Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $\frac{8}{3} x^{3} - 2 x + \frac{1}{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{8}{3} x^{3}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$ .

$\frac{d}{d x} [\frac{8}{3} x^{3}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2

Evaluate $\frac{d}{d x} [\frac{8}{3} x^{3}]$ .

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Step 1.2.1

Since $\frac{8}{3}$ is constant with respect to $x$ , the derivative of $\frac{8}{3} x^{3}$ with respect to $x$ is $\frac{8}{3} \frac{d}{d x} [x^{3}]$ .

$\frac{8}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.2

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

$\frac{8}{3} (3 x^{2}) + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.3

Combine $3$ and $\frac{8}{3}$ .

$\frac{3 \cdot 8}{3} x^{2} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.4

Multiply $3$ by $8$ .

$\frac{24}{3} x^{2} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.5

Combine $\frac{24}{3}$ and $x^{2}$ .

$\frac{24 x^{2}}{3} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.6

Cancel the common factor of $24$ and $3$ .

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Step 1.2.6.1

Factor $3$ out of $24 x^{2}$ .

$\frac{3 (8 x^{2})}{3} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.6.2

Cancel the common factors.

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Step 1.2.6.2.1

Factor $3$ out of $3$ .

$\frac{3 (8 x^{2})}{3 (1)} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.6.2.2

Cancel the common factor.

$\frac{3 (8 x^{2})}{3 \cdot 1} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.6.2.3

Rewrite the expression.

$\frac{8 x^{2}}{1} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.2.6.2.4

Divide $8 x^{2}$ by $1$ .

$8 x^{2} + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.3

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

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Step 1.3.1

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

$8 x^{2} - 2 \frac{d}{d x} [x] + \frac{d}{d x} [\frac{1}{3}]$

Step 1.3.2

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

$8 x^{2} - 2 \cdot 1 + \frac{d}{d x} [\frac{1}{3}]$

Step 1.3.3

Multiply $- 2$ by $1$ .

$8 x^{2} - 2 + \frac{d}{d x} [\frac{1}{3}]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3}$ with respect to $x$ is $0$ .

$8 x^{2} - 2 + 0$

Step 1.4.2

Add $8 x^{2} - 2$ and $0$ .

$8 x^{2} - 2$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $8 x^{2} - 2$ with respect to $x$ is $\frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [- 2]$ .

$f'' (x) = \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (- 2)$

Step 2.2

Evaluate $\frac{d}{d x} [8 x^{2}]$ .

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Step 2.2.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 8 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 2)$

Step 2.2.2

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

$f'' (x) = 8 (2 x) + \frac{d}{d x} (- 2)$

Step 2.2.3

Multiply $2$ by $8$ .

$f'' (x) = 16 x + \frac{d}{d x} (- 2)$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$f'' (x) = 16 x + 0$

Step 2.3.2

Add $16 x$ and $0$ .

$f'' (x) = 16 x$

Step 3

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

$8 x^{2} - 2 = 0$

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

By the Sum Rule, the derivative of $\frac{8}{3} x^{3} - 2 x + \frac{1}{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{8}{3} x^{3}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [\frac{1}{3}]$ .

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

Step 4.1.2

Evaluate $\frac{d}{d x} [\frac{8}{3} x^{3}]$ .

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Step 4.1.2.1

Since $\frac{8}{3}$ is constant with respect to $x$ , the derivative of $\frac{8}{3} x^{3}$ with respect to $x$ is $\frac{8}{3} \frac{d}{d x} [x^{3}]$ .

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Combine $3$ and $\frac{8}{3}$ .

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

Step 4.1.2.4

Multiply $3$ by $8$ .

$f' (x) = \frac{24}{3} x^{2} + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (\frac{1}{3})$

Step 4.1.2.5

Combine $\frac{24}{3}$ and $x^{2}$ .

$f' (x) = \frac{24 x^{2}}{3} + \frac{d}{d x} (- 2 x) + \frac{d}{d x} (\frac{1}{3})$

Step 4.1.2.6

Cancel the common factor of $24$ and $3$ .

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Step 4.1.2.6.1

Factor $3$ out of $24 x^{2}$ .

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

Step 4.1.2.6.2

Cancel the common factors.

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Step 4.1.2.6.2.1

Factor $3$ out of $3$ .

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

Step 4.1.2.6.2.2

Cancel the common factor.

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

Step 4.1.2.6.2.3

Rewrite the expression.

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

Step 4.1.2.6.2.4

Divide $8 x^{2}$ by $1$ .

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

Step 4.1.3

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

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Step 4.1.3.1

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

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

Step 4.1.3.2

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

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

Step 4.1.3.3

Multiply $- 2$ by $1$ .

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

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{1}{3}$ with respect to $x$ is $0$ .

$f' (x) = 8 x^{2} - 2 + 0$

Step 4.1.4.2

Add $8 x^{2} - 2$ and $0$ .

$f' (x) = 8 x^{2} - 2$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $8 x^{2} - 2$ .

$8 x^{2} - 2$

Step 5

Set the first derivative equal to $0$ then solve the equation $8 x^{2} - 2 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$8 x^{2} - 2 = 0$

Step 5.2

Add $2$ to both sides of the equation.

$8 x^{2} = 2$

Step 5.3

Divide each term in $8 x^{2} = 2$ by $8$ and simplify.

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Step 5.3.1

Divide each term in $8 x^{2} = 2$ by $8$ .

$\frac{8 x^{2}}{8} = \frac{2}{8}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $8$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{8 x^{2}}{8} = \frac{2}{8}$

Step 5.3.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{2}{8}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Cancel the common factor of $2$ and $8$ .

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Step 5.3.3.1.1

Factor $2$ out of $2$ .

$x^{2} = \frac{2 (1)}{8}$

Step 5.3.3.1.2

Cancel the common factors.

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Step 5.3.3.1.2.1

Factor $2$ out of $8$ .

$x^{2} = \frac{2 \cdot 1}{2 \cdot 4}$

Step 5.3.3.1.2.2

Cancel the common factor.

$x^{2} = \frac{2 \cdot 1}{2 \cdot 4}$

Step 5.3.3.1.2.3

Rewrite the expression.

$x^{2} = \frac{1}{4}$

Step 5.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\frac{1}{4}}$

Step 5.5

Simplify $\pm \sqrt{\frac{1}{4}}$ .

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Step 5.5.1

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Step 5.5.2

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{4}}$

Step 5.5.3

Simplify the denominator.

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Step 5.5.3.1

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

$x = \pm \frac{1}{\sqrt{2^{2}}}$

Step 5.5.3.2

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

$x = \pm \frac{1}{2}$

Step 5.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.6.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{1}{2}$

Step 5.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{1}{2}$

Step 5.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{1}{2}, - \frac{1}{2}$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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

Step 7

Critical points to evaluate.

$x = \frac{1}{2}, - \frac{1}{2}$

Step 8

Evaluate the second derivative at $x = \frac{1}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$16 (\frac{1}{2})$

Step 9

Cancel the common factor of $2$ .

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Step 9.1

Factor $2$ out of $16$ .

$2 (8) \frac{1}{2}$

Step 9.2

Cancel the common factor.

$2 \cdot 8 \frac{1}{2}$

Step 9.3

Rewrite the expression.

$8$

Step 10

$x = \frac{1}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{1}{2}$ is a local minimum

Step 11

Find the y-value when $x = \frac{1}{2}$ .

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Step 11.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

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

Apply the product rule to $\frac{1}{2}$ .

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

Step 11.2.1.2

Combine.

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

Step 11.2.1.3

One to any power is one.

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

Step 11.2.1.4

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

$f (\frac{1}{2}) = \frac{8 \cdot 1}{3 \cdot 8} - 2 (\frac{1}{2}) + \frac{1}{3}$

Step 11.2.1.5

Multiply $8$ by $1$ .

$f (\frac{1}{2}) = \frac{8}{3 \cdot 8} - 2 (\frac{1}{2}) + \frac{1}{3}$

Step 11.2.1.6

Multiply $3$ by $8$ .

$f (\frac{1}{2}) = \frac{8}{24} - 2 (\frac{1}{2}) + \frac{1}{3}$

Step 11.2.1.7

Cancel the common factor of $8$ and $24$ .

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Step 11.2.1.7.1

Factor $8$ out of $8$ .

$f (\frac{1}{2}) = \frac{8 (1)}{24} - 2 (\frac{1}{2}) + \frac{1}{3}$

Step 11.2.1.7.2

Cancel the common factors.

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Step 11.2.1.7.2.1

Factor $8$ out of $24$ .

$f (\frac{1}{2}) = \frac{8 \cdot 1}{8 \cdot 3} - 2 (\frac{1}{2}) + \frac{1}{3}$

Step 11.2.1.7.2.2

Cancel the common factor.

$f (\frac{1}{2}) = \frac{8 \cdot 1}{8 \cdot 3} - 2 (\frac{1}{2}) + \frac{1}{3}$

Step 11.2.1.7.2.3

Rewrite the expression.

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

Step 11.2.1.8

Cancel the common factor of $2$ .

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Step 11.2.1.8.1

Factor $2$ out of $- 2$ .

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

Step 11.2.1.8.2

Cancel the common factor.

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

Step 11.2.1.8.3

Rewrite the expression.

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

Step 11.2.2

Combine fractions.

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Step 11.2.2.1

Combine the numerators over the common denominator.

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

Step 11.2.2.2

Add $1$ and $1$ .

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

Step 11.2.3

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

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

Step 11.2.4

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

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

Step 11.2.5

Combine the numerators over the common denominator.

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

Step 11.2.6

Simplify the numerator.

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Step 11.2.6.1

Multiply $- 1$ by $3$ .

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

Step 11.2.6.2

Add $- 3$ and $2$ .

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

Step 11.2.7

Move the negative in front of the fraction.

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

Step 11.2.8

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

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

Step 12

Evaluate the second derivative at $x = - \frac{1}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$16 (- \frac{1}{2})$

Step 13

Evaluate the second derivative.

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Step 13.1

Cancel the common factor of $2$ .

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Step 13.1.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$16 (\frac{- 1}{2})$

Step 13.1.2

Factor $2$ out of $16$ .

$2 (8) \frac{- 1}{2}$

Step 13.1.3

Cancel the common factor.

$2 \cdot 8 \frac{- 1}{2}$

Step 13.1.4

Rewrite the expression.

$8 \cdot - 1$

Step 13.2

Multiply $8$ by $- 1$ .

$- 8$

Step 14

$x = - \frac{1}{2}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \frac{1}{2}$ is a local maximum

Step 15

Find the y-value when $x = - \frac{1}{2}$ .

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Step 15.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

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

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 15.2.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

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

Step 15.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

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

Step 15.2.1.2

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

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

Step 15.2.1.3

One to any power is one.

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

Step 15.2.1.4

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

$f (- \frac{1}{2}) = \frac{8}{3} \cdot (- \frac{1}{8}) - 2 (- \frac{1}{2}) + \frac{1}{3}$

Step 15.2.1.5

Cancel the common factor of $8$ .

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Step 15.2.1.5.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$f (- \frac{1}{2}) = \frac{8}{3} \cdot \frac{- 1}{8} - 2 (- \frac{1}{2}) + \frac{1}{3}$

Step 15.2.1.5.2

Cancel the common factor.

$f (- \frac{1}{2}) = \frac{8}{3} \cdot \frac{- 1}{8} - 2 (- \frac{1}{2}) + \frac{1}{3}$

Step 15.2.1.5.3

Rewrite the expression.

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

Step 15.2.1.6

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

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

Step 15.2.1.7

Move the negative in front of the fraction.

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

Step 15.2.1.8

Cancel the common factor of $2$ .

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Step 15.2.1.8.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 15.2.1.8.2

Factor $2$ out of $- 2$ .

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

Step 15.2.1.8.3

Cancel the common factor.

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

Step 15.2.1.8.4

Rewrite the expression.

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

Step 15.2.1.9

Multiply $- 1$ by $- 1$ .

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

Step 15.2.2

Combine fractions.

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Step 15.2.2.1

Combine the numerators over the common denominator.

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

Step 15.2.2.2

Simplify the expression.

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Step 15.2.2.2.1

Add $- 1$ and $1$ .

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

Step 15.2.2.2.2

Divide $0$ by $3$ .

$f (- \frac{1}{2}) = 1 + 0$

Step 15.2.2.2.3

Add $1$ and $0$ .

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

Step 15.2.3

The final answer is $1$ .

$y = 1$

Step 16

These are the local extrema for $f (x) = \frac{8}{3} x^{3} - 2 x + \frac{1}{3}$ .

$(\frac{1}{2}, - \frac{1}{3})$ is a local minima

$(- \frac{1}{2}, 1)$ is a local maxima

Step 17