

Set-Builder Notation:

${x | x \geq - 9}$

Step 3

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

$[- 9, \infty)$

Step 4

Substitute any number from the interval $[- 9, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

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

$f'' (0) = \frac{3 ((0) + 12)}{4 {((0) + 9)}^{\frac{3}{2}}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Factor $4$ out of $3 ((0) + 12)$ .

$f'' (0) = \frac{4 (3 (0 + 3))}{4 {((0) + 9)}^{\frac{3}{2}}}$

Step 4.2.2

Cancel the common factors.

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Step 4.2.2.1

Factor $4$ out of $4 {((0) + 9)}^{\frac{3}{2}}$ .

$f'' (0) = \frac{4 (3 (0 + 3))}{4 ({((0) + 9)}^{\frac{3}{2}})}$

Step 4.2.2.2

Cancel the common factor.

$f'' (0) = \frac{4 (3 (0 + 3))}{4 {((0) + 9)}^{\frac{3}{2}}}$

Step 4.2.2.3

Rewrite the expression.

$f'' (0) = \frac{3 (0 + 3)}{{((0) + 9)}^{\frac{3}{2}}}$

Step 4.2.3

Add $0$ and $3$ .

$f'' (0) = \frac{3 \cdot 3}{{(0 + 9)}^{\frac{3}{2}}}$

Step 4.2.4

Simplify the denominator.

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Step 4.2.4.1

Add $0$ and $9$ .

$f'' (0) = \frac{3 \cdot 3}{9^{\frac{3}{2}}}$

Step 4.2.4.2

Rewrite $9$ as $3^{2}$ .

$f'' (0) = \frac{3 \cdot 3}{{(3^{2})}^{\frac{3}{2}}}$

Step 4.2.4.3

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

$f'' (0) = \frac{3 \cdot 3}{3^{2 (\frac{3}{2})}}$

Step 4.2.4.4

Cancel the common factor of $2$ .

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Step 4.2.4.4.1

Cancel the common factor.

$f'' (0) = \frac{3 \cdot 3}{3^{2 (\frac{3}{2})}}$

Step 4.2.4.4.2

Rewrite the expression.

$f'' (0) = \frac{3 \cdot 3}{3^{3}}$

Step 4.2.4.5

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

$f'' (0) = \frac{3 \cdot 3}{27}$

Step 4.2.5

Reduce the expression by cancelling the common factors.

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Step 4.2.5.1

Multiply $3$ by $3$ .

$f'' (0) = \frac{9}{27}$

Step 4.2.5.2

Cancel the common factor of $9$ and $27$ .

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Step 4.2.5.2.1

Factor $9$ out of $9$ .

$f'' (0) = \frac{9 (1)}{27}$

Step 4.2.5.2.2

Cancel the common factors.

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Step 4.2.5.2.2.1

Factor $9$ out of $27$ .

$f'' (0) = \frac{9 \cdot 1}{9 \cdot 3}$

Step 4.2.5.2.2.2

Cancel the common factor.

$f'' (0) = \frac{9 \cdot 1}{9 \cdot 3}$

Step 4.2.5.2.2.3

Rewrite the expression.

$f'' (0) = \frac{1}{3}$

Step 4.2.6

The final answer is $\frac{1}{3}$ .

$\frac{1}{3}$

Step 4.3

The graph is concave up on the interval $[- 9, \infty)$ because $f'' (0)$ is positive.

Concave up on $[- 9, \infty)$ since $f'' (x)$ is positive

Step 5