Calculus Examples

$\sqrt{x^{2} + 2} - x$

Step 1

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

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

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

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

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

Step 2.1.1.2

Evaluate $\frac{d}{d x} [\sqrt{x^{2} + 2}]$ .

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Step 2.1.1.2.1

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

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

Step 2.1.1.2.2

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

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Step 2.1.1.2.2.1

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

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

Step 2.1.1.2.2.2

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

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

Step 2.1.1.2.2.3

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

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

Step 2.1.1.2.3

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

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

Step 2.1.1.2.4

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

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

Step 2.1.1.2.5

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

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

Step 2.1.1.2.6

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

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

Step 2.1.1.2.7

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

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

Step 2.1.1.2.8

Combine the numerators over the common denominator.

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

Step 2.1.1.2.9

Simplify the numerator.

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Step 2.1.1.2.9.1

Multiply $- 1$ by $2$ .

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

Step 2.1.1.2.9.2

Subtract $2$ from $1$ .

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

Step 2.1.1.2.10

Move the negative in front of the fraction.

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

Step 2.1.1.2.11

Add $2 x$ and $0$ .

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

Step 2.1.1.2.12

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

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

Step 2.1.1.2.13

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

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

Step 2.1.1.2.14

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

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

Step 2.1.1.2.15

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

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

Step 2.1.1.2.16

Cancel the common factor.

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

Step 2.1.1.2.17

Rewrite the expression.

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

Step 2.1.1.3

Evaluate $\frac{d}{d x} [- x]$ .

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Step 2.1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 2.1.1.3.2

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

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

Step 2.1.1.3.3

Multiply $- 1$ by $1$ .

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

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

By the Sum Rule, the derivative of $\frac{x}{{(x^{2} + 2)}^{\frac{1}{2}}} - 1$ with respect to $x$ is $\frac{d}{d x} [\frac{x}{{(x^{2} + 2)}^{\frac{1}{2}}}] + \frac{d}{d x} [- 1]$ .

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

Step 2.1.2.2

Evaluate $\frac{d}{d x} [\frac{x}{{(x^{2} + 2)}^{\frac{1}{2}}}]$ .

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Step 2.1.2.2.1

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

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

Step 2.1.2.2.2

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

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

Step 2.1.2.2.3

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

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Step 2.1.2.2.3.1

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

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

Step 2.1.2.2.3.2

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

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

Step 2.1.2.2.3.3

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

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

Step 2.1.2.2.4

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

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

Step 2.1.2.2.5

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

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

Step 2.1.2.2.6

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

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

Step 2.1.2.2.7

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

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

Step 2.1.2.2.8

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

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

Step 2.1.2.2.9

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

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

Step 2.1.2.2.10

Combine the numerators over the common denominator.

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

Step 2.1.2.2.11

Simplify the numerator.

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Step 2.1.2.2.11.1

Multiply $- 1$ by $2$ .

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

Step 2.1.2.2.11.2

Subtract $2$ from $1$ .

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

Step 2.1.2.2.12

Move the negative in front of the fraction.

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

Step 2.1.2.2.13

Add $2 x$ and $0$ .

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

Step 2.1.2.2.14

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

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

Step 2.1.2.2.15

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

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

Step 2.1.2.2.16

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

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

Step 2.1.2.2.17

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

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

Step 2.1.2.2.18

Cancel the common factor.

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

Step 2.1.2.2.19

Rewrite the expression.

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

Step 2.1.2.2.20

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

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

Step 2.1.2.2.21

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

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

Step 2.1.2.2.22

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

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

Step 2.1.2.2.23

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

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

Step 2.1.2.2.24

Add $1$ and $1$ .

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

Step 2.1.2.2.25

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

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

Step 2.1.2.2.26

Combine the numerators over the common denominator.

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

Step 2.1.2.2.27

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

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Step 2.1.2.2.27.1

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

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

Step 2.1.2.2.27.2

Combine the numerators over the common denominator.

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

Step 2.1.2.2.27.3

Add $1$ and $1$ .

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

Step 2.1.2.2.27.4

Divide $2$ by $2$ .

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

Step 2.1.2.2.28

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

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

Step 2.1.2.2.29

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

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

Step 2.1.2.2.30

Add $0$ and $2$ .

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

Step 2.1.2.2.31

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

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Step 2.1.2.2.31.1

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

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

Step 2.1.2.2.31.2

Cancel the common factor of $2$ .

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Step 2.1.2.2.31.2.1

Cancel the common factor.

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

Step 2.1.2.2.31.2.2

Rewrite the expression.

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

Step 2.1.2.2.32

Simplify.

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

Step 2.1.2.2.33

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

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

Step 2.1.2.2.34

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

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

Step 2.1.2.2.35

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

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Step 2.1.2.2.35.1

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

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Step 2.1.2.2.35.1.1

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

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

Step 2.1.2.2.35.1.2

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

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

Step 2.1.2.2.35.2

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

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

Step 2.1.2.2.35.3

Combine the numerators over the common denominator.

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

Step 2.1.2.2.35.4

Add $1$ and $2$ .

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

Step 2.1.2.3

Differentiate using the Constant Rule.

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Step 2.1.2.3.1

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

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

Step 2.1.2.3.2

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

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

Step 2.1.3

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

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

Step 2.2

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

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Step 2.2.1

Set the second derivative equal to $0$ .

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

Step 2.2.2

Set the numerator equal to zero.

$2 = 0$

Step 2.2.3

Since $2 \neq 0$ , there are no solutions.

No solution

Step 3

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

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Step 3.1

Set the radicand in $\sqrt{x^{2} + 2}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} + 2 \geq 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Subtract $2$ from both sides of the inequality.

$x^{2} \geq - 2$

Step 3.2.2

Since the left side has an even power, it is always positive for all real numbers.

All real numbers

Step 3.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

The graph is concave up because the second derivative is positive.

The graph is concave up

Step 5