Calculus Examples

$f (x) = \sqrt{2 x - 5}$

Step 1

Find the first derivative of the function.

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Step 1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 5}$ as ${(2 x - 5)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [{(2 x - 5)}^{\frac{1}{2}}]$

Step 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 $g (x) = 2 x - 5$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $2 x - 5$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [2 x - 5]$

Step 1.2.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [2 x - 5]$

Step 1.2.3

Replace all occurrences of $u$ with $2 x - 5$ .

$\frac{1}{2} {(2 x - 5)}^{\frac{1}{2} - 1} \frac{d}{d x} [2 x - 5]$

Step 1.3

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

$\frac{1}{2} {(2 x - 5)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [2 x - 5]$

Step 1.4

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

$\frac{1}{2} {(2 x - 5)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [2 x - 5]$

Step 1.5

Combine the numerators over the common denominator.

$\frac{1}{2} {(2 x - 5)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [2 x - 5]$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(2 x - 5)}^{\frac{1 - 2}{2}} \frac{d}{d x} [2 x - 5]$

Step 1.6.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(2 x - 5)}^{\frac{- 1}{2}} \frac{d}{d x} [2 x - 5]$

Step 1.7

Combine fractions.

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Step 1.7.1

Move the negative in front of the fraction.

$\frac{1}{2} {(2 x - 5)}^{- \frac{1}{2}} \frac{d}{d x} [2 x - 5]$

Step 1.7.2

Combine $\frac{1}{2}$ and ${(2 x - 5)}^{- \frac{1}{2}}$ .

$\frac{{(2 x - 5)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [2 x - 5]$

Step 1.7.3

Move ${(2 x - 5)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} \frac{d}{d x} [2 x - 5]$

Step 1.8

By the Sum Rule, the derivative of $2 x - 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5]$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5])$

Step 1.9

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

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 5])$

Step 1.10

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

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 \cdot 1 + \frac{d}{d x} [- 5])$

Step 1.11

Multiply $2$ by $1$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 + \frac{d}{d x} [- 5])$

Step 1.12

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 + 0)$

Step 1.13

Simplify terms.

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Step 1.13.1

Add $2$ and $0$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} \cdot 2$

Step 1.13.2

Combine $\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}}$ and $2$ .

$\frac{2}{2 {(2 x - 5)}^{\frac{1}{2}}}$

Step 1.13.3

Cancel the common factor.

$\frac{2}{2 {(2 x - 5)}^{\frac{1}{2}}}$

Step 1.13.4

Rewrite the expression.

$\frac{1}{{(2 x - 5)}^{\frac{1}{2}}}$

Step 2

Find the second derivative of the function.

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Step 2.1

Apply basic rules of exponents.

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Step 2.1.1

Rewrite $\frac{1}{{(2 x - 5)}^{\frac{1}{2}}}$ as ${({(2 x - 5)}^{\frac{1}{2}})}^{- 1}$ .

$f'' (x) = \frac{d}{d x} ({({(2 x - 5)}^{\frac{1}{2}})}^{- 1})$

Step 2.1.2

Multiply the exponents in ${({(2 x - 5)}^{\frac{1}{2}})}^{- 1}$ .

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Step 2.1.2.1

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

$f'' (x) = \frac{d}{d x} ({(2 x - 5)}^{\frac{1}{2} \cdot - 1})$

Step 2.1.2.2

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

$f'' (x) = \frac{d}{d x} ({(2 x - 5)}^{\frac{- 1}{2}})$

Step 2.1.2.3

Move the negative in front of the fraction.

$f'' (x) = \frac{d}{d x} ({(2 x - 5)}^{- \frac{1}{2}})$

Step 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 $g (x) = 2 x - 5$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $2 x - 5$ .

$f'' (x) = \frac{d}{d u} (u^{- \frac{1}{2}}) \frac{d}{d x} (2 x - 5)$

Step 2.2.2

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

$f'' (x) = - \frac{1}{2} u^{- \frac{1}{2} - 1} \frac{d}{d x} (2 x - 5)$

Step 2.2.3

Replace all occurrences of $u$ with $2 x - 5$ .

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{- \frac{1}{2} - 1} \frac{d}{d x} (2 x - 5)$

Step 2.3

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

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{- \frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (2 x - 5)$

Step 2.4

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

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{- \frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (2 x - 5)$

Step 2.5

Combine the numerators over the common denominator.

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{\frac{- 1 - 1 \cdot 2}{2}} \frac{d}{d x} (2 x - 5)$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Multiply $- 1$ by $2$ .

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{\frac{- 1 - 2}{2}} \frac{d}{d x} (2 x - 5)$

Step 2.6.2

Subtract $2$ from $- 1$ .

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{\frac{- 3}{2}} \frac{d}{d x} (2 x - 5)$

Step 2.7

Combine fractions.

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Step 2.7.1

Move the negative in front of the fraction.

$f'' (x) = - \frac{1}{2} \cdot {(2 x - 5)}^{- \frac{3}{2}} \frac{d}{d x} (2 x - 5)$

Step 2.7.2

Combine ${(2 x - 5)}^{- \frac{3}{2}}$ and $\frac{1}{2}$ .

$f'' (x) = - \frac{{(2 x - 5)}^{- \frac{3}{2}}}{2} \cdot \frac{d}{d x} (2 x - 5)$

Step 2.7.3

Move ${(2 x - 5)}^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (2 x - 5)$

Step 2.8

By the Sum Rule, the derivative of $2 x - 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5]$ .

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot (\frac{d}{d x} (2 x) + \frac{d}{d x} (- 5))$

Step 2.9

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

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot (2 \frac{d}{d x} (x) + \frac{d}{d x} (- 5))$

Step 2.10

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

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot (2 \cdot 1 + \frac{d}{d x} (- 5))$

Step 2.11

Multiply $2$ by $1$ .

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot (2 + \frac{d}{d x} (- 5))$

Step 2.12

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot (2 + 0)$

Step 2.13

Simplify terms.

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Step 2.13.1

Add $2$ and $0$ .

$f'' (x) = - \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}} \cdot 2$

Step 2.13.2

Multiply $2$ by $- 1$ .

$f'' (x) = - 2 \frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}}$

Step 2.13.3

Combine $- 2$ and $\frac{1}{2 {(2 x - 5)}^{\frac{3}{2}}}$ .

$f'' (x) = \frac{- 2}{2 {(2 x - 5)}^{\frac{3}{2}}}$

Step 2.13.4

Factor $2$ out of $- 2$ .

$f'' (x) = \frac{2 \cdot - 1}{2 {(2 x - 5)}^{\frac{3}{2}}}$

Step 2.14

Cancel the common factors.

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Step 2.14.1

Factor $2$ out of $2 {(2 x - 5)}^{\frac{3}{2}}$ .

$f'' (x) = \frac{2 \cdot - 1}{2 ({(2 x - 5)}^{\frac{3}{2}})}$

Step 2.14.2

Cancel the common factor.

$f'' (x) = \frac{2 \cdot - 1}{2 {(2 x - 5)}^{\frac{3}{2}}}$

Step 2.14.3

Rewrite the expression.

$f'' (x) = \frac{- 1}{{(2 x - 5)}^{\frac{3}{2}}}$

Step 2.15

Move the negative in front of the fraction.

$f'' (x) = - \frac{1}{{(2 x - 5)}^{\frac{3}{2}}}$

Step 3

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

$\frac{1}{{(2 x - 5)}^{\frac{1}{2}}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 5}$ as ${(2 x - 5)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [{(2 x - 5)}^{\frac{1}{2}}]$

Step 4.1.2

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 $g (x) = 2 x - 5$ .

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Step 4.1.2.1

To apply the Chain Rule, set $u$ as $2 x - 5$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [2 x - 5]$

Step 4.1.2.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [2 x - 5]$

Step 4.1.2.3

Replace all occurrences of $u$ with $2 x - 5$ .

$\frac{1}{2} {(2 x - 5)}^{\frac{1}{2} - 1} \frac{d}{d x} [2 x - 5]$

Step 4.1.3

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

$\frac{1}{2} {(2 x - 5)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [2 x - 5]$

Step 4.1.4

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

$\frac{1}{2} {(2 x - 5)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [2 x - 5]$

Step 4.1.5

Combine the numerators over the common denominator.

$\frac{1}{2} {(2 x - 5)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [2 x - 5]$

Step 4.1.6

Simplify the numerator.

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Step 4.1.6.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(2 x - 5)}^{\frac{1 - 2}{2}} \frac{d}{d x} [2 x - 5]$

Step 4.1.6.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(2 x - 5)}^{\frac{- 1}{2}} \frac{d}{d x} [2 x - 5]$

Step 4.1.7

Combine fractions.

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Step 4.1.7.1

Move the negative in front of the fraction.

$\frac{1}{2} {(2 x - 5)}^{- \frac{1}{2}} \frac{d}{d x} [2 x - 5]$

Step 4.1.7.2

Combine $\frac{1}{2}$ and ${(2 x - 5)}^{- \frac{1}{2}}$ .

$\frac{{(2 x - 5)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [2 x - 5]$

Step 4.1.7.3

Move ${(2 x - 5)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} \frac{d}{d x} [2 x - 5]$

Step 4.1.8

By the Sum Rule, the derivative of $2 x - 5$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5]$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 5])$

Step 4.1.9

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

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 5])$

Step 4.1.10

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

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 \cdot 1 + \frac{d}{d x} [- 5])$

Step 4.1.11

Multiply $2$ by $1$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 + \frac{d}{d x} [- 5])$

Step 4.1.12

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} (2 + 0)$

Step 4.1.13

Simplify terms.

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Step 4.1.13.1

Add $2$ and $0$ .

$\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}} \cdot 2$

Step 4.1.13.2

Combine $\frac{1}{2 {(2 x - 5)}^{\frac{1}{2}}}$ and $2$ .

$\frac{2}{2 {(2 x - 5)}^{\frac{1}{2}}}$

Step 4.1.13.3

Cancel the common factor.

$\frac{2}{2 {(2 x - 5)}^{\frac{1}{2}}}$

Step 4.1.13.4

Rewrite the expression.

$f' (x) = \frac{1}{{(2 x - 5)}^{\frac{1}{2}}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{1}{{(2 x - 5)}^{\frac{1}{2}}}$ .

$\frac{1}{{(2 x - 5)}^{\frac{1}{2}}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{1}{{(2 x - 5)}^{\frac{1}{2}}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{1}{{(2 x - 5)}^{\frac{1}{2}}} = 0$

Step 5.2

Set the numerator equal to zero.

$1 = 0$

Step 5.3

Since $1 \neq 0$ , there are no solutions.

No solution

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Convert expressions with fractional exponents to radicals.

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Step 6.1.1

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

$\frac{1}{\sqrt{{(2 x - 5)}^{1}}}$

Step 6.1.2

Anything raised to $1$ is the base itself.

$\frac{1}{\sqrt{2 x - 5}}$

Step 6.2

Set the denominator in $\frac{1}{\sqrt{2 x - 5}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{2 x - 5} = 0$

Step 6.3

Solve for $x$ .

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Step 6.3.1

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

${\sqrt{2 x - 5}}^{2} = 0^{2}$

Step 6.3.2

Simplify each side of the equation.

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Step 6.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 5}$ as ${(2 x - 5)}^{\frac{1}{2}}$ .

${({(2 x - 5)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Simplify ${({(2 x - 5)}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.2.2.1.1

Multiply the exponents in ${({(2 x - 5)}^{\frac{1}{2}})}^{2}$ .

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Step 6.3.2.2.1.1.1

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

${(2 x - 5)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 6.3.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.3.2.2.1.1.2.1

Cancel the common factor.

${(2 x - 5)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 6.3.2.2.1.1.2.2

Rewrite the expression.

${(2 x - 5)}^{1} = 0^{2}$

Step 6.3.2.2.1.2

Simplify.

$2 x - 5 = 0^{2}$

Step 6.3.2.3

Simplify the right side.

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Step 6.3.2.3.1

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

$2 x - 5 = 0$

Step 6.3.3

Solve for $x$ .

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Step 6.3.3.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 6.3.3.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 6.3.3.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 6.3.3.2.2

Simplify the left side.

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Step 6.3.3.2.2.1

Cancel the common factor of $2$ .

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Step 6.3.3.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 6.3.3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 6.4

Set the radicand in $\sqrt{2 x - 5}$ less than $0$ to find where the expression is undefined.

$2 x - 5 < 0$

Step 6.5

Solve for $x$ .

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Step 6.5.1

Add $5$ to both sides of the inequality.

$2 x < 5$

Step 6.5.2

Divide each term in $2 x < 5$ by $2$ and simplify.

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Step 6.5.2.1

Divide each term in $2 x < 5$ by $2$ .

$\frac{2 x}{2} < \frac{5}{2}$

Step 6.5.2.2

Simplify the left side.

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Step 6.5.2.2.1

Cancel the common factor of $2$ .

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Step 6.5.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} < \frac{5}{2}$

Step 6.5.2.2.1.2

Divide $x$ by $1$ .

$x < \frac{5}{2}$

Step 6.6

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 $0$ .

$x \leq \frac{5}{2}$

$(- \infty, \frac{5}{2}]$

$x \leq \frac{5}{2}$

$(- \infty, \frac{5}{2}]$

Step 7

Critical points to evaluate.

$x = \frac{5}{2}$

Step 8

Evaluate the second derivative at $x = \frac{5}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{1}{{(2 (\frac{5}{2}) - 5)}^{\frac{3}{2}}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Cancel the common factor of $2$ .

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Step 9.1.1

Cancel the common factor.

$- \frac{1}{{(2 (\frac{5}{2}) - 5)}^{\frac{3}{2}}}$

Step 9.1.2

Rewrite the expression.

$- \frac{1}{{(5 - 5)}^{\frac{3}{2}}}$

Step 9.2

Simplify the expression.

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Step 9.2.1

Subtract $5$ from $5$ .

$- \frac{1}{0^{\frac{3}{2}}}$

Step 9.2.2

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

$- \frac{1}{{(0^{2})}^{\frac{3}{2}}}$

Step 9.2.3

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

$- \frac{1}{0^{2 (\frac{3}{2})}}$

Step 9.3

Cancel the common factor of $2$ .

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Step 9.3.1

Cancel the common factor.

$- \frac{1}{0^{2 (\frac{3}{2})}}$

Step 9.3.2

Rewrite the expression.

$- \frac{1}{0^{3}}$

Step 9.4

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

$- \frac{1}{0}$

Step 9.5

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

Undefined

Step 10

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

No Local Extrema

Step 11