Find the Concavity y=x square root of 2-x^2

$y = x \sqrt{2 - x^{2}}$

Step 1

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

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

Step 2

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

Find the second derivative.

Find the first derivative.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Combine fractions.

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

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

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

Combine fractions.

Multiply $2$ by $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

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

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

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

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

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

Reorder terms.

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

Find the second derivative.

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Differentiate.

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

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

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

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

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

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

Multiply $2$ by $- 2$ .

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

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

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

Add $- 4 x$ and $0$ .

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Combine fractions.

Move the negative in front of the fraction.

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

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

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

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

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

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

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

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

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

Multiply $2$ by $- 1$ .

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

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

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

Simplify terms.

Add $- 2 x$ and $0$ .

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

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

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

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

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply $- 1$ by $- 1$ .

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

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

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

Simplify.

Simplify the numerator.

Rewrite using the commutative property of multiplication.

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

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

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

Simplify the numerator.

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

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

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

Factor $2$ out of $2$ .

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

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

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

Rewrite $1$ as $1^{2}$ .

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

Reorder $- x^{2}$ and $1^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

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

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

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

Factor $2 x$ out of $2 (1 + x) (1 - x) x$ .

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

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

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

Combine exponents.

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

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

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

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 1$ by $- 2$ .

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

Multiply $- 2$ by $2$ .

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

Expand $(1 + x) (1 - x)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

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

Multiply $- x$ by $1$ .

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

Multiply $x$ by $1$ .

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

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Add $- x$ and $x$ .

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

Add $1$ and $0$ .

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

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

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

Add $- 4$ and $1$ .

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

Combine terms.

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $2$ .

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

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

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

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

Set the second derivative equal to $0$ .

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

Set the numerator equal to zero.

$2 x (x^{2} - 3) = 0$

Solve the equation for $x$ .

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

$x = 0$

$x^{2} - 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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

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

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

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

The final solution is all the values that make $2 x (x^{2} - 3) = 0$ true.

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

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

$x = 0$

Step 3

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

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

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

Solve for $x$ .

Subtract $2$ from both sides of the inequality.

$- x^{2} \geq - 2$

Divide each term in $- x^{2} \geq - 2$ by $- 1$ and simplify.

Divide each term in $- x^{2} \geq - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x^{2}}{- 1} \leq \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} \leq \frac{- 2}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} \leq \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x^{2} \leq 2$

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

$\sqrt{x^{2}} \leq \sqrt{2}$

Simplify the left side.

Pull terms out from under the radical.

$| x | \leq \sqrt{2}$

Write $| x | \leq \sqrt{2}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq \sqrt{2}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq \sqrt{2}$

Write as a piecewise.

${\begin{cases} x \leq \sqrt{2} & x \geq 0 \\ - x \leq \sqrt{2} & x < 0 \end{cases}$

Find the intersection of $x \leq \sqrt{2}$ and $x \geq 0$ .

$0 \leq x \leq \sqrt{2}$

Solve $- x \leq \sqrt{2}$ when $x < 0$ .

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ and simplify.

Divide each term in $- x \leq \sqrt{2}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{\sqrt{2}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{\sqrt{2}}{- 1}$

Divide $x$ by $1$ .

$x \geq \frac{\sqrt{2}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{2}}{- 1}$ .

$x \geq - 1 \cdot \sqrt{2}$

Rewrite $- 1 \cdot \sqrt{2}$ as $- \sqrt{2}$ .

$x \geq - \sqrt{2}$

Find the intersection of $x \geq - \sqrt{2}$ and $x < 0$ .

$- \sqrt{2} \leq x < 0$

Find the union of the solutions.

$- \sqrt{2} \leq x \leq \sqrt{2}$

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

Interval Notation:

$[- \sqrt{2}, \sqrt{2}]$

Set-Builder Notation:

${x | - \sqrt{2} \leq x \leq \sqrt{2}}$

Interval Notation:

$[- \sqrt{2}, \sqrt{2}]$

Set-Builder Notation:

${x | - \sqrt{2} \leq x \leq \sqrt{2}}$

Step 4

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

$[- \sqrt{2}, 0) \cup (0, \sqrt{2}]$

Step 5

Substitute any number from the interval $[- \sqrt{2}, 0)$ into the second derivative and evaluate to determine the concavity.

Replace the variable $x$ with $- 1$ in the expression.

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

Simplify the result.

Multiply $2$ by $- 1$ .

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

Simplify the denominator.

Simplify each term.

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

Multiply $- 1$ by ${(- 1)}^{2}$ .

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

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

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

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

Add $1$ and $2$ .

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

Raise $- 1$ to the power of $3$ .

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

Add $- 1$ and $2$ .

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

One to any power is one.

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

Simplify the numerator.

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

$f'' (- 1) = \frac{- 2 (1 - 3)}{1}$

Subtract $3$ from $1$ .

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

Simplify the expression.

Multiply $- 2$ by $- 2$ .

$f'' (- 1) = \frac{4}{1}$

Divide $4$ by $1$ .

$f'' (- 1) = 4$

The final answer is $4$ .

$4$

The graph is concave up on the interval $[- \sqrt{2}, 0)$ because $f'' (- 1)$ is positive.

Concave up on $[- \sqrt{2}, 0)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(0, \sqrt{2}]$ into the second derivative and evaluate to determine the concavity.

Replace the variable $x$ with $1$ in the expression.

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

Simplify the result.

Multiply $2$ by $1$ .

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

Simplify the denominator.

Simplify each term.

One to any power is one.

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

Multiply $- 1$ by $1$ .

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

Add $- 1$ and $2$ .

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

One to any power is one.

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

Simplify the numerator.

One to any power is one.

$f'' (1) = \frac{2 (1 - 3)}{1}$

Subtract $3$ from $1$ .

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

Simplify the expression.

Multiply $2$ by $- 2$ .

$f'' (1) = \frac{- 4}{1}$

Divide $- 4$ by $1$ .

$f'' (1) = - 4$

The final answer is $- 4$ .

$- 4$

The graph is concave down on the interval $(0, \sqrt{2}]$ because $f'' (1)$ is negative.

Concave down on $(0, \sqrt{2}]$ since $f'' (x)$ is negative

Step 7

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $[- \sqrt{2}, 0)$ since $f'' (x)$ is positive

Concave down on $(0, \sqrt{2}]$ since $f'' (x)$ is negative

Step 8

Find the Concavity y=x square root of 2-x^2

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