Find the Maximum/Minimum Value x square root of 4-x

$x \sqrt{4 - x}$

Find the first derivative of the function.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 - x}$ as ${(4 - x)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [x {(4 - x)}^{\frac{1}{2}}]$

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 $g (x) = {(4 - x)}^{\frac{1}{2}}$ .

$x \frac{d}{d x} [{(4 - x)}^{\frac{1}{2}}] + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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 $g (x) = 4 - x$ .

To apply the Chain Rule, set $u$ as $4 - x$ .

$x (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$x (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Replace all occurrences of $u$ with $4 - x$ .

$x (\frac{1}{2} {(4 - x)}^{\frac{1}{2} - 1} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$x (\frac{1}{2} {(4 - x)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$x (\frac{1}{2} {(4 - x)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Combine the numerators over the common denominator.

$x (\frac{1}{2} {(4 - x)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$x (\frac{1}{2} {(4 - x)}^{\frac{1 - 2}{2}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Subtract $2$ from $1$ .

$x (\frac{1}{2} {(4 - x)}^{\frac{- 1}{2}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Combine fractions.

Move the negative in front of the fraction.

$x (\frac{1}{2} {(4 - x)}^{- \frac{1}{2}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$x (\frac{{(4 - x)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Move ${(4 - x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x (\frac{1}{2 {(4 - x)}^{\frac{1}{2}}} \frac{d}{d x} [4 - x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} \frac{d}{d x} [4 - x] + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} (\frac{d}{d x} [4] + \frac{d}{d x} [- x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} (0 + \frac{d}{d x} [- x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} \frac{d}{d x} [- x] + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} (- \frac{d}{d x} [x]) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} (- 1 \cdot 1) + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Combine fractions.

Multiply $- 1$ by $1$ .

$\frac{x}{2 {(4 - x)}^{\frac{1}{2}}} \cdot - 1 + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

$\frac{x \cdot - 1}{2 {(4 - x)}^{\frac{1}{2}}} + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Simplify the expression.

Move $- 1$ to the left of $x$ .

$\frac{- 1 \cdot x}{2 {(4 - x)}^{\frac{1}{2}}} + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Rewrite $- 1 x$ as $- x$ .

$\frac{- x}{2 {(4 - x)}^{\frac{1}{2}}} + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Move the negative in front of the fraction.

$- \frac{x}{2 {(4 - x)}^{\frac{1}{2}}} + {(4 - x)}^{\frac{1}{2}} \frac{d}{d x} [x]$

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

Move ${(4 - x)}^{\frac{1}{2}}$ .

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

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

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

Simplify ${(4 - x)}^{1} \cdot 2$ .

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

Move $2$ to the left of $4 - x$ .

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

Simplify.

Apply the distributive property.

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

Simplify the numerator.

Simplify each term.

Multiply $2$ by $4$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $2 x$ from $- x$ .

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

Factor $- 1$ out of $- 3 x$ .

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

Rewrite $8$ as $- 1 (- 8)$ .

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

Factor $- 1$ out of $- (3 x) - 1 (- 8)$ .

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

Rewrite $- (3 x - 8)$ as $- 1 (3 x - 8)$ .

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

Move the negative in front of the fraction.

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

Find the second derivative of the function.

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{3 x - 8}{2 {(4 - x)}^{\frac{1}{2}}}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [\frac{3 x - 8}{{(4 - x)}^{\frac{1}{2}}}]$ .

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

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 $f (x) = 3 x - 8$ and $g (x) = {(4 - x)}^{\frac{1}{2}}$ .

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

Multiply the exponents in ${({(4 - x)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Differentiate.

By the Sum Rule, the derivative of $3 x - 8$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [- 8]$ .

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

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

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

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

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

Multiply $3$ by $1$ .

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

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

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

Simplify the expression.

Add $3$ and $0$ .

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

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

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

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 $g (x) = 4 - x$ .

To apply the Chain Rule, set $u_{1}$ as $4 - x$ .

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

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

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

Replace all occurrences of $u_{1}$ with $4 - x$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Combine fractions.

Move the negative in front of the fraction.

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

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

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

Move ${(4 - x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

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

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

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

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

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

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

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

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $3 x - 8$ by $1$ .

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

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

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

Combine fractions.

Multiply $3 x - 8$ by $1$ .

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

Multiply $\frac{3 {(4 - x)}^{\frac{1}{2}} + (3 x - 8) \frac{1}{2 {(4 - x)}^{\frac{1}{2}}}}{4 - x}$ by $\frac{1}{2}$ .

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

Move $2$ to the left of $4 - x$ .

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

Simplify.

Apply the distributive property.

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

Simplify the numerator.

Let $u_{2} = {(4 - x)}^{\frac{1}{2}}$ . Substitute $u_{2}$ for all occurrences of ${(4 - x)}^{\frac{1}{2}}$ .

Rewrite using the commutative property of multiplication.

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

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

Move $u_{2}$ .

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

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

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

Multiply $3$ by $2$ .

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

Replace all occurrences of $u_{2}$ with ${(4 - x)}^{\frac{1}{2}}$ .

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

Simplify.

Simplify each term.

Multiply the exponents in ${({(4 - x)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $6$ by $4$ .

$f'' (x) = - \frac{\frac{24 + 6 (- x) + 3 x - 8}{2 {(4 - x)}^{\frac{1}{2}}}}{2 \cdot 4 + 2 (- x)}$

Multiply $- 1$ by $6$ .

$f'' (x) = - \frac{\frac{24 - 6 x + 3 x - 8}{2 {(4 - x)}^{\frac{1}{2}}}}{2 \cdot 4 + 2 (- x)}$

Subtract $8$ from $24$ .

$f'' (x) = - \frac{\frac{- 6 x + 3 x + 16}{2 {(4 - x)}^{\frac{1}{2}}}}{2 \cdot 4 + 2 (- x)}$

Add $- 6 x$ and $3 x$ .

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

Combine terms.

Multiply $2$ by $4$ .

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

Multiply $- 1$ by $2$ .

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

Rewrite $\frac{\frac{- 3 x + 16}{2 {(4 - x)}^{\frac{1}{2}}}}{8 - 2 x}$ as a product.

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

Multiply $\frac{- 3 x + 16}{2 {(4 - x)}^{\frac{1}{2}}}$ by $\frac{1}{8 - 2 x}$ .

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

Simplify the denominator.

Factor $2$ out of $8 - 2 x$ .

Factor $2$ out of $8$ .

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

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

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

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

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

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

Combine exponents.

Multiply $2$ by $2$ .

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

Raise $4 - x$ to the power of $1$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Add $2$ and $1$ .

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

Factor $- 1$ out of $- 3 x$ .

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

Rewrite $16$ as $- 1 (- 16)$ .

$f'' (x) = - \frac{- (3 x) - 1 \cdot - 16}{4 {(4 - x)}^{\frac{3}{2}}}$

Factor $- 1$ out of $- (3 x) - 1 (- 16)$ .

$f'' (x) = - \frac{- (3 x - 16)}{4 {(4 - x)}^{\frac{3}{2}}}$

Rewrite $- (3 x - 16)$ as $- 1 (3 x - 16)$ .

$f'' (x) = - \frac{- 1 (3 x - 16)}{4 {(4 - x)}^{\frac{3}{2}}}$

Move the negative in front of the fraction.

$f'' (x) = \frac{3 x - 16}{4 {(4 - x)}^{\frac{3}{2}}}$

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (\frac{3 x - 16}{4 {(4 - x)}^{\frac{3}{2}}})$

Multiply $\frac{3 x - 16}{4 {(4 - x)}^{\frac{3}{2}}}$ by $1$ .

$f'' (x) = \frac{3 x - 16}{4 {(4 - x)}^{\frac{3}{2}}}$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

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

Find the first derivative.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 - x}$ as ${(4 - x)}^{\frac{1}{2}}$ .

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

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

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 $g (x) = 4 - x$ .

To apply the Chain Rule, set $u$ as $4 - x$ .

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

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

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

Replace all occurrences of $u$ with $4 - x$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Combine fractions.

Move the negative in front of the fraction.

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

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

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

Move ${(4 - x)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine fractions.

Multiply $- 1$ by $1$ .

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

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

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

Simplify the expression.

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

Move ${(4 - x)}^{\frac{1}{2}}$ .

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

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

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

Combine the numerators over the common denominator.

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

Add $1$ and $1$ .

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

Divide $2$ by $2$ .

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

Simplify ${(4 - x)}^{1} \cdot 2$ .

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

Move $2$ to the left of $4 - x$ .

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

Simplify.

Apply the distributive property.

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

Simplify the numerator.

Simplify each term.

Multiply $2$ by $4$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $2 x$ from $- x$ .

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

Factor $- 1$ out of $- 3 x$ .

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

Rewrite $8$ as $- 1 (- 8)$ .

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

Factor $- 1$ out of $- (3 x) - 1 (- 8)$ .

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

Rewrite $- (3 x - 8)$ as $- 1 (3 x - 8)$ .

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

Move the negative in front of the fraction.

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

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

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

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

Set the first derivative equal to $0$ .

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

Set the numerator equal to zero.

$3 x - 8 = 0$

Solve the equation for $x$ .

Add $8$ to both sides of the equation.

$3 x = 8$

Divide each term in $3 x = 8$ by $3$ and simplify.

Divide each term in $3 x = 8$ by $3$ .

$\frac{3 x}{3} = \frac{8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{8}{3}$

Divide $x$ by $1$ .

$x = \frac{8}{3}$

Find the values where the derivative is undefined.

Convert expressions with fractional exponents to radicals.

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$- \frac{3 x - 8}{2 \sqrt{{(4 - x)}^{1}}}$

Anything raised to $1$ is the base itself.

$- \frac{3 x - 8}{2 \sqrt{4 - x}}$

Set the denominator in $\frac{3 x - 8}{2 \sqrt{4 - x}}$ equal to $0$ to find where the expression is undefined.

$2 \sqrt{4 - x} = 0$

Solve for $x$ .

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

${(2 \sqrt{4 - x})}^{2} = 0^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 - x}$ as ${(4 - x)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${(2 {(4 - x)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(4 - x)}^{\frac{1}{2}}$ .

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

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

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

Multiply the exponents in ${({(4 - x)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$4 {(4 - x)}^{1} = 0^{2}$

Simplify.

$4 (4 - x) = 0^{2}$

Apply the distributive property.

$4 \cdot 4 + 4 (- x) = 0^{2}$

Multiply.

Multiply $4$ by $4$ .

$16 + 4 (- x) = 0^{2}$

Multiply $- 1$ by $4$ .

$16 - 4 x = 0^{2}$

Simplify the right side.

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

$16 - 4 x = 0$

Solve for $x$ .

Subtract $16$ from both sides of the equation.

$- 4 x = - 16$

Divide each term in $- 4 x = - 16$ by $- 4$ and simplify.

Divide each term in $- 4 x = - 16$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{- 16}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{- 16}{- 4}$

Divide $x$ by $1$ .

$x = \frac{- 16}{- 4}$

Simplify the right side.

Divide $- 16$ by $- 4$ .

$x = 4$

Set the radicand in $\sqrt{4 - x}$ less than $0$ to find where the expression is undefined.

$4 - x < 0$

Solve for $x$ .

Subtract $4$ from both sides of the inequality.

$- x < - 4$

Divide each term in $- x < - 4$ by $- 1$ and simplify.

Divide each term in $- x < - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{- 4}{- 1}$

Divide $x$ by $1$ .

$x > \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x > 4$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x \geq 4$

$[4, \infty)$

$x \geq 4$

$[4, \infty)$

Critical points to evaluate.

$x = \frac{8}{3}, 4$

Evaluate the second derivative at $x = \frac{8}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{3 (\frac{8}{3}) - 16}{4 {(4 - (\frac{8}{3}))}^{\frac{3}{2}}}$

Evaluate the second derivative.

Simplify the numerator.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (\frac{8}{3}) - 16}{4 {(4 - \frac{8}{3})}^{\frac{3}{2}}}$

Rewrite the expression.

$\frac{8 - 16}{4 {(4 - \frac{8}{3})}^{\frac{3}{2}}}$

Subtract $16$ from $8$ .

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

Simplify the denominator.

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Combine $4$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $4$ by $3$ .

$\frac{- 8}{4 {(\frac{12 - 8}{3})}^{\frac{3}{2}}}$

Subtract $8$ from $12$ .

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

Apply the product rule to $\frac{4}{3}$ .

$\frac{- 8}{4 \frac{4^{\frac{3}{2}}}{3^{\frac{3}{2}}}}$

Simplify the numerator.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{- 8}{4 \frac{2^{3}}{3^{\frac{3}{2}}}}$

Raise $2$ to the power of $3$ .

$\frac{- 8}{4 \frac{8}{3^{\frac{3}{2}}}}$

Combine fractions.

Combine $4$ and $\frac{8}{3^{\frac{3}{2}}}$ .

$\frac{- 8}{\frac{4 \cdot 8}{3^{\frac{3}{2}}}}$

Multiply $4$ by $8$ .

$\frac{- 8}{\frac{32}{3^{\frac{3}{2}}}}$

Multiply the numerator by the reciprocal of the denominator.

$- 8 \frac{3^{\frac{3}{2}}}{32}$

Cancel the common factor of $8$ .

Factor $8$ out of $- 8$ .

$8 (- 1) \frac{3^{\frac{3}{2}}}{32}$

Factor $8$ out of $32$ .

$8 \cdot - 1 \frac{3^{\frac{3}{2}}}{8 \cdot 4}$

Cancel the common factor.

$8 \cdot - 1 \frac{3^{\frac{3}{2}}}{8 \cdot 4}$

Rewrite the expression.

$- \frac{3^{\frac{3}{2}}}{4}$

$x = \frac{8}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{8}{3}$ is a local maximum

Find the y-value when $x = \frac{8}{3}$ .

Replace the variable $x$ with $\frac{8}{3}$ in the expression.

$f (\frac{8}{3}) = (\frac{8}{3}) \sqrt{4 - (\frac{8}{3})}$

Simplify the result.

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \sqrt{4 \cdot \frac{3}{3} - \frac{8}{3}}$

Combine $4$ and $\frac{3}{3}$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \sqrt{\frac{4 \cdot 3}{3} - \frac{8}{3}}$

Combine the numerators over the common denominator.

$f (\frac{8}{3}) = \frac{8}{3} \cdot \sqrt{\frac{4 \cdot 3 - 8}{3}}$

Simplify the numerator.

Multiply $4$ by $3$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \sqrt{\frac{12 - 8}{3}}$

Subtract $8$ from $12$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \sqrt{\frac{4}{3}}$

Rewrite $\sqrt{\frac{4}{3}}$ as $\frac{\sqrt{4}}{\sqrt{3}}$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{\sqrt{4}}{\sqrt{3}}$

Simplify the numerator.

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

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{\sqrt{2^{2}}}{\sqrt{3}}$

Pull terms out from under the radical, assuming positive real numbers.

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2}{\sqrt{3}}$

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot (\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}})$

Combine and simplify the denominator.

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{\sqrt{3} \sqrt{3}}$

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

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

Add $1$ and $1$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

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

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

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

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{3}$

Evaluate the exponent.

$f (\frac{8}{3}) = \frac{8}{3} \cdot \frac{2 \sqrt{3}}{3}$

Multiply $\frac{8}{3} \cdot \frac{2 \sqrt{3}}{3}$ .

Multiply $\frac{8}{3}$ by $\frac{2 \sqrt{3}}{3}$ .

$f (\frac{8}{3}) = \frac{8 (2 \sqrt{3})}{3 \cdot 3}$

Multiply $2$ by $8$ .

$f (\frac{8}{3}) = \frac{16 \sqrt{3}}{3 \cdot 3}$

Multiply $3$ by $3$ .

$f (\frac{8}{3}) = \frac{16 \sqrt{3}}{9}$

The final answer is $\frac{16 \sqrt{3}}{9}$ .

$y = \frac{16 \sqrt{3}}{9}$

Evaluate the second derivative at $x = 4$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{3 (4) - 16}{4 {(4 - (4))}^{\frac{3}{2}}}$

Evaluate the second derivative.

Simplify the expression.

Multiply $- 1$ by $4$ .

$\frac{3 (4) - 16}{4 {(4 - 4)}^{\frac{3}{2}}}$

Subtract $4$ from $4$ .

$\frac{3 (4) - 16}{4 \cdot 0^{\frac{3}{2}}}$

Rewrite $0$ as $0^{2}$ .

$\frac{3 (4) - 16}{4 \cdot {(0^{2})}^{\frac{3}{2}}}$

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

$\frac{3 (4) - 16}{4 \cdot 0^{2 (\frac{3}{2})}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 (4) - 16}{4 \cdot 0^{2 (\frac{3}{2})}}$

Rewrite the expression.

$\frac{3 (4) - 16}{4 \cdot 0^{3}}$

Simplify the expression.

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

$\frac{3 (4) - 16}{4 \cdot 0}$

Multiply $4$ by $0$ .

$\frac{3 (4) - 16}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{3 (4) - 16}{0}$

The expression contains a division by $0$ . The expression is undefined.

Undefined

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Find the Maximum/Minimum Value x square root of 4-x

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