Calculus Examples

$g (x) = \sqrt[3]{x + 4}$

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 $\sqrt[3]{x + 4}$ as ${(x + 4)}^{\frac{1}{3}}$ .

$\frac{d}{d x} [{(x + 4)}^{\frac{1}{3}}]$

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

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Step 1.2.1

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

$\frac{d}{d u} [u^{\frac{1}{3}}] \frac{d}{d x} [x + 4]$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{3}$ .

$\frac{1}{3} u^{\frac{1}{3} - 1} \frac{d}{d x} [x + 4]$

Step 1.2.3

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

$\frac{1}{3} {(x + 4)}^{\frac{1}{3} - 1} \frac{d}{d x} [x + 4]$

Step 1.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{3} {(x + 4)}^{\frac{1}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [x + 4]$

Step 1.4

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{1}{3} {(x + 4)}^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [x + 4]$

Step 1.5

Combine the numerators over the common denominator.

$\frac{1}{3} {(x + 4)}^{\frac{1 - 1 \cdot 3}{3}} \frac{d}{d x} [x + 4]$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $- 1$ by $3$ .

$\frac{1}{3} {(x + 4)}^{\frac{1 - 3}{3}} \frac{d}{d x} [x + 4]$

Step 1.6.2

Subtract $3$ from $1$ .

$\frac{1}{3} {(x + 4)}^{\frac{- 2}{3}} \frac{d}{d x} [x + 4]$

Step 1.7

Combine fractions.

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Step 1.7.1

Move the negative in front of the fraction.

$\frac{1}{3} {(x + 4)}^{- \frac{2}{3}} \frac{d}{d x} [x + 4]$

Step 1.7.2

Combine $\frac{1}{3}$ and ${(x + 4)}^{- \frac{2}{3}}$ .

$\frac{{(x + 4)}^{- \frac{2}{3}}}{3} \frac{d}{d x} [x + 4]$

Step 1.7.3

Move ${(x + 4)}^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} \frac{d}{d x} [x + 4]$

Step 1.8

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

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} (\frac{d}{d x} [x] + \frac{d}{d x} [4])$

Step 1.9

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

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} (1 + \frac{d}{d x} [4])$

Step 1.10

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

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} (1 + 0)$

Step 1.11

Simplify the expression.

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Step 1.11.1

Add $1$ and $0$ .

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

Step 1.11.2

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

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

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Constant Multiple Rule.

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Step 2.1.1

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

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

Step 2.1.2

Apply basic rules of exponents.

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Step 2.1.2.1

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

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

Step 2.1.2.2

Multiply the exponents in ${({(x + 4)}^{\frac{2}{3}})}^{- 1}$ .

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Step 2.1.2.2.1

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

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

Step 2.1.2.2.2

Multiply $\frac{2}{3} \cdot - 1$ .

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Step 2.1.2.2.2.1

Combine $\frac{2}{3}$ and $- 1$ .

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

Step 2.1.2.2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.1.2.2.3

Move the negative in front of the fraction.

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

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

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Step 2.2.1

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

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

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{2}{3}$ .

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

Step 2.2.3

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

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

Step 2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 2.4

Combine $- 1$ and $\frac{3}{3}$ .

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

Step 2.5

Combine the numerators over the common denominator.

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

Step 2.6

Simplify the numerator.

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Step 2.6.1

Multiply $- 1$ by $3$ .

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

Step 2.6.2

Subtract $3$ from $- 2$ .

$g'' (x) = \frac{1}{3} \cdot (- \frac{2}{3} \cdot {(x + 4)}^{\frac{- 5}{3}} \frac{d}{d x} (x + 4))$

Step 2.7

Combine fractions.

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Step 2.7.1

Move the negative in front of the fraction.

$g'' (x) = \frac{1}{3} \cdot (- \frac{2}{3} \cdot {(x + 4)}^{- \frac{5}{3}} \frac{d}{d x} (x + 4))$

Step 2.7.2

Combine ${(x + 4)}^{- \frac{5}{3}}$ and $\frac{2}{3}$ .

$g'' (x) = \frac{1}{3} \cdot (- \frac{{(x + 4)}^{- \frac{5}{3}} \cdot 2}{3} \cdot \frac{d}{d x} (x + 4))$

Step 2.7.3

Simplify the expression.

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Step 2.7.3.1

Move $2$ to the left of ${(x + 4)}^{- \frac{5}{3}}$ .

$g'' (x) = \frac{1}{3} \cdot (- \frac{2 \cdot {(x + 4)}^{- \frac{5}{3}}}{3} \cdot \frac{d}{d x} (x + 4))$

Step 2.7.3.2

Move ${(x + 4)}^{- \frac{5}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$g'' (x) = \frac{1}{3} \cdot (- \frac{2}{3 {(x + 4)}^{\frac{5}{3}}} \cdot \frac{d}{d x} (x + 4))$

Step 2.7.4

Multiply $\frac{1}{3}$ by $\frac{2}{3 {(x + 4)}^{\frac{5}{3}}}$ .

$g'' (x) = - \frac{2}{3 (3 {(x + 4)}^{\frac{5}{3}})} \cdot \frac{d}{d x} (x + 4)$

Step 2.7.5

Multiply $3$ by $3$ .

$g'' (x) = - \frac{2}{9 {(x + 4)}^{\frac{5}{3}}} \cdot \frac{d}{d x} (x + 4)$

Step 2.8

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

$g'' (x) = - \frac{2}{9 {(x + 4)}^{\frac{5}{3}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (4))$

Step 2.9

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

$g'' (x) = - \frac{2}{9 {(x + 4)}^{\frac{5}{3}}} \cdot (1 + \frac{d}{d x} (4))$

Step 2.10

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

$g'' (x) = - \frac{2}{9 {(x + 4)}^{\frac{5}{3}}} \cdot (1 + 0)$

Step 2.11

Simplify the expression.

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Step 2.11.1

Add $1$ and $0$ .

$g'' (x) = - \frac{2}{9 {(x + 4)}^{\frac{5}{3}}} \cdot 1$

Step 2.11.2

Multiply $- 1$ by $1$ .

$g'' (x) = - \frac{2}{9 {(x + 4)}^{\frac{5}{3}}}$

Step 3

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

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

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 $\sqrt[3]{x + 4}$ as ${(x + 4)}^{\frac{1}{3}}$ .

$\frac{d}{d x} [{(x + 4)}^{\frac{1}{3}}]$

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

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Step 4.1.2.1

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

$\frac{d}{d u} [u^{\frac{1}{3}}] \frac{d}{d x} [x + 4]$

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 $n = \frac{1}{3}$ .

$\frac{1}{3} u^{\frac{1}{3} - 1} \frac{d}{d x} [x + 4]$

Step 4.1.2.3

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

$\frac{1}{3} {(x + 4)}^{\frac{1}{3} - 1} \frac{d}{d x} [x + 4]$

Step 4.1.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{3} {(x + 4)}^{\frac{1}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [x + 4]$

Step 4.1.4

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{1}{3} {(x + 4)}^{\frac{1}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [x + 4]$

Step 4.1.5

Combine the numerators over the common denominator.

$\frac{1}{3} {(x + 4)}^{\frac{1 - 1 \cdot 3}{3}} \frac{d}{d x} [x + 4]$

Step 4.1.6

Simplify the numerator.

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Step 4.1.6.1

Multiply $- 1$ by $3$ .

$\frac{1}{3} {(x + 4)}^{\frac{1 - 3}{3}} \frac{d}{d x} [x + 4]$

Step 4.1.6.2

Subtract $3$ from $1$ .

$\frac{1}{3} {(x + 4)}^{\frac{- 2}{3}} \frac{d}{d x} [x + 4]$

Step 4.1.7

Combine fractions.

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Step 4.1.7.1

Move the negative in front of the fraction.

$\frac{1}{3} {(x + 4)}^{- \frac{2}{3}} \frac{d}{d x} [x + 4]$

Step 4.1.7.2

Combine $\frac{1}{3}$ and ${(x + 4)}^{- \frac{2}{3}}$ .

$\frac{{(x + 4)}^{- \frac{2}{3}}}{3} \frac{d}{d x} [x + 4]$

Step 4.1.7.3

Move ${(x + 4)}^{- \frac{2}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} \frac{d}{d x} [x + 4]$

Step 4.1.8

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

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} (\frac{d}{d x} [x] + \frac{d}{d x} [4])$

Step 4.1.9

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

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} (1 + \frac{d}{d x} [4])$

Step 4.1.10

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

$\frac{1}{3 {(x + 4)}^{\frac{2}{3}}} (1 + 0)$

Step 4.1.11

Simplify the expression.

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Step 4.1.11.1

Add $1$ and $0$ .

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

Step 4.1.11.2

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

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

Step 4.2

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

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

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

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

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

$\frac{1}{3 \sqrt[3]{{(x + 4)}^{2}}}$

Step 6.2

Set the denominator in $\frac{1}{3 \sqrt[3]{{(x + 4)}^{2}}}$ equal to $0$ to find where the expression is undefined.

$3 \sqrt[3]{{(x + 4)}^{2}} = 0$

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, cube both sides of the equation.

${(3 \sqrt[3]{{(x + 4)}^{2}})}^{3} = 0^{3}$

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 $\sqrt[3]{{(x + 4)}^{2}}$ as ${(x + 4)}^{\frac{2}{3}}$ .

${(3 {(x + 4)}^{\frac{2}{3}})}^{3} = 0^{3}$

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

Simplify ${(3 {(x + 4)}^{\frac{2}{3}})}^{3}$ .

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Step 6.3.2.2.1.1

Apply the product rule to $3 {(x + 4)}^{\frac{2}{3}}$ .

$3^{3} {({(x + 4)}^{\frac{2}{3}})}^{3} = 0^{3}$

Step 6.3.2.2.1.2

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

$27 {({(x + 4)}^{\frac{2}{3}})}^{3} = 0^{3}$

Step 6.3.2.2.1.3

Multiply the exponents in ${({(x + 4)}^{\frac{2}{3}})}^{3}$ .

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Step 6.3.2.2.1.3.1

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

$27 {(x + 4)}^{\frac{2}{3} \cdot 3} = 0^{3}$

Step 6.3.2.2.1.3.2

Cancel the common factor of $3$ .

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Step 6.3.2.2.1.3.2.1

Cancel the common factor.

$27 {(x + 4)}^{\frac{2}{3} \cdot 3} = 0^{3}$

Step 6.3.2.2.1.3.2.2

Rewrite the expression.

$27 {(x + 4)}^{2} = 0^{3}$

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 $0$ .

$27 {(x + 4)}^{2} = 0$

Step 6.3.3

Solve for $x$ .

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Step 6.3.3.1

Divide each term in $27 {(x + 4)}^{2} = 0$ by $27$ and simplify.

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Step 6.3.3.1.1

Divide each term in $27 {(x + 4)}^{2} = 0$ by $27$ .

$\frac{27 {(x + 4)}^{2}}{27} = \frac{0}{27}$

Step 6.3.3.1.2

Simplify the left side.

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Step 6.3.3.1.2.1

Cancel the common factor of $27$ .

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Step 6.3.3.1.2.1.1

Cancel the common factor.

$\frac{27 {(x + 4)}^{2}}{27} = \frac{0}{27}$

Step 6.3.3.1.2.1.2

Divide ${(x + 4)}^{2}$ by $1$ .

${(x + 4)}^{2} = \frac{0}{27}$

Step 6.3.3.1.3

Simplify the right side.

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Step 6.3.3.1.3.1

Divide $0$ by $27$ .

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

Step 6.3.3.2

Set the $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 6.3.3.3

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 7

Critical points to evaluate.

$x = - 4$

Step 8

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.

$- \frac{2}{9 {((- 4) + 4)}^{\frac{5}{3}}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the expression.

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Step 9.1.1

Add $- 4$ and $4$ .

$- \frac{2}{9 \cdot 0^{\frac{5}{3}}}$

Step 9.1.2

Rewrite $0$ as $0^{3}$ .

$- \frac{2}{9 \cdot {(0^{3})}^{\frac{5}{3}}}$

Step 9.1.3

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

$- \frac{2}{9 \cdot 0^{3 (\frac{5}{3})}}$

Step 9.2

Cancel the common factor of $3$ .

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Step 9.2.1

Cancel the common factor.

$- \frac{2}{9 \cdot 0^{3 (\frac{5}{3})}}$

Step 9.2.2

Rewrite the expression.

$- \frac{2}{9 \cdot 0^{5}}$

Step 9.3

Simplify the expression.

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Step 9.3.1

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

$- \frac{2}{9 \cdot 0}$

Step 9.3.2

Multiply $9$ by $0$ .

$- \frac{2}{0}$

Step 9.3.3

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

Undefined

$- \frac{2}{0}$

Step 9.4

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

Undefined

Step 10

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 10.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 4) \cup (- 4, \infty)$

Step 10.2

Substitute any number, such as $- 6$ , from the interval $(- \infty, - 4)$ in the first derivative $\frac{1}{3 {(x + 4)}^{\frac{2}{3}}}$ to check if the result is negative or positive.

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Step 10.2.1

Replace the variable $x$ with $- 6$ in the expression.

$g' (- 6) = \frac{1}{3 {((- 6) + 4)}^{\frac{2}{3}}}$

Step 10.2.2

Simplify the result.

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Step 10.2.2.1

Add $- 6$ and $4$ .

$g' (- 6) = \frac{1}{3 {(- 2)}^{\frac{2}{3}}}$

Step 10.2.2.2

The final answer is $\frac{1}{3 {(- 2)}^{\frac{2}{3}}}$ .

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

Step 10.3

Substitute any number, such as $0$ , from the interval $(- 4, \infty)$ in the first derivative $\frac{1}{3 {(x + 4)}^{\frac{2}{3}}}$ to check if the result is negative or positive.

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Step 10.3.1

Replace the variable $x$ with $0$ in the expression.

$g' (0) = \frac{1}{3 {((0) + 4)}^{\frac{2}{3}}}$

Step 10.3.2

Simplify the result.

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Step 10.3.2.1

Add $0$ and $4$ .

$g' (0) = \frac{1}{3 \cdot 4^{\frac{2}{3}}}$

Step 10.3.2.2

The final answer is $\frac{1}{3 \cdot 4^{\frac{2}{3}}}$ .

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

Step 10.4

Since the first derivative did not change signs around $x = - 4$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 10.5

No local maxima or minima found for $g (x) = \sqrt[3]{x + 4}$ .

No local maxima or minima

Step 11