Calculus Examples

$f (x) = 6 {(x - 1)}^{\frac{2}{3}} - 2 {(x - 1)}^{2}$

Step 1

Find the first derivative of the function.

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Step 1.1

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 ((x - 1) (x - 1))]$

Step 1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x (x - 1) - 1 (x - 1))]$

Step 1.2.2

Apply the distributive property.

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x \cdot x + x \cdot - 1 - 1 (x - 1))]$

Step 1.2.3

Apply the distributive property.

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1)]$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1)]$

Step 1.3.1.2

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1)]$

Step 1.3.1.3

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - x - 1 x - 1 \cdot - 1)]$

Step 1.3.1.4

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - x - x - 1 \cdot - 1)]$

Step 1.3.1.5

Multiply $- 1$ by $- 1$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - x - x + 1)]$

Step 1.3.2

Subtract $x$ from $- x$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - 2 x + 1)]$

Step 1.4

By the Sum Rule, the derivative of $6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - 2 x + 1)$ with respect to $x$ is $\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}}] + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}}] + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5

Evaluate $\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}}]$ .

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Step 1.5.1

Since $6$ is constant with respect to $x$ , the derivative of $6 {(x - 1)}^{\frac{2}{3}}$ with respect to $x$ is $6 \frac{d}{d x} [{(x - 1)}^{\frac{2}{3}}]$ .

$6 \frac{d}{d x} [{(x - 1)}^{\frac{2}{3}}] + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.2

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 $g (x) = x - 1$ .

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Step 1.5.2.1

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

$6 (\frac{d}{d u} [u^{\frac{2}{3}}] \frac{d}{d x} [x - 1]) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.2.2

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

$6 (\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} [x - 1]) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.2.3

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} \frac{d}{d x} [x - 1]) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.3

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [- 1])) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.4

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} (1 + \frac{d}{d x} [- 1])) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.5

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.6

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.7

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.8

Combine the numerators over the common denominator.

$6 (\frac{2}{3} {(x - 1)}^{\frac{2 - 1 \cdot 3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.9

Simplify the numerator.

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Step 1.5.9.1

Multiply $- 1$ by $3$ .

$6 (\frac{2}{3} {(x - 1)}^{\frac{2 - 3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.9.2

Subtract $3$ from $2$ .

$6 (\frac{2}{3} {(x - 1)}^{\frac{- 1}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.10

Move the negative in front of the fraction.

$6 (\frac{2}{3} {(x - 1)}^{- \frac{1}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.11

Add $1$ and $0$ .

$6 (\frac{2}{3} {(x - 1)}^{- \frac{1}{3}} \cdot 1) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.12

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

$6 (\frac{2 {(x - 1)}^{- \frac{1}{3}}}{3} \cdot 1) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.13

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

$6 \frac{2 {(x - 1)}^{- \frac{1}{3}}}{3} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.14

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

$6 \frac{2}{3 {(x - 1)}^{\frac{1}{3}}} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.15

Combine $6$ and $\frac{2}{3 {(x - 1)}^{\frac{1}{3}}}$ .

$\frac{6 \cdot 2}{3 {(x - 1)}^{\frac{1}{3}}} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.16

Multiply $6$ by $2$ .

$\frac{12}{3 {(x - 1)}^{\frac{1}{3}}} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 1.5.17

Factor $3$ out of $12$ .

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

Step 1.5.18

Cancel the common factors.

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Step 1.5.18.1

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

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

Step 1.5.18.2

Cancel the common factor.

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

Step 1.5.18.3

Rewrite the expression.

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

Step 1.6

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

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Step 1.6.1

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

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

Step 1.6.2

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

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

Step 1.6.3

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

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

Step 1.6.4

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

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

Step 1.6.5

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

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

Step 1.6.6

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

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

Step 1.6.7

Multiply $- 2$ by $1$ .

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

Step 1.6.8

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

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

Step 1.7

Simplify.

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Step 1.7.1

Apply the distributive property.

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

Step 1.7.2

Combine terms.

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Step 1.7.2.1

Multiply $2$ by $- 2$ .

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

Step 1.7.2.2

Multiply $- 2$ by $- 2$ .

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

Step 1.7.3

Reorder terms.

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

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- 4 x + \frac{4}{{(x - 1)}^{\frac{1}{3}}} + 4$ with respect to $x$ is $\frac{d}{d x} [- 4 x] + \frac{d}{d x} [\frac{4}{{(x - 1)}^{\frac{1}{3}}}] + \frac{d}{d x} [4]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 2.2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $- 4$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

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

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

Step 2.3.4

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 $g (x) = x - 1$ .

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Step 2.3.4.1

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

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

Step 2.3.4.2

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

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

Step 2.3.4.3

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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

Step 2.3.8

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

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Step 2.3.8.1

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

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

Step 2.3.8.2

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

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

Step 2.3.8.3

Move the negative in front of the fraction.

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

Step 2.3.9

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

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

Step 2.3.10

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

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

Step 2.3.11

Combine the numerators over the common denominator.

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

Step 2.3.12

Simplify the numerator.

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Step 2.3.12.1

Multiply $- 1$ by $3$ .

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

Step 2.3.12.2

Subtract $3$ from $1$ .

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

Step 2.3.13

Move the negative in front of the fraction.

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

Step 2.3.14

Add $1$ and $0$ .

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

Step 2.3.15

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

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

Step 2.3.16

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

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

Step 2.3.17

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

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

Step 2.3.18

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

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

Step 2.3.19

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

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

Step 2.3.20

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

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Step 2.3.20.1

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

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

Step 2.3.20.2

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

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

Step 2.3.20.3

Combine the numerators over the common denominator.

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

Step 2.3.20.4

Add $2$ and $2$ .

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

Step 2.3.21

Multiply $- 1$ by $4$ .

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

Step 2.3.22

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

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

Step 2.3.23

Move the negative in front of the fraction.

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

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

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

Step 2.4.2

Add $- 4 - \frac{4}{3 {(x - 1)}^{\frac{4}{3}}}$ and $0$ .

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

Step 3

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

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

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

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 ((x - 1) (x - 1))]$

Step 4.1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 4.1.2.1

Apply the distributive property.

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x (x - 1) - 1 (x - 1))]$

Step 4.1.2.2

Apply the distributive property.

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x \cdot x + x \cdot - 1 - 1 (x - 1))]$

Step 4.1.2.3

Apply the distributive property.

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1)]$

Step 4.1.3

Simplify and combine like terms.

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Step 4.1.3.1

Simplify each term.

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Step 4.1.3.1.1

Multiply $x$ by $x$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1)]$

Step 4.1.3.1.2

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1)]$

Step 4.1.3.1.3

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - x - 1 x - 1 \cdot - 1)]$

Step 4.1.3.1.4

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

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - x - x - 1 \cdot - 1)]$

Step 4.1.3.1.5

Multiply $- 1$ by $- 1$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - x - x + 1)]$

Step 4.1.3.2

Subtract $x$ from $- x$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - 2 x + 1)]$

Step 4.1.4

By the Sum Rule, the derivative of $6 {(x - 1)}^{\frac{2}{3}} - 2 (x^{2} - 2 x + 1)$ with respect to $x$ is $\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}}] + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$ .

$\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}}] + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5

Evaluate $\frac{d}{d x} [6 {(x - 1)}^{\frac{2}{3}}]$ .

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Step 4.1.5.1

Since $6$ is constant with respect to $x$ , the derivative of $6 {(x - 1)}^{\frac{2}{3}}$ with respect to $x$ is $6 \frac{d}{d x} [{(x - 1)}^{\frac{2}{3}}]$ .

$6 \frac{d}{d x} [{(x - 1)}^{\frac{2}{3}}] + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.2

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 $g (x) = x - 1$ .

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Step 4.1.5.2.1

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

$6 (\frac{d}{d u} [u^{\frac{2}{3}}] \frac{d}{d x} [x - 1]) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.2.2

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

$6 (\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} [x - 1]) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.2.3

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} \frac{d}{d x} [x - 1]) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.3

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [- 1])) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.4

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} (1 + \frac{d}{d x} [- 1])) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.5

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.6

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.7

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

$6 (\frac{2}{3} {(x - 1)}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.8

Combine the numerators over the common denominator.

$6 (\frac{2}{3} {(x - 1)}^{\frac{2 - 1 \cdot 3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.9

Simplify the numerator.

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Step 4.1.5.9.1

Multiply $- 1$ by $3$ .

$6 (\frac{2}{3} {(x - 1)}^{\frac{2 - 3}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.9.2

Subtract $3$ from $2$ .

$6 (\frac{2}{3} {(x - 1)}^{\frac{- 1}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.10

Move the negative in front of the fraction.

$6 (\frac{2}{3} {(x - 1)}^{- \frac{1}{3}} (1 + 0)) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.11

Add $1$ and $0$ .

$6 (\frac{2}{3} {(x - 1)}^{- \frac{1}{3}} \cdot 1) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.12

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

$6 (\frac{2 {(x - 1)}^{- \frac{1}{3}}}{3} \cdot 1) + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.13

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

$6 \frac{2 {(x - 1)}^{- \frac{1}{3}}}{3} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.14

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

$6 \frac{2}{3 {(x - 1)}^{\frac{1}{3}}} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.15

Combine $6$ and $\frac{2}{3 {(x - 1)}^{\frac{1}{3}}}$ .

$\frac{6 \cdot 2}{3 {(x - 1)}^{\frac{1}{3}}} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.16

Multiply $6$ by $2$ .

$\frac{12}{3 {(x - 1)}^{\frac{1}{3}}} + \frac{d}{d x} [- 2 (x^{2} - 2 x + 1)]$

Step 4.1.5.17

Factor $3$ out of $12$ .

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

Step 4.1.5.18

Cancel the common factors.

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Step 4.1.5.18.1

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

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

Step 4.1.5.18.2

Cancel the common factor.

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

Step 4.1.5.18.3

Rewrite the expression.

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

Step 4.1.6

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

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Step 4.1.6.1

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

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

Step 4.1.6.2

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

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

Step 4.1.6.3

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

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

Step 4.1.6.4

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

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

Step 4.1.6.5

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

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

Step 4.1.6.6

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

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

Step 4.1.6.7

Multiply $- 2$ by $1$ .

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

Step 4.1.6.8

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

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

Step 4.1.7

Simplify.

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Step 4.1.7.1

Apply the distributive property.

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

Step 4.1.7.2

Combine terms.

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Step 4.1.7.2.1

Multiply $2$ by $- 2$ .

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

Step 4.1.7.2.2

Multiply $- 2$ by $- 2$ .

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

Step 4.1.7.3

Reorder terms.

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

Step 4.2

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

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

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = 0, 2$

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.

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

Step 6.1.2

Anything raised to $1$ is the base itself.

$- 4 x + \frac{4}{\sqrt[3]{x - 1}} + 4$

Step 6.2

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

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

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.

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

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

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

Step 6.3.2.2

Simplify the left side.

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Step 6.3.2.2.1

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

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Step 6.3.2.2.1.1

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

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Step 6.3.2.2.1.1.1

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

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

Step 6.3.2.2.1.1.2

Cancel the common factor of $3$ .

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Step 6.3.2.2.1.1.2.1

Cancel the common factor.

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

Step 6.3.2.2.1.1.2.2

Rewrite the expression.

${(x - 1)}^{1} = 0^{3}$

Step 6.3.2.2.1.2

Simplify.

$x - 1 = 0^{3}$

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 $0$ .

$x - 1 = 0$

Step 6.3.3

Add $1$ to both sides of the equation.

$x = 1$

Step 7

Critical points to evaluate.

$x = 0, 2, 1$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 - \frac{4}{3 {((0) - 1)}^{\frac{4}{3}}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Simplify the denominator.

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Step 9.1.1.1

Subtract $1$ from $0$ .

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

Step 9.1.1.2

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

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

Step 9.1.1.3

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

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

Step 9.1.1.4

Cancel the common factor of $3$ .

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Step 9.1.1.4.1

Cancel the common factor.

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

Step 9.1.1.4.2

Rewrite the expression.

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

Step 9.1.1.5

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

$- 4 - \frac{4}{3 \cdot 1}$

Step 9.1.2

Multiply $3$ by $1$ .

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

Step 9.2

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

$- 4 \cdot \frac{3}{3} - \frac{4}{3}$

Step 9.3

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

$\frac{- 4 \cdot 3}{3} - \frac{4}{3}$

Step 9.4

Combine the numerators over the common denominator.

$\frac{- 4 \cdot 3 - 4}{3}$

Step 9.5

Simplify the numerator.

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Step 9.5.1

Multiply $- 4$ by $3$ .

$\frac{- 12 - 4}{3}$

Step 9.5.2

Subtract $4$ from $- 12$ .

$\frac{- 16}{3}$

Step 9.6

Move the negative in front of the fraction.

$- \frac{16}{3}$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

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

$f (0) = 6 {((0) - 1)}^{\frac{2}{3}} - 2 {((0) - 1)}^{2}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Subtract $1$ from $0$ .

$f (0) = 6 {(- 1)}^{\frac{2}{3}} - 2 {((0) - 1)}^{2}$

Step 11.2.1.2

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

$f (0) = 6 {({(- 1)}^{3})}^{\frac{2}{3}} - 2 {((0) - 1)}^{2}$

Step 11.2.1.3

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

$f (0) = 6 {(- 1)}^{3 (\frac{2}{3})} - 2 {((0) - 1)}^{2}$

Step 11.2.1.4

Cancel the common factor of $3$ .

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Step 11.2.1.4.1

Cancel the common factor.

$f (0) = 6 {(- 1)}^{3 (\frac{2}{3})} - 2 {((0) - 1)}^{2}$

Step 11.2.1.4.2

Rewrite the expression.

$f (0) = 6 {(- 1)}^{2} - 2 {((0) - 1)}^{2}$

Step 11.2.1.5

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

$f (0) = 6 \cdot 1 - 2 {((0) - 1)}^{2}$

Step 11.2.1.6

Multiply $6$ by $1$ .

$f (0) = 6 - 2 {((0) - 1)}^{2}$

Step 11.2.1.7

Subtract $1$ from $0$ .

$f (0) = 6 - 2 {(- 1)}^{2}$

Step 11.2.1.8

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

$f (0) = 6 - 2 \cdot 1$

Step 11.2.1.9

Multiply $- 2$ by $1$ .

$f (0) = 6 - 2$

Step 11.2.2

Subtract $2$ from $6$ .

$f (0) = 4$

Step 11.2.3

The final answer is $4$ .

$y = 4$

Step 12

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.

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

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

Simplify the denominator.

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Step 13.1.1.1

Subtract $1$ from $2$ .

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

Step 13.1.1.2

One to any power is one.

$- 4 - \frac{4}{3 \cdot 1}$

Step 13.1.2

Multiply $3$ by $1$ .

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

Step 13.2

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

$- 4 \cdot \frac{3}{3} - \frac{4}{3}$

Step 13.3

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

$\frac{- 4 \cdot 3}{3} - \frac{4}{3}$

Step 13.4

Combine the numerators over the common denominator.

$\frac{- 4 \cdot 3 - 4}{3}$

Step 13.5

Simplify the numerator.

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Step 13.5.1

Multiply $- 4$ by $3$ .

$\frac{- 12 - 4}{3}$

Step 13.5.2

Subtract $4$ from $- 12$ .

$\frac{- 16}{3}$

Step 13.6

Move the negative in front of the fraction.

$- \frac{16}{3}$

Step 14

$x = 2$ 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

Step 15

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

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Step 15.1

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

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

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

Subtract $1$ from $2$ .

$f (2) = 6 \cdot 1^{\frac{2}{3}} - 2 {((2) - 1)}^{2}$

Step 15.2.1.2

One to any power is one.

$f (2) = 6 \cdot 1 - 2 {((2) - 1)}^{2}$

Step 15.2.1.3

Multiply $6$ by $1$ .

$f (2) = 6 - 2 {((2) - 1)}^{2}$

Step 15.2.1.4

Subtract $1$ from $2$ .

$f (2) = 6 - 2 \cdot 1^{2}$

Step 15.2.1.5

One to any power is one.

$f (2) = 6 - 2 \cdot 1$

Step 15.2.1.6

Multiply $- 2$ by $1$ .

$f (2) = 6 - 2$

Step 15.2.2

Subtract $2$ from $6$ .

$f (2) = 4$

Step 15.2.3

The final answer is $4$ .

$y = 4$

Step 16

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.

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

Step 17

Evaluate the second derivative.

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Step 17.1

Simplify the expression.

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Step 17.1.1

Subtract $1$ from $1$ .

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

Step 17.1.2

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

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

Step 17.1.3

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

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

Step 17.2

Cancel the common factor of $3$ .

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Step 17.2.1

Cancel the common factor.

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

Step 17.2.2

Rewrite the expression.

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

Step 17.3

Simplify the expression.

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Step 17.3.1

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

$- 4 - \frac{4}{3 \cdot 0}$

Step 17.3.2

Multiply $3$ by $0$ .

$- 4 - \frac{4}{0}$

Step 17.3.3

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

Undefined

$- 4 - \frac{4}{0}$

Step 17.4

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

Undefined

Step 18

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

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Step 18.1

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

$(- \infty, 0) \cup (0, 1) \cup (1, 2) \cup (2, \infty)$

Step 18.2

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

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Step 18.2.1

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

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

Step 18.2.2

Simplify the result.

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Step 18.2.2.1

Simplify each term.

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Step 18.2.2.1.1

Multiply $- 4$ by $- 2$ .

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

Step 18.2.2.1.2

Subtract $1$ from $- 2$ .

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

Step 18.2.2.2

Add $8$ and $4$ .

$f' (- 2) = 12 + \frac{4}{{(- 3)}^{\frac{1}{3}}}$

Step 18.2.2.3

The final answer is $12 + \frac{4}{{(- 3)}^{\frac{1}{3}}}$ .

$12 + \frac{4}{{(- 3)}^{\frac{1}{3}}}$

Step 18.3

Substitute any number, such as $0.5$ , from the interval $(0, 1)$ in the first derivative $- 4 x + \frac{4}{{(x - 1)}^{\frac{1}{3}}} + 4$ to check if the result is negative or positive.

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Step 18.3.1

Replace the variable $x$ with $0.5$ in the expression.

$f' (0.5) = - 4 \cdot 0.5 + \frac{4}{{((0.5) - 1)}^{\frac{1}{3}}} + 4$

Step 18.3.2

Simplify the result.

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Step 18.3.2.1

Simplify each term.

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Step 18.3.2.1.1

Multiply $- 4$ by $0.5$ .

$f' (0.5) = - 2 + \frac{4}{{((0.5) - 1)}^{\frac{1}{3}}} + 4$

Step 18.3.2.1.2

Subtract $1$ from $0.5$ .

$f' (0.5) = - 2 + \frac{4}{{(- 0.5)}^{\frac{1}{3}}} + 4$

Step 18.3.2.2

Add $- 2$ and $4$ .

$f' (0.5) = 2 + \frac{4}{{(- 0.5)}^{\frac{1}{3}}}$

Step 18.3.2.3

The final answer is $2 + \frac{4}{{(- 0.5)}^{\frac{1}{3}}}$ .

$2 + \frac{4}{{(- 0.5)}^{\frac{1}{3}}}$

Step 18.4

Substitute any number, such as $1.5$ , from the interval $(1, 2)$ in the first derivative $- 4 x + \frac{4}{{(x - 1)}^{\frac{1}{3}}} + 4$ to check if the result is negative or positive.

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Step 18.4.1

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

$f' (1.5) = - 4 \cdot 1.5 + \frac{4}{{((1.5) - 1)}^{\frac{1}{3}}} + 4$

Step 18.4.2

Simplify the result.

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Step 18.4.2.1

Simplify each term.

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Step 18.4.2.1.1

Multiply $- 4$ by $1.5$ .

$f' (1.5) = - 6 + \frac{4}{{((1.5) - 1)}^{\frac{1}{3}}} + 4$

Step 18.4.2.1.2

Simplify the denominator.

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Step 18.4.2.1.2.1

Subtract $1$ from $1.5$ .

$f' (1.5) = - 6 + \frac{4}{{0.5}^{\frac{1}{3}}} + 4$

Step 18.4.2.1.2.2

Raise $0.5$ to the power of $\frac{1}{3}$ .

$f' (1.5) = - 6 + \frac{4}{0.79370052} + 4$

Step 18.4.2.1.3

Divide $4$ by $0.79370052$ .

$f' (1.5) = - 6 + 5.03968419 + 4$

Step 18.4.2.2

Simplify by adding numbers.

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Step 18.4.2.2.1

Add $- 6$ and $5.03968419$ .

$f' (1.5) = - 0.9603158 + 4$

Step 18.4.2.2.2

Add $- 0.9603158$ and $4$ .

$f' (1.5) = 3.03968419$

Step 18.4.2.3

The final answer is $3.03968419$ .

$3.03968419$

Step 18.5

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

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Step 18.5.1

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

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

Step 18.5.2

Simplify the result.

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Step 18.5.2.1

Simplify each term.

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Step 18.5.2.1.1

Multiply $- 4$ by $4$ .

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

Step 18.5.2.1.2

Subtract $1$ from $4$ .

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

Step 18.5.2.2

Add $- 16$ and $4$ .

$f' (4) = - 12 + \frac{4}{3^{\frac{1}{3}}}$

Step 18.5.2.3

The final answer is $- 12 + \frac{4}{3^{\frac{1}{3}}}$ .

$- 12 + \frac{4}{3^{\frac{1}{3}}}$

Step 18.6

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 18.7

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

Step 18.8

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

Step 18.9

These are the local extrema for $f (x) = 6 {(x - 1)}^{\frac{2}{3}} - 2 {(x - 1)}^{2}$ .

$x = 0$ is a local maximum

$x = 1$ is a local minimum

$x = 2$ is a local maximum

$x = 0$ is a local maximum

$x = 1$ is a local minimum

$x = 2$ is a local maximum

Step 19