Find the Maximum/Minimum Value (2x)/3+(x+1)^(2/3)

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

Find the first derivative of the function.

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

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

Evaluate $\frac{d}{d x} [\frac{2 x}{3}]$ .

Since $\frac{2}{3}$ is constant with respect to $x$ , the derivative of $\frac{2 x}{3}$ with respect to $x$ is $\frac{2}{3} \frac{d}{d x} [x]$ .

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

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

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

Multiply $\frac{2}{3}$ by $1$ .

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

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

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 $g (x) = x + 1$ .

To apply the Chain Rule, set $u$ as $x + 1$ .

$\frac{2}{3} + \frac{d}{d u} [u^{\frac{2}{3}}] \frac{d}{d x} [x + 1]$

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

$\frac{2}{3} + \frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} [x + 1]$

Replace all occurrences of $u$ with $x + 1$ .

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

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

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

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

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $3$ .

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

Subtract $3$ from $2$ .

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

Move the negative in front of the fraction.

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

Add $1$ and $0$ .

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

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

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

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

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

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

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

Find the second derivative of the function.

Differentiate.

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

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

Since $\frac{2}{3}$ is constant with respect to $x$ , the derivative of $\frac{2}{3}$ with respect to $x$ is $0$ .

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

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

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

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

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

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

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = {(x + 1)}^{\frac{1}{3}}$ .

To apply the Chain Rule, set $u_{1}$ as ${(x + 1)}^{\frac{1}{3}}$ .

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

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

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

Replace all occurrences of $u_{1}$ with ${(x + 1)}^{\frac{1}{3}}$ .

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

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 $g (x) = x + 1$ .

To apply the Chain Rule, set $u_{2}$ as $x + 1$ .

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

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

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

Replace all occurrences of $u_{2}$ with $x + 1$ .

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

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

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

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

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

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

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

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

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

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

Move the negative in front of the fraction.

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $3$ .

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

Subtract $3$ from $1$ .

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

Move the negative in front of the fraction.

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

Add $1$ and $0$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Add $2$ and $2$ .

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

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

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

Multiply $3$ by $3$ .

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

Subtract $\frac{2}{9 {(x + 1)}^{\frac{4}{3}}}$ from $0$ .

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

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

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

Find the first derivative.

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

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

Evaluate $\frac{d}{d x} [\frac{2 x}{3}]$ .

Since $\frac{2}{3}$ is constant with respect to $x$ , the derivative of $\frac{2 x}{3}$ with respect to $x$ is $\frac{2}{3} \frac{d}{d x} [x]$ .

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

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

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

Multiply $\frac{2}{3}$ by $1$ .

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

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

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 $g (x) = x + 1$ .

To apply the Chain Rule, set $u$ as $x + 1$ .

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

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

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

Replace all occurrences of $u$ with $x + 1$ .

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

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

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

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

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $3$ .

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

Subtract $3$ from $2$ .

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

Move the negative in front of the fraction.

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

Add $1$ and $0$ .

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

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

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

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

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

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

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

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

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

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

Set the first derivative equal to $0$ .

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

Subtract $\frac{2}{3}$ from both sides of the equation.

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

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

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

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

The LCM of $3, 3$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$3$

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

The LCM of ${(x + 1)}^{\frac{1}{3}}$ is the result of multiplying all factors the greatest number of times they occur in either term.

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

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

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

Multiply each term in $\frac{2}{3 {(x + 1)}^{\frac{1}{3}}} = - \frac{2}{3}$ by $3 {(x + 1)}^{\frac{1}{3}}$ to eliminate the fractions.

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

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

Simplify the left side.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of ${(x + 1)}^{\frac{1}{3}}$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Factor $3$ out of $3 {(x + 1)}^{\frac{1}{3}}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Solve the equation.

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

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

Divide each term in $- 2 {(x + 1)}^{\frac{1}{3}} = 2$ by $- 2$ and simplify.

Divide each term in $- 2 {(x + 1)}^{\frac{1}{3}} = 2$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor.

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

Divide ${(x + 1)}^{\frac{1}{3}}$ by $1$ .

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

Simplify the right side.

Divide $2$ by $- 2$ .

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

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

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

Simplify the exponent.

Simplify the left side.

Simplify ${({(x + 1)}^{\frac{1}{3}})}^{3}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

${(x + 1)}^{1} = {(- 1)}^{3}$

Simplify.

$x + 1 = {(- 1)}^{3}$

Simplify the right side.

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

$x + 1 = - 1$

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$x = - 1 - 1$

Subtract $1$ from $- 1$ .

$x = - 2$

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.

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

Anything raised to $1$ is the base itself.

$\frac{2}{3} + \frac{2}{3 \sqrt[3]{x + 1}}$

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

$3 \sqrt[3]{x + 1} = 0$

Solve for $x$ .

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

${(3 \sqrt[3]{x + 1})}^{3} = 0^{3}$

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${(3 {(x + 1)}^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $3 {(x + 1)}^{\frac{1}{3}}$ .

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

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

$27 {({(x + 1)}^{\frac{1}{3}})}^{3} = 0^{3}$

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

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

$27 {(x + 1)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$27 {(x + 1)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Rewrite the expression.

$27 {(x + 1)}^{1} = 0^{3}$

Simplify.

$27 (x + 1) = 0^{3}$

Apply the distributive property.

$27 x + 27 \cdot 1 = 0^{3}$

Multiply $27$ by $1$ .

$27 x + 27 = 0^{3}$

Simplify the right side.

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

$27 x + 27 = 0$

Solve for $x$ .

Subtract $27$ from both sides of the equation.

$27 x = - 27$

Divide each term in $27 x = - 27$ by $27$ and simplify.

Divide each term in $27 x = - 27$ by $27$ .

$\frac{27 x}{27} = \frac{- 27}{27}$

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$\frac{27 x}{27} = \frac{- 27}{27}$

Divide $x$ by $1$ .

$x = \frac{- 27}{27}$

Simplify the right side.

Divide $- 27$ by $27$ .

$x = - 1$

Critical points to evaluate.

$x = - 2, - 1$

Evaluate the second derivative at $x = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2}{9 {((- 2) + 1)}^{\frac{4}{3}}}$

Evaluate the second derivative.

Simplify the denominator.

Add $- 2$ and $1$ .

$- \frac{2}{9 {(- 1)}^{\frac{4}{3}}}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$- \frac{2}{9 {({(- 1)}^{3})}^{\frac{4}{3}}}$

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

$- \frac{2}{9 {(- 1)}^{3 (\frac{4}{3})}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$- \frac{2}{9 {(- 1)}^{3 (\frac{4}{3})}}$

Rewrite the expression.

$- \frac{2}{9 {(- 1)}^{4}}$

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

$- \frac{2}{9 \cdot 1}$

Multiply $9$ by $1$ .

$- \frac{2}{9}$

$x = - 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 2$ is a local maximum

Find the y-value when $x = - 2$ .

Replace the variable $x$ with $- 2$ in the expression.

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

Simplify the result.

Multiply $2$ by $- 2$ .

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

Simplify each term.

Move the negative in front of the fraction.

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

Add $- 2$ and $1$ .

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

Rewrite $- 1$ as ${(- 1)}^{3}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

$f (- 2) = - \frac{4}{3} + 1$

Simplify the expression.

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

$f (- 2) = - \frac{4}{3} + \frac{3}{3}$

Combine the numerators over the common denominator.

$f (- 2) = \frac{- 4 + 3}{3}$

Add $- 4$ and $3$ .

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

Move the negative in front of the fraction.

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

The final answer is $- \frac{1}{3}$ .

$y = - \frac{1}{3}$

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2}{9 {((- 1) + 1)}^{\frac{4}{3}}}$

Evaluate the second derivative.

Simplify the expression.

Add $- 1$ and $1$ .

$- \frac{2}{9 \cdot 0^{\frac{4}{3}}}$

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

$- \frac{2}{9 \cdot {(0^{3})}^{\frac{4}{3}}}$

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

$- \frac{2}{9 \cdot 0^{3 (\frac{4}{3})}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$- \frac{2}{9 \cdot 0^{3 (\frac{4}{3})}}$

Rewrite the expression.

$- \frac{2}{9 \cdot 0^{4}}$

Simplify the expression.

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

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

Multiply $9$ by $0$ .

$- \frac{2}{0}$

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

Undefined

$- \frac{2}{0}$

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

Undefined

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

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

$(- \infty, - 2) \cup (- 2, - 1) \cup (- 1, \infty)$

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

Replace the variable $x$ with $- 4$ in the expression.

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

Simplify the result.

Add $- 4$ and $1$ .

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

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

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

Substitute any number, such as $- 1.5$ , from the interval $(- 2, - 1)$ in the first derivative $\frac{2}{3} + \frac{2}{3 {(x + 1)}^{\frac{1}{3}}}$ to check if the result is negative or positive.

Replace the variable $x$ with $- 1.5$ in the expression.

$f' (- 1.5) = \frac{2}{3} + \frac{2}{3 {((- 1.5) + 1)}^{\frac{1}{3}}}$

Simplify the result.

Add $- 1.5$ and $1$ .

$f' (- 1.5) = \frac{2}{3} + \frac{2}{3 {(- 0.5)}^{\frac{1}{3}}}$

The final answer is $\frac{2}{3} + \frac{2}{3 {(- 0.5)}^{\frac{1}{3}}}$ .

$\frac{2}{3} + \frac{2}{3 {(- 0.5)}^{\frac{1}{3}}}$

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

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

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

Simplify the result.

Simplify each term.

Simplify the denominator.

Add $0$ and $1$ .

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

One to any power is one.

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

Multiply $3$ by $1$ .

$f' (0) = \frac{2}{3} + \frac{2}{3}$

Combine fractions.

Combine the numerators over the common denominator.

$f' (0) = \frac{2 + 2}{3}$

Add $2$ and $2$ .

$f' (0) = \frac{4}{3}$

The final answer is $\frac{4}{3}$ .

$\frac{4}{3}$

Since the first derivative changed signs from positive to negative around $x = - 2$ , then $x = - 2$ is a local maximum.

$x = - 2$ is a local maximum

Since the first derivative changed signs from negative to positive around $x = - 1$ , then $x = - 1$ is a local minimum.

$x = - 1$ is a local minimum

These are the local extrema for $f (x) = \frac{2 x}{3} + {(x + 1)}^{\frac{2}{3}}$ .

$x = - 2$ is a local maximum

$x = - 1$ is a local minimum

$x = - 2$ is a local maximum

$x = - 1$ is a local minimum

Find the Maximum/Minimum Value (2x)/3+(x+1)^(2/3)

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