Find the Maximum/Minimum Value square root of x^2-9x+25

$\sqrt{x^{2} - 9 x + 25}$

Step 1

Find the first derivative of the function.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 9 x + 25}$ as ${(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [{(x^{2} - 9 x + 25)}^{\frac{1}{2}}]$

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 $g (x) = x^{2} - 9 x + 25$ .

To apply the Chain Rule, set $u$ as $x^{2} - 9 x + 25$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x^{2} - 9 x + 25]$

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} - 9 x + 25]$

Replace all occurrences of $u$ with $x^{2} - 9 x + 25$ .

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} - 9 x + 25]$

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

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x^{2} - 9 x + 25]$

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

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x^{2} - 9 x + 25]$

Combine the numerators over the common denominator.

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x^{2} - 9 x + 25]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{\frac{1 - 2}{2}} \frac{d}{d x} [x^{2} - 9 x + 25]$

Subtract $2$ from $1$ .

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{\frac{- 1}{2}} \frac{d}{d x} [x^{2} - 9 x + 25]$

Combine fractions.

Move the negative in front of the fraction.

$\frac{1}{2} {(x^{2} - 9 x + 25)}^{- \frac{1}{2}} \frac{d}{d x} [x^{2} - 9 x + 25]$

Combine $\frac{1}{2}$ and ${(x^{2} - 9 x + 25)}^{- \frac{1}{2}}$ .

$\frac{{(x^{2} - 9 x + 25)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [x^{2} - 9 x + 25]$

Move ${(x^{2} - 9 x + 25)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \frac{d}{d x} [x^{2} - 9 x + 25]$

By the Sum Rule, the derivative of $x^{2} - 9 x + 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [25]$ .

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [25])$

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

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [25])$

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9 \frac{d}{d x} [x] + \frac{d}{d x} [25])$

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

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9 \cdot 1 + \frac{d}{d x} [25])$

Multiply $- 9$ by $1$ .

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9 + \frac{d}{d x} [25])$

Since $25$ is constant with respect to $x$ , the derivative of $25$ with respect to $x$ is $0$ .

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9 + 0)$

Add $2 x - 9$ and $0$ .

$\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9)$

Simplify.

Reorder the factors of $\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9)$ .

$(2 x - 9) \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$

Multiply $2 x - 9$ by $\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ .

$\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$

Step 2

Find the second derivative of the function.

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\frac{2 x - 9}{{(x^{2} - 9 x + 25)}^{\frac{1}{2}}}]$ .

$f'' (x) = \frac{1}{2} \cdot \frac{d}{d x} (\frac{2 x - 9}{{(x^{2} - 9 x + 25)}^{\frac{1}{2}}})$

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 $f (x) = 2 x - 9$ and $g (x) = {(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} \frac{d}{d x} (2 x - 9) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{{({(x^{2} - 9 x + 25)}^{\frac{1}{2}})}^{2}}$

Multiply the exponents in ${({(x^{2} - 9 x + 25)}^{\frac{1}{2}})}^{2}$ .

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

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} \frac{d}{d x} (2 x - 9) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{{(x^{2} - 9 x + 25)}^{\frac{1}{2} \cdot 2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

Rewrite the expression.

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} \frac{d}{d x} (2 x - 9) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

Simplify.

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} \frac{d}{d x} (2 x - 9) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

Differentiate.

By the Sum Rule, the derivative of $2 x - 9$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 9]$ .

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} (\frac{d}{d x} (2 x) + \frac{d}{d x} (- 9)) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} (2 \frac{d}{d x} (x) + \frac{d}{d x} (- 9)) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} (2 \cdot 1 + \frac{d}{d x} (- 9)) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

Multiply $2$ by $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} (2 + \frac{d}{d x} (- 9)) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} (2 + 0) - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

Simplify the expression.

Add $2$ and $0$ .

$f'' (x) = \frac{1}{2} \cdot \frac{{(x^{2} - 9 x + 25)}^{\frac{1}{2}} \cdot 2 - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

Move $2$ to the left of ${(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) \frac{d}{d x} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{x^{2} - 9 x + 25}$

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 $g (x) = x^{2} - 9 x + 25$ .

To apply the Chain Rule, set $u$ as $x^{2} - 9 x + 25$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Replace all occurrences of $u$ with $x^{2} - 9 x + 25$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Combine the numerators over the common denominator.

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1 - 2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Subtract $2$ from $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{- 1}{2}} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Combine fractions.

Move the negative in front of the fraction.

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{- \frac{1}{2}} \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Combine $\frac{1}{2}$ and ${(x^{2} - 9 x + 25)}^{- \frac{1}{2}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{{(x^{2} - 9 x + 25)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

Move ${(x^{2} - 9 x + 25)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x^{2} - 9 x + 25))}{x^{2} - 9 x + 25}$

By the Sum Rule, the derivative of $x^{2} - 9 x + 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [25]$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (25)))}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (25)))}{x^{2} - 9 x + 25}$

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 \frac{d}{d x} x + \frac{d}{d x} (25)))}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 \cdot 1 + \frac{d}{d x} (25)))}{x^{2} - 9 x + 25}$

Multiply $- 9$ by $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 + \frac{d}{d x} (25)))}{x^{2} - 9 x + 25}$

Since $25$ is constant with respect to $x$ , the derivative of $25$ with respect to $x$ is $0$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 + 0))}{x^{2} - 9 x + 25}$

Add $2 x - 9$ and $0$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9))}{x^{2} - 9 x + 25}$

Raise $2 x - 9$ to the power of $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (2 x - 9) \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Raise $2 x - 9$ to the power of $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - (2 x - 9) (2 x - 9) \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - {(2 x - 9)}^{1 + 1} \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Add $1$ and $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - {(2 x - 9)}^{2} \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Combine $\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ and ${(2 x - 9)}^{2}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - \frac{{(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

To write $2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} \cdot \frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} - \frac{{(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Combine $2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ and $\frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} (2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}})}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} - \frac{{(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Combine the numerators over the common denominator.

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} (2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Multiply $2$ by $2$ .

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} {(x^{2} - 9 x + 25)}^{\frac{1}{2}} - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Multiply ${(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ by ${(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ by adding the exponents.

Move ${(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 ({(x^{2} - 9 x + 25)}^{\frac{1}{2}} {(x^{2} - 9 x + 25)}^{\frac{1}{2}}) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 {(x^{2} - 9 x + 25)}^{\frac{1}{2} + \frac{1}{2}} - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Combine the numerators over the common denominator.

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 {(x^{2} - 9 x + 25)}^{\frac{1 + 1}{2}} - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Add $1$ and $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 {(x^{2} - 9 x + 25)}^{\frac{2}{2}} - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Divide $2$ by $2$ .

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Simplify $4 {(x^{2} - 9 x + 25)}^{1}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{\frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$

Rewrite $\frac{\frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}}{x^{2} - 9 x + 25}$ as a product.

$f'' (x) = \frac{1}{2} \cdot (\frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot \frac{1}{x^{2} - 9 x + 25})$

Multiply $\frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ by $\frac{1}{x^{2} - 9 x + 25}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}} (x^{2} - 9 x + 25)}$

Raise $x^{2} - 9 x + 25$ to the power of $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 ((x^{2} - 9 x + 25) {(x^{2} - 9 x + 25)}^{\frac{1}{2}})}$

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

$f'' (x) = \frac{1}{2} \cdot \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{1 + \frac{1}{2}}}$

Simplify the expression.

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

$f'' (x) = \frac{1}{2} \cdot \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{2}{2} + \frac{1}{2}}}$

Combine the numerators over the common denominator.

$f'' (x) = \frac{1}{2} \cdot \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{2 + 1}{2}}}$

Add $2$ and $1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $\frac{1}{2}$ by $\frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$ .

$f'' (x) = \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{2 (2 {(x^{2} - 9 x + 25)}^{\frac{3}{2}})}$

Multiply $2$ by $2$ .

$f'' (x) = \frac{4 (x^{2} - 9 x + 25) - {(2 x - 9)}^{2}}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Simplify.

Apply the distributive property.

$f'' (x) = \frac{4 x^{2} + 4 (- 9 x) + 4 \cdot 25 - {(2 x - 9)}^{2}}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Simplify the numerator.

Simplify each term.

Multiply $- 9$ by $4$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 4 \cdot 25 - {(2 x - 9)}^{2}}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $4$ by $25$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - {(2 x - 9)}^{2}}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Rewrite ${(2 x - 9)}^{2}$ as $(2 x - 9) (2 x - 9)$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - ((2 x - 9) (2 x - 9))}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Expand $(2 x - 9) (2 x - 9)$ using the FOIL Method.

Apply the distributive property.

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (2 x (2 x - 9) - 9 (2 x - 9))}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Apply the distributive property.

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (2 x (2 x) + 2 x \cdot - 9 - 9 (2 x - 9))}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Apply the distributive property.

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (2 x (2 x) + 2 x \cdot - 9 - 9 (2 x) - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (2 \cdot (2 x \cdot x) + 2 x \cdot - 9 - 9 (2 x) - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

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

Move $x$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (2 \cdot (2 (x \cdot x)) + 2 x \cdot - 9 - 9 (2 x) - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $x$ by $x$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (2 \cdot (2 x^{2}) + 2 x \cdot - 9 - 9 (2 x) - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $2$ by $2$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (4 x^{2} + 2 x \cdot - 9 - 9 (2 x) - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $- 9$ by $2$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (4 x^{2} - 18 x - 9 (2 x) - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $2$ by $- 9$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (4 x^{2} - 18 x - 18 x - 9 \cdot - 9)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $- 9$ by $- 9$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (4 x^{2} - 18 x - 18 x + 81)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Subtract $18 x$ from $- 18 x$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (4 x^{2} - 36 x + 81)}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Apply the distributive property.

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - (4 x^{2}) - (- 36 x) - 1 \cdot 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Simplify.

Multiply $4$ by $- 1$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - 4 x^{2} - (- 36 x) - 1 \cdot 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $- 36$ by $- 1$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - 4 x^{2} + 36 x - 1 \cdot 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Multiply $- 1$ by $81$ .

$f'' (x) = \frac{4 x^{2} - 36 x + 100 - 4 x^{2} + 36 x - 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Combine the opposite terms in $4 x^{2} - 36 x + 100 - 4 x^{2} + 36 x - 81$ .

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

$f'' (x) = \frac{- 36 x + 100 + 0 + 36 x - 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Add $- 36 x + 100$ and $0$ .

$f'' (x) = \frac{- 36 x + 100 + 36 x - 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Add $- 36 x$ and $36 x$ .

$f'' (x) = \frac{0 + 100 - 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Add $0$ and $100$ .

$f'' (x) = \frac{100 - 81}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Subtract $81$ from $100$ .

$f'' (x) = \frac{19}{4 {(x^{2} - 9 x + 25)}^{\frac{3}{2}}}$

Step 3

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

$\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} = 0$

Step 4

Find the first derivative.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} - 9 x + 25}$ as ${(x^{2} - 9 x + 25)}^{\frac{1}{2}}$ .

$f' (x) = \frac{d}{d x} ({(x^{2} - 9 x + 25)}^{\frac{1}{2}})$

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 $g (x) = x^{2} - 9 x + 25$ .

To apply the Chain Rule, set $u$ as $x^{2} - 9 x + 25$ .

$f' (x) = \frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (x^{2} - 9 x + 25)$

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

$f' (x) = \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} - 9 x + 25)$

Replace all occurrences of $u$ with $x^{2} - 9 x + 25$ .

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} - 9 x + 25)$

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

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25)$

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

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25)$

Combine the numerators over the common denominator.

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25)$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{1 - 2}{2}} \frac{d}{d x} (x^{2} - 9 x + 25)$

Subtract $2$ from $1$ .

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{\frac{- 1}{2}} \frac{d}{d x} (x^{2} - 9 x + 25)$

Combine fractions.

Move the negative in front of the fraction.

$f' (x) = \frac{1}{2} \cdot {(x^{2} - 9 x + 25)}^{- \frac{1}{2}} \frac{d}{d x} (x^{2} - 9 x + 25)$

Combine $\frac{1}{2}$ and ${(x^{2} - 9 x + 25)}^{- \frac{1}{2}}$ .

$f' (x) = \frac{{(x^{2} - 9 x + 25)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (x^{2} - 9 x + 25)$

Move ${(x^{2} - 9 x + 25)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x^{2} - 9 x + 25)$

By the Sum Rule, the derivative of $x^{2} - 9 x + 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 9 x] + \frac{d}{d x} [25]$ .

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (25))$

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

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x + \frac{d}{d x} (- 9 x) + \frac{d}{d x} (25))$

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 \frac{d}{d x} x + \frac{d}{d x} (25))$

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

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 \cdot 1 + \frac{d}{d x} (25))$

Multiply $- 9$ by $1$ .

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 + \frac{d}{d x} (25))$

Since $25$ is constant with respect to $x$ , the derivative of $25$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9 + 0)$

Add $2 x - 9$ and $0$ .

$f' (x) = \frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} \cdot (2 x - 9)$

Simplify.

Reorder the factors of $\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} (2 x - 9)$ .

$f' (x) = (2 x - 9) (\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}})$

Multiply $2 x - 9$ by $\frac{1}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ .

$f' (x) = \frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$

The first derivative of $f (x)$ with respect to $x$ is $\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$ .

$\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} = 0$ .

Set the first derivative equal to $0$ .

$\frac{2 x - 9}{2 {(x^{2} - 9 x + 25)}^{\frac{1}{2}}} = 0$

Set the numerator equal to zero.

$2 x - 9 = 0$

Solve the equation for $x$ .

Add $9$ to both sides of the equation.

$2 x = 9$

Divide each term in $2 x = 9$ by $2$ and simplify.

Divide each term in $2 x = 9$ by $2$ .

$\frac{2 x}{2} = \frac{9}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{9}{2}$

Divide $x$ by $1$ .

$x = \frac{9}{2}$

Step 6

Find the values where the derivative is undefined.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = \frac{9}{2}$

Step 8

Evaluate the second derivative at $x = \frac{9}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{19}{4 {({(\frac{9}{2})}^{2} - 9 (\frac{9}{2}) + 25)}^{\frac{3}{2}}}$

Step 9

Evaluate the second derivative.

Simplify the denominator.

Simplify each term.

Apply the product rule to $\frac{9}{2}$ .

$\frac{19}{4 {(\frac{9^{2}}{2^{2}} - 9 (\frac{9}{2}) + 25)}^{\frac{3}{2}}}$

Raise $9$ to the power of $2$ .

$\frac{19}{4 {(\frac{81}{2^{2}} - 9 (\frac{9}{2}) + 25)}^{\frac{3}{2}}}$

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

$\frac{19}{4 {(\frac{81}{4} - 9 (\frac{9}{2}) + 25)}^{\frac{3}{2}}}$

Multiply $- 9 (\frac{9}{2})$ .

Combine $- 9$ and $\frac{9}{2}$ .

$\frac{19}{4 {(\frac{81}{4} + \frac{- 9 \cdot 9}{2} + 25)}^{\frac{3}{2}}}$

Multiply $- 9$ by $9$ .

$\frac{19}{4 {(\frac{81}{4} + \frac{- 81}{2} + 25)}^{\frac{3}{2}}}$

Move the negative in front of the fraction.

$\frac{19}{4 {(\frac{81}{4} - \frac{81}{2} + 25)}^{\frac{3}{2}}}$

Find the common denominator.

Multiply $\frac{81}{2}$ by $\frac{2}{2}$ .

$\frac{19}{4 {(\frac{81}{4} - (\frac{81}{2} \cdot \frac{2}{2}) + 25)}^{\frac{3}{2}}}$

Multiply $\frac{81}{2}$ by $\frac{2}{2}$ .

$\frac{19}{4 {(\frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + 25)}^{\frac{3}{2}}}$

Write $25$ as a fraction with denominator $1$ .

$\frac{19}{4 {(\frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + \frac{25}{1})}^{\frac{3}{2}}}$

Multiply $\frac{25}{1}$ by $\frac{4}{4}$ .

$\frac{19}{4 {(\frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + \frac{25}{1} \cdot \frac{4}{4})}^{\frac{3}{2}}}$

Multiply $\frac{25}{1}$ by $\frac{4}{4}$ .

$\frac{19}{4 {(\frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + \frac{25 \cdot 4}{4})}^{\frac{3}{2}}}$

Multiply $2$ by $2$ .

$\frac{19}{4 {(\frac{81}{4} - \frac{81 \cdot 2}{4} + \frac{25 \cdot 4}{4})}^{\frac{3}{2}}}$

Combine the numerators over the common denominator.

$\frac{19}{4 {(\frac{81 - 81 \cdot 2 + 25 \cdot 4}{4})}^{\frac{3}{2}}}$

Simplify each term.

Multiply $- 81$ by $2$ .

$\frac{19}{4 {(\frac{81 - 162 + 25 \cdot 4}{4})}^{\frac{3}{2}}}$

Multiply $25$ by $4$ .

$\frac{19}{4 {(\frac{81 - 162 + 100}{4})}^{\frac{3}{2}}}$

Subtract $162$ from $81$ .

$\frac{19}{4 {(\frac{- 81 + 100}{4})}^{\frac{3}{2}}}$

Add $- 81$ and $100$ .

$\frac{19}{4 {(\frac{19}{4})}^{\frac{3}{2}}}$

Apply the product rule to $\frac{19}{4}$ .

$\frac{19}{4 \frac{19^{\frac{3}{2}}}{4^{\frac{3}{2}}}}$

Simplify the denominator.

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

$\frac{19}{4 \frac{19^{\frac{3}{2}}}{{(2^{2})}^{\frac{3}{2}}}}$

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

$\frac{19}{4 \frac{19^{\frac{3}{2}}}{2^{2 (\frac{3}{2})}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{19}{4 \frac{19^{\frac{3}{2}}}{2^{2 (\frac{3}{2})}}}$

Rewrite the expression.

$\frac{19}{4 \frac{19^{\frac{3}{2}}}{2^{3}}}$

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

$\frac{19}{4 \frac{19^{\frac{3}{2}}}{8}}$

Simplify terms.

Combine $4$ and $\frac{19^{\frac{3}{2}}}{8}$ .

$\frac{19}{\frac{4 \cdot 19^{\frac{3}{2}}}{8}}$

Reduce the expression $\frac{4 \cdot 19^{\frac{3}{2}}}{8}$ by cancelling the common factors.

Factor $4$ out of $4 \cdot 19^{\frac{3}{2}}$ .

$\frac{19}{\frac{4 (19^{\frac{3}{2}})}{8}}$

Factor $4$ out of $8$ .

$\frac{19}{\frac{4 \cdot 19^{\frac{3}{2}}}{4 \cdot 2}}$

Cancel the common factor.

$\frac{19}{\frac{4 \cdot 19^{\frac{3}{2}}}{4 \cdot 2}}$

Rewrite the expression.

$\frac{19}{\frac{19^{\frac{3}{2}}}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$19 \frac{2}{19^{\frac{3}{2}}}$

Multiply $19 \frac{2}{19^{\frac{3}{2}}}$ .

Combine $19$ and $\frac{2}{19^{\frac{3}{2}}}$ .

$\frac{19 \cdot 2}{19^{\frac{3}{2}}}$

Multiply $19$ by $2$ .

$\frac{38}{19^{\frac{3}{2}}}$

Step 10

$x = \frac{9}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{9}{2}$ is a local minimum

Step 11

Find the y-value when $x = \frac{9}{2}$ .

Replace the variable $x$ with $\frac{9}{2}$ in the expression.

$f (\frac{9}{2}) = \sqrt{{(\frac{9}{2})}^{2} - 9 (\frac{9}{2}) + 25}$

Simplify the result.

Apply the product rule to $\frac{9}{2}$ .

$f (\frac{9}{2}) = \sqrt{\frac{9^{2}}{2^{2}} - 9 (\frac{9}{2}) + 25}$

Raise $9$ to the power of $2$ .

$f (\frac{9}{2}) = \sqrt{\frac{81}{2^{2}} - 9 (\frac{9}{2}) + 25}$

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

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} - 9 (\frac{9}{2}) + 25}$

Multiply $- 9 (\frac{9}{2})$ .

Combine $- 9$ and $\frac{9}{2}$ .

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} + \frac{- 9 \cdot 9}{2} + 25}$

Multiply $- 9$ by $9$ .

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} + \frac{- 81}{2} + 25}$

Move the negative in front of the fraction.

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} - \frac{81}{2} + 25}$

To write $- \frac{81}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} - \frac{81}{2} \cdot \frac{2}{2} + 25}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{81}{2}$ by $\frac{2}{2}$ .

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} - \frac{81 \cdot 2}{2 \cdot 2} + 25}$

Multiply $2$ by $2$ .

$f (\frac{9}{2}) = \sqrt{\frac{81}{4} - \frac{81 \cdot 2}{4} + 25}$

Combine the numerators over the common denominator.

$f (\frac{9}{2}) = \sqrt{\frac{81 - 81 \cdot 2}{4} + 25}$

Simplify the numerator.

Multiply $- 81$ by $2$ .

$f (\frac{9}{2}) = \sqrt{\frac{81 - 162}{4} + 25}$

Subtract $162$ from $81$ .

$f (\frac{9}{2}) = \sqrt{\frac{- 81}{4} + 25}$

To write $25$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (\frac{9}{2}) = \sqrt{\frac{- 81}{4} + 25 \cdot \frac{4}{4}}$

Combine $25$ and $\frac{4}{4}$ .

$f (\frac{9}{2}) = \sqrt{\frac{- 81}{4} + \frac{25 \cdot 4}{4}}$

Combine the numerators over the common denominator.

$f (\frac{9}{2}) = \sqrt{\frac{- 81 + 25 \cdot 4}{4}}$

Simplify the numerator.

Multiply $25$ by $4$ .

$f (\frac{9}{2}) = \sqrt{\frac{- 81 + 100}{4}}$

Add $- 81$ and $100$ .

$f (\frac{9}{2}) = \sqrt{\frac{19}{4}}$

Rewrite $\sqrt{\frac{19}{4}}$ as $\frac{\sqrt{19}}{\sqrt{4}}$ .

$f (\frac{9}{2}) = \frac{\sqrt{19}}{\sqrt{4}}$

Simplify the denominator.

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

$f (\frac{9}{2}) = \frac{\sqrt{19}}{\sqrt{2^{2}}}$

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

$f (\frac{9}{2}) = \frac{\sqrt{19}}{2}$

The final answer is $\frac{\sqrt{19}}{2}$ .

$y = \frac{\sqrt{19}}{2}$

Step 12

These are the local extrema for $f (x) = \sqrt{x^{2} - 9 x + 25}$ .

$(\frac{9}{2}, \frac{\sqrt{19}}{2})$ is a local minima

Step 13

Find the Maximum/Minimum Value square root of x^2-9x+25

Related Questions

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