Calculus Examples

$y = \frac{2 x^{5}}{5} - \frac{3 x^{4}}{4} - x^{3} + x^{2}$

Step 1

Write $y = \frac{2 x^{5}}{5} - \frac{3 x^{4}}{4} - x^{3} + x^{2}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $\frac{2 x^{5}}{5} - \frac{3 x^{4}}{4} - x^{3} + x^{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{2 x^{5}}{5}] + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$ .

$\frac{d}{d x} [\frac{2 x^{5}}{5}] + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{2 x^{5}}{5}]$ .

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Step 2.2.1

Since $\frac{2}{5}$ is constant with respect to $x$ , the derivative of $\frac{2 x^{5}}{5}$ with respect to $x$ is $\frac{2}{5} \frac{d}{d x} [x^{5}]$ .

$\frac{2}{5} \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{2}{5} (5 x^{4}) + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.3

Combine $5$ and $\frac{2}{5}$ .

$\frac{5 \cdot 2}{5} x^{4} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.4

Multiply $5$ by $2$ .

$\frac{10}{5} x^{4} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.5

Combine $\frac{10}{5}$ and $x^{4}$ .

$\frac{10 x^{4}}{5} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.6

Cancel the common factor of $10$ and $5$ .

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Step 2.2.6.1

Factor $5$ out of $10 x^{4}$ .

$\frac{5 (2 x^{4})}{5} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.6.2

Cancel the common factors.

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Step 2.2.6.2.1

Factor $5$ out of $5$ .

$\frac{5 (2 x^{4})}{5 (1)} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.6.2.2

Cancel the common factor.

$\frac{5 (2 x^{4})}{5 \cdot 1} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.2.6.2.3

Rewrite the expression.

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

Step 2.2.6.2.4

Divide $2 x^{4}$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 2.3.3

Multiply $4$ by $- 1$ .

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $- 4$ by $3$ .

$2 x^{4} + \frac{- 12}{4} x^{3} + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.3.6

Combine $\frac{- 12}{4}$ and $x^{3}$ .

$2 x^{4} + \frac{- 12 x^{3}}{4} + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 2.3.7

Cancel the common factor of $- 12$ and $4$ .

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Step 2.3.7.1

Factor $4$ out of $- 12 x^{3}$ .

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

Step 2.3.7.2

Cancel the common factors.

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Step 2.3.7.2.1

Factor $4$ out of $4$ .

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

Step 2.3.7.2.2

Cancel the common factor.

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

Step 2.3.7.2.3

Rewrite the expression.

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

Step 2.3.7.2.4

Divide $- 3 x^{3}$ by $1$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $3$ by $- 1$ .

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

Step 2.5

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

$2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x$ with respect to $x$ is $\frac{d}{d x} [2 x^{4}] + \frac{d}{d x} [- 3 x^{3}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [2 x]$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [2 x^{4}]$ .

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Step 3.2.1

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

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

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 3.2.3

Multiply $4$ by $2$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Multiply $3$ by $- 3$ .

$f'' (x) = 8 x^{3} - 9 x^{2} + \frac{d}{d x} (- 3 x^{2}) + \frac{d}{d x} (2 x)$

Step 3.4

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

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Step 3.4.1

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

$f'' (x) = 8 x^{3} - 9 x^{2} - 3 \frac{d}{d x} x^{2} + \frac{d}{d x} (2 x)$

Step 3.4.2

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

$f'' (x) = 8 x^{3} - 9 x^{2} - 3 (2 x) + \frac{d}{d x} (2 x)$

Step 3.4.3

Multiply $2$ by $- 3$ .

$f'' (x) = 8 x^{3} - 9 x^{2} - 6 x + \frac{d}{d x} (2 x)$

Step 3.5

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

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Step 3.5.1

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

$f'' (x) = 8 x^{3} - 9 x^{2} - 6 x + 2 \frac{d}{d x} (x)$

Step 3.5.2

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

$f'' (x) = 8 x^{3} - 9 x^{2} - 6 x + 2 \cdot 1$

Step 3.5.3

Multiply $2$ by $1$ .

$f'' (x) = 8 x^{3} - 9 x^{2} - 6 x + 2$

Step 4

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

$2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

$\frac{d}{d x} [\frac{2 x^{5}}{5}] + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2

Evaluate $\frac{d}{d x} [\frac{2 x^{5}}{5}]$ .

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Step 5.1.2.1

Since $\frac{2}{5}$ is constant with respect to $x$ , the derivative of $\frac{2 x^{5}}{5}$ with respect to $x$ is $\frac{2}{5} \frac{d}{d x} [x^{5}]$ .

$\frac{2}{5} \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{2}{5} (5 x^{4}) + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.3

Combine $5$ and $\frac{2}{5}$ .

$\frac{5 \cdot 2}{5} x^{4} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.4

Multiply $5$ by $2$ .

$\frac{10}{5} x^{4} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.5

Combine $\frac{10}{5}$ and $x^{4}$ .

$\frac{10 x^{4}}{5} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.6

Cancel the common factor of $10$ and $5$ .

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Step 5.1.2.6.1

Factor $5$ out of $10 x^{4}$ .

$\frac{5 (2 x^{4})}{5} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.6.2

Cancel the common factors.

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Step 5.1.2.6.2.1

Factor $5$ out of $5$ .

$\frac{5 (2 x^{4})}{5 (1)} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.6.2.2

Cancel the common factor.

$\frac{5 (2 x^{4})}{5 \cdot 1} + \frac{d}{d x} [- \frac{3 x^{4}}{4}] + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.2.6.2.3

Rewrite the expression.

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

Step 5.1.2.6.2.4

Divide $2 x^{4}$ by $1$ .

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

Step 5.1.3

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

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Step 5.1.3.1

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

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

Step 5.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 5.1.3.3

Multiply $4$ by $- 1$ .

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

Step 5.1.3.4

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

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

Step 5.1.3.5

Multiply $- 4$ by $3$ .

$2 x^{4} + \frac{- 12}{4} x^{3} + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.3.6

Combine $\frac{- 12}{4}$ and $x^{3}$ .

$2 x^{4} + \frac{- 12 x^{3}}{4} + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x^{2}]$

Step 5.1.3.7

Cancel the common factor of $- 12$ and $4$ .

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Step 5.1.3.7.1

Factor $4$ out of $- 12 x^{3}$ .

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

Step 5.1.3.7.2

Cancel the common factors.

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Step 5.1.3.7.2.1

Factor $4$ out of $4$ .

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

Step 5.1.3.7.2.2

Cancel the common factor.

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

Step 5.1.3.7.2.3

Rewrite the expression.

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

Step 5.1.3.7.2.4

Divide $- 3 x^{3}$ by $1$ .

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

Step 5.1.4

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

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Step 5.1.4.1

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

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

Step 5.1.4.2

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

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

Step 5.1.4.3

Multiply $3$ by $- 1$ .

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

Step 5.1.5

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

$f' (x) = 2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x$ .

$2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x$

Step 6

Set the first derivative equal to $0$ then solve the equation $2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x = 0$

Step 6.2

Factor the left side of the equation.

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Step 6.2.1

Factor $x$ out of $2 x^{4} - 3 x^{3} - 3 x^{2} + 2 x$ .

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Step 6.2.1.1

Factor $x$ out of $2 x^{4}$ .

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

Step 6.2.1.2

Factor $x$ out of $- 3 x^{3}$ .

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

Step 6.2.1.3

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

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

Step 6.2.1.4

Factor $x$ out of $2 x$ .

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

Step 6.2.1.5

Factor $x$ out of $x (2 x^{3}) + x (- 3 x^{2})$ .

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

Step 6.2.1.6

Factor $x$ out of $x (2 x^{3} - 3 x^{2}) + x (- 3 x)$ .

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

Step 6.2.1.7

Factor $x$ out of $x (2 x^{3} - 3 x^{2} - 3 x) + x \cdot 2$ .

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

Step 6.2.2

Factor $2 x^{3} - 3 x^{2} - 3 x + 2$ using the rational roots test.

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Step 6.2.2.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2$

Step 6.2.2.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.5, \pm 2$

Step 6.2.2.3

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

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Step 6.2.2.3.1

Substitute $- 1$ into the polynomial.

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

Step 6.2.2.3.2

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

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

Step 6.2.2.3.3

Multiply $2$ by $- 1$ .

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

Step 6.2.2.3.4

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

$- 2 - 3 \cdot 1 - 3 \cdot - 1 + 2$

Step 6.2.2.3.5

Multiply $- 3$ by $1$ .

$- 2 - 3 - 3 \cdot - 1 + 2$

Step 6.2.2.3.6

Subtract $3$ from $- 2$ .

$- 5 - 3 \cdot - 1 + 2$

Step 6.2.2.3.7

Multiply $- 3$ by $- 1$ .

$- 5 + 3 + 2$

Step 6.2.2.3.8

Add $- 5$ and $3$ .

$- 2 + 2$

Step 6.2.2.3.9

Add $- 2$ and $2$ .

$0$

Step 6.2.2.4

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

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

Step 6.2.2.5

Divide $2 x^{3} - 3 x^{2} - 3 x + 2$ by $x + 1$ .

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Step 6.2.2.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$2 x^{3}$

$3 x^{2}$

$3 x$

$2$

Step 6.2.2.5.2

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x$ .

					$2 x^{2}$
	$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$

Step 6.2.2.5.3

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			+	$2 x^{3}$	+	$2 x^{2}$

Step 6.2.2.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} + 2 x^{2}$

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$

Step 6.2.2.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$

Step 6.2.2.5.6

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$

Step 6.2.2.5.7

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$5 x$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$

Step 6.2.2.5.8

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$5 x$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					-	$5 x^{2}$	-	$5 x$

Step 6.2.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} - 5 x$

				$2 x^{2}$	-	$5 x$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$

Step 6.2.2.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	-	$5 x$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$
							+	$2 x$

Step 6.2.2.5.11

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$	-	$5 x$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$
							+	$2 x$	+	$2$

Step 6.2.2.5.12

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	-	$5 x$	+	$2$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$
							+	$2 x$	+	$2$

Step 6.2.2.5.13

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$5 x$	+	$2$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$
							+	$2 x$	+	$2$
							+	$2 x$	+	$2$

Step 6.2.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $2 x + 2$

				$2 x^{2}$	-	$5 x$	+	$2$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$
							+	$2 x$	+	$2$
							-	$2 x$	-	$2$

Step 6.2.2.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	-	$5 x$	+	$2$
$x$	+	$1$		$2 x^{3}$	-	$3 x^{2}$	-	$3 x$	+	$2$
			-	$2 x^{3}$	-	$2 x^{2}$
					-	$5 x^{2}$	-	$3 x$
					+	$5 x^{2}$	+	$5 x$
							+	$2 x$	+	$2$
							-	$2 x$	-	$2$
										$0$

Step 6.2.2.5.16

Since the remander is $0$ , the final answer is the quotient.

$2 x^{2} - 5 x + 2$

Step 6.2.2.6

Write $2 x^{3} - 3 x^{2} - 3 x + 2$ as a set of factors.

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

Step 6.2.3

Factor.

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Step 6.2.3.1

Factor by grouping.

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Step 6.2.3.1.1

Factor by grouping.

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Step 6.2.3.1.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

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Step 6.2.3.1.1.1.1

Factor $- 5$ out of $- 5 x$ .

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

Step 6.2.3.1.1.1.2

Rewrite $- 5$ as $- 1$ plus $- 4$

$x ((x + 1) (2 x^{2} + (- 1 - 4) x + 2)) = 0$

Step 6.2.3.1.1.1.3

Apply the distributive property.

$x ((x + 1) (2 x^{2} - 1 x - 4 x + 2)) = 0$

Step 6.2.3.1.1.2

Factor out the greatest common factor from each group.

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Step 6.2.3.1.1.2.1

Group the first two terms and the last two terms.

$x ((x + 1) ((2 x^{2} - 1 x) - 4 x + 2)) = 0$

Step 6.2.3.1.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x ((x + 1) (x (2 x - 1) - 2 (2 x - 1))) = 0$

Step 6.2.3.1.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$x ((x + 1) ((2 x - 1) (x - 2))) = 0$

Step 6.2.3.1.2

Remove unnecessary parentheses.

$x ((x + 1) (2 x - 1) (x - 2)) = 0$

Step 6.2.3.2

Remove unnecessary parentheses.

$x (x + 1) (2 x - 1) (x - 2) = 0$

Step 6.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 1 = 0$

$2 x - 1 = 0$

$x - 2 = 0$

Step 6.4

Set $x$ equal to $0$ .

$x = 0$

Step 6.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 6.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 6.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.6

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 6.6.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 6.6.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 6.6.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 6.6.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 6.6.2.2.1

Divide each term in $2 x = 1$ by $2$ .

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

Step 6.6.2.2.2

Simplify the left side.

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Step 6.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.6.2.2.2.1.1

Cancel the common factor.

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

Step 6.6.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 6.7

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 6.7.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 6.7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 6.8

The final solution is all the values that make $x (x + 1) (2 x - 1) (x - 2) = 0$ true.

$x = 0, - 1, \frac{1}{2}, 2$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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 8

Critical points to evaluate.

$x = 0, - 1, \frac{1}{2}, 2$

Step 9

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.

$8 {(0)}^{3} - 9 {(0)}^{2} - 6 \cdot 0 + 2$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

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

$8 \cdot 0 - 9 {(0)}^{2} - 6 \cdot 0 + 2$

Step 10.1.2

Multiply $8$ by $0$ .

$0 - 9 {(0)}^{2} - 6 \cdot 0 + 2$

Step 10.1.3

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

$0 - 9 \cdot 0 - 6 \cdot 0 + 2$

Step 10.1.4

Multiply $- 9$ by $0$ .

$0 + 0 - 6 \cdot 0 + 2$

Step 10.1.5

Multiply $- 6$ by $0$ .

$0 + 0 + 0 + 2$

Step 10.2

Simplify by adding numbers.

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Step 10.2.1

Add $0$ and $0$ .

$0 + 0 + 2$

Step 10.2.2

Add $0$ and $0$ .

$0 + 2$

Step 10.2.3

Add $0$ and $2$ .

$2$

Step 11

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

$x = 0$ is a local minimum

Step 12

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

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Step 12.1

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

$f (0) = \frac{2 {(0)}^{5}}{5} - \frac{3 {(0)}^{4}}{4} - {(0)}^{3} + {(0)}^{2}$

Step 12.2

Simplify the result.

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Step 12.2.1

Find the common denominator.

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Step 12.2.1.1

Multiply $\frac{2 {(0)}^{5}}{5}$ by $\frac{4}{4}$ .

$f (0) = \frac{2 {(0)}^{5}}{5} \cdot \frac{4}{4} - \frac{3 {(0)}^{4}}{4} - {(0)}^{3} + {(0)}^{2}$

Step 12.2.1.2

Multiply $\frac{2 {(0)}^{5}}{5}$ by $\frac{4}{4}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4}}{4} - {(0)}^{3} + {(0)}^{2}$

Step 12.2.1.3

Multiply $\frac{3 {(0)}^{4}}{4}$ by $\frac{5}{5}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - (\frac{3 {(0)}^{4}}{4} \cdot \frac{5}{5}) - {(0)}^{3} + {(0)}^{2}$

Step 12.2.1.4

Multiply $\frac{3 {(0)}^{4}}{4}$ by $\frac{5}{5}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} - {(0)}^{3} + {(0)}^{2}$

Step 12.2.1.5

Write $- {(0)}^{3}$ as a fraction with denominator $1$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3}}{1} + {(0)}^{2}$

Step 12.2.1.6

Multiply $\frac{- {(0)}^{3}}{1}$ by $\frac{20}{20}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3}}{1} \cdot \frac{20}{20} + {(0)}^{2}$

Step 12.2.1.7

Multiply $\frac{- {(0)}^{3}}{1}$ by $\frac{20}{20}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3} \cdot 20}{20} + {(0)}^{2}$

Step 12.2.1.8

Write ${(0)}^{2}$ as a fraction with denominator $1$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3} \cdot 20}{20} + \frac{{(0)}^{2}}{1}$

Step 12.2.1.9

Multiply $\frac{{(0)}^{2}}{1}$ by $\frac{20}{20}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3} \cdot 20}{20} + \frac{{(0)}^{2}}{1} \cdot \frac{20}{20}$

Step 12.2.1.10

Multiply $\frac{{(0)}^{2}}{1}$ by $\frac{20}{20}$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3} \cdot 20}{20} + \frac{{(0)}^{2} \cdot 20}{20}$

Step 12.2.1.11

Reorder the factors of $5 \cdot 4$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{4 \cdot 5} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3} \cdot 20}{20} + \frac{{(0)}^{2} \cdot 20}{20}$

Step 12.2.1.12

Multiply $4$ by $5$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{20} - \frac{3 {(0)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(0)}^{3} \cdot 20}{20} + \frac{{(0)}^{2} \cdot 20}{20}$

Step 12.2.1.13

Multiply $4$ by $5$ .

$f (0) = \frac{2 {(0)}^{5} \cdot 4}{20} - \frac{3 {(0)}^{4} \cdot 5}{20} + \frac{- {(0)}^{3} \cdot 20}{20} + \frac{{(0)}^{2} \cdot 20}{20}$

Step 12.2.2

Combine the numerators over the common denominator.

$f (0) = \frac{2 {(0)}^{5} \cdot 4 - 3 {(0)}^{4} \cdot 5 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3

Simplify each term.

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Step 12.2.3.1

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

$f (0) = \frac{2 \cdot 0 \cdot 4 - 3 {(0)}^{4} \cdot 5 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.2

Multiply $2 \cdot 0 \cdot 4$ .

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Step 12.2.3.2.1

Multiply $2$ by $0$ .

$f (0) = \frac{0 \cdot 4 - 3 {(0)}^{4} \cdot 5 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.2.2

Multiply $0$ by $4$ .

$f (0) = \frac{0 - 3 {(0)}^{4} \cdot 5 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.3

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

$f (0) = \frac{0 - 3 \cdot 0 \cdot 5 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.4

Multiply $- 3 \cdot 0 \cdot 5$ .

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Step 12.2.3.4.1

Multiply $- 3$ by $0$ .

$f (0) = \frac{0 + 0 \cdot 5 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.4.2

Multiply $0$ by $5$ .

$f (0) = \frac{0 + 0 - {(0)}^{3} \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.5

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

$f (0) = \frac{0 + 0 + 0 \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.6

Multiply $- 0 \cdot 20$ .

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Step 12.2.3.6.1

Multiply $- 1$ by $0$ .

$f (0) = \frac{0 + 0 + 0 \cdot 20 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.6.2

Multiply $0$ by $20$ .

$f (0) = \frac{0 + 0 + 0 + {(0)}^{2} \cdot 20}{20}$

Step 12.2.3.7

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

$f (0) = \frac{0 + 0 + 0 + 0 \cdot 20}{20}$

Step 12.2.3.8

Multiply $0$ by $20$ .

$f (0) = \frac{0 + 0 + 0 + 0}{20}$

Step 12.2.4

Simplify the expression.

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Step 12.2.4.1

Add $0$ and $0$ .

$f (0) = \frac{0 + 0 + 0}{20}$

Step 12.2.4.2

Add $0$ and $0$ .

$f (0) = \frac{0 + 0}{20}$

Step 12.2.4.3

Add $0$ and $0$ .

$f (0) = \frac{0}{20}$

Step 12.2.4.4

Divide $0$ by $20$ .

$f (0) = 0$

Step 12.2.5

The final answer is $0$ .

$y = 0$

Step 13

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.

$8 {(- 1)}^{3} - 9 {(- 1)}^{2} - 6 \cdot - 1 + 2$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

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

$8 \cdot - 1 - 9 {(- 1)}^{2} - 6 \cdot - 1 + 2$

Step 14.1.2

Multiply $8$ by $- 1$ .

$- 8 - 9 {(- 1)}^{2} - 6 \cdot - 1 + 2$

Step 14.1.3

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

$- 8 - 9 \cdot 1 - 6 \cdot - 1 + 2$

Step 14.1.4

Multiply $- 9$ by $1$ .

$- 8 - 9 - 6 \cdot - 1 + 2$

Step 14.1.5

Multiply $- 6$ by $- 1$ .

$- 8 - 9 + 6 + 2$

Step 14.2

Simplify by adding and subtracting.

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Step 14.2.1

Subtract $9$ from $- 8$ .

$- 17 + 6 + 2$

Step 14.2.2

Add $- 17$ and $6$ .

$- 11 + 2$

Step 14.2.3

Add $- 11$ and $2$ .

$- 9$

Step 15

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

$x = - 1$ is a local maximum

Step 16

Find the y-value when $x = - 1$ .

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Step 16.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \frac{2 {(- 1)}^{5}}{5} - \frac{3 {(- 1)}^{4}}{4} - {(- 1)}^{3} + {(- 1)}^{2}$

Step 16.2

Simplify the result.

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Step 16.2.1

Find the common denominator.

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Step 16.2.1.1

Multiply $\frac{2 {(- 1)}^{5}}{5}$ by $\frac{4}{4}$ .

$f (- 1) = \frac{2 {(- 1)}^{5}}{5} \cdot \frac{4}{4} - \frac{3 {(- 1)}^{4}}{4} - {(- 1)}^{3} + {(- 1)}^{2}$

Step 16.2.1.2

Multiply $\frac{2 {(- 1)}^{5}}{5}$ by $\frac{4}{4}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4}}{4} - {(- 1)}^{3} + {(- 1)}^{2}$

Step 16.2.1.3

Multiply $\frac{3 {(- 1)}^{4}}{4}$ by $\frac{5}{5}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - (\frac{3 {(- 1)}^{4}}{4} \cdot \frac{5}{5}) - {(- 1)}^{3} + {(- 1)}^{2}$

Step 16.2.1.4

Multiply $\frac{3 {(- 1)}^{4}}{4}$ by $\frac{5}{5}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} - {(- 1)}^{3} + {(- 1)}^{2}$

Step 16.2.1.5

Write $- {(- 1)}^{3}$ as a fraction with denominator $1$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3}}{1} + {(- 1)}^{2}$

Step 16.2.1.6

Multiply $\frac{- {(- 1)}^{3}}{1}$ by $\frac{20}{20}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3}}{1} \cdot \frac{20}{20} + {(- 1)}^{2}$

Step 16.2.1.7

Multiply $\frac{- {(- 1)}^{3}}{1}$ by $\frac{20}{20}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3} \cdot 20}{20} + {(- 1)}^{2}$

Step 16.2.1.8

Write ${(- 1)}^{2}$ as a fraction with denominator $1$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3} \cdot 20}{20} + \frac{{(- 1)}^{2}}{1}$

Step 16.2.1.9

Multiply $\frac{{(- 1)}^{2}}{1}$ by $\frac{20}{20}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3} \cdot 20}{20} + \frac{{(- 1)}^{2}}{1} \cdot \frac{20}{20}$

Step 16.2.1.10

Multiply $\frac{{(- 1)}^{2}}{1}$ by $\frac{20}{20}$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3} \cdot 20}{20} + \frac{{(- 1)}^{2} \cdot 20}{20}$

Step 16.2.1.11

Reorder the factors of $5 \cdot 4$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{4 \cdot 5} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3} \cdot 20}{20} + \frac{{(- 1)}^{2} \cdot 20}{20}$

Step 16.2.1.12

Multiply $4$ by $5$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{20} - \frac{3 {(- 1)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(- 1)}^{3} \cdot 20}{20} + \frac{{(- 1)}^{2} \cdot 20}{20}$

Step 16.2.1.13

Multiply $4$ by $5$ .

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4}{20} - \frac{3 {(- 1)}^{4} \cdot 5}{20} + \frac{- {(- 1)}^{3} \cdot 20}{20} + \frac{{(- 1)}^{2} \cdot 20}{20}$

Step 16.2.2

Combine the numerators over the common denominator.

$f (- 1) = \frac{2 {(- 1)}^{5} \cdot 4 - 3 {(- 1)}^{4} \cdot 5 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3

Simplify each term.

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Step 16.2.3.1

Raise $- 1$ to the power of $5$ .

$f (- 1) = \frac{2 \cdot - 1 \cdot 4 - 3 {(- 1)}^{4} \cdot 5 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.2

Multiply $2 \cdot - 1 \cdot 4$ .

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Step 16.2.3.2.1

Multiply $2$ by $- 1$ .

$f (- 1) = \frac{- 2 \cdot 4 - 3 {(- 1)}^{4} \cdot 5 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.2.2

Multiply $- 2$ by $4$ .

$f (- 1) = \frac{- 8 - 3 {(- 1)}^{4} \cdot 5 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.3

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

$f (- 1) = \frac{- 8 - 3 \cdot 1 \cdot 5 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.4

Multiply $- 3 \cdot 1 \cdot 5$ .

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Step 16.2.3.4.1

Multiply $- 3$ by $1$ .

$f (- 1) = \frac{- 8 - 3 \cdot 5 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.4.2

Multiply $- 3$ by $5$ .

$f (- 1) = \frac{- 8 - 15 - {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.5

Multiply $- 1$ by ${(- 1)}^{3}$ by adding the exponents.

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Step 16.2.3.5.1

Multiply $- 1$ by ${(- 1)}^{3}$ .

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Step 16.2.3.5.1.1

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

$f (- 1) = \frac{- 8 - 15 + (- 1) {(- 1)}^{3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.5.1.2

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

$f (- 1) = \frac{- 8 - 15 + {(- 1)}^{1 + 3} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.5.2

Add $1$ and $3$ .

$f (- 1) = \frac{- 8 - 15 + {(- 1)}^{4} \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.6

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

$f (- 1) = \frac{- 8 - 15 + 1 \cdot 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.7

Multiply $20$ by $1$ .

$f (- 1) = \frac{- 8 - 15 + 20 + {(- 1)}^{2} \cdot 20}{20}$

Step 16.2.3.8

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

$f (- 1) = \frac{- 8 - 15 + 20 + 1 \cdot 20}{20}$

Step 16.2.3.9

Multiply $20$ by $1$ .

$f (- 1) = \frac{- 8 - 15 + 20 + 20}{20}$

Step 16.2.4

Simplify by adding and subtracting.

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Step 16.2.4.1

Subtract $15$ from $- 8$ .

$f (- 1) = \frac{- 23 + 20 + 20}{20}$

Step 16.2.4.2

Add $- 23$ and $20$ .

$f (- 1) = \frac{- 3 + 20}{20}$

Step 16.2.4.3

Add $- 3$ and $20$ .

$f (- 1) = \frac{17}{20}$

Step 16.2.5

The final answer is $\frac{17}{20}$ .

$y = \frac{17}{20}$

Step 17

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.

$8 {(\frac{1}{2})}^{3} - 9 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 2$

Step 18

Evaluate the second derivative.

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Step 18.1

Simplify each term.

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Step 18.1.1

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

$8 \frac{1^{3}}{2^{3}} - 9 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 2$

Step 18.1.2

One to any power is one.

$8 \frac{1}{2^{3}} - 9 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 2$

Step 18.1.3

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

$8 (\frac{1}{8}) - 9 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 2$

Step 18.1.4

Cancel the common factor of $8$ .

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Step 18.1.4.1

Cancel the common factor.

$8 (\frac{1}{8}) - 9 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 2$

Step 18.1.4.2

Rewrite the expression.

$1 - 9 {(\frac{1}{2})}^{2} - 6 (\frac{1}{2}) + 2$

Step 18.1.5

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

$1 - 9 \frac{1^{2}}{2^{2}} - 6 (\frac{1}{2}) + 2$

Step 18.1.6

One to any power is one.

$1 - 9 \frac{1}{2^{2}} - 6 (\frac{1}{2}) + 2$

Step 18.1.7

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

$1 - 9 (\frac{1}{4}) - 6 (\frac{1}{2}) + 2$

Step 18.1.8

Combine $- 9$ and $\frac{1}{4}$ .

$1 + \frac{- 9}{4} - 6 (\frac{1}{2}) + 2$

Step 18.1.9

Move the negative in front of the fraction.

$1 - \frac{9}{4} - 6 (\frac{1}{2}) + 2$

Step 18.1.10

Cancel the common factor of $2$ .

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Step 18.1.10.1

Factor $2$ out of $- 6$ .

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

Step 18.1.10.2

Cancel the common factor.

$1 - \frac{9}{4} + 2 \cdot - 3 \frac{1}{2} + 2$

Step 18.1.10.3

Rewrite the expression.

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

Step 18.2

Find the common denominator.

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Step 18.2.1

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

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

Step 18.2.2

Multiply $\frac{1}{1}$ by $\frac{4}{4}$ .

$\frac{1}{1} \cdot \frac{4}{4} - \frac{9}{4} - 3 + 2$

Step 18.2.3

Multiply $\frac{1}{1}$ by $\frac{4}{4}$ .

$\frac{4}{4} - \frac{9}{4} - 3 + 2$

Step 18.2.4

Write $- 3$ as a fraction with denominator $1$ .

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

Step 18.2.5

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$\frac{4}{4} - \frac{9}{4} + \frac{- 3}{1} \cdot \frac{4}{4} + 2$

Step 18.2.6

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$\frac{4}{4} - \frac{9}{4} + \frac{- 3 \cdot 4}{4} + 2$

Step 18.2.7

Write $2$ as a fraction with denominator $1$ .

$\frac{4}{4} - \frac{9}{4} + \frac{- 3 \cdot 4}{4} + \frac{2}{1}$

Step 18.2.8

Multiply $\frac{2}{1}$ by $\frac{4}{4}$ .

$\frac{4}{4} - \frac{9}{4} + \frac{- 3 \cdot 4}{4} + \frac{2}{1} \cdot \frac{4}{4}$

Step 18.2.9

Multiply $\frac{2}{1}$ by $\frac{4}{4}$ .

$\frac{4}{4} - \frac{9}{4} + \frac{- 3 \cdot 4}{4} + \frac{2 \cdot 4}{4}$

Step 18.3

Combine the numerators over the common denominator.

$\frac{4 - 9 - 3 \cdot 4 + 2 \cdot 4}{4}$

Step 18.4

Simplify each term.

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Step 18.4.1

Multiply $- 3$ by $4$ .

$\frac{4 - 9 - 12 + 2 \cdot 4}{4}$

Step 18.4.2

Multiply $2$ by $4$ .

$\frac{4 - 9 - 12 + 8}{4}$

Step 18.5

Simplify the expression.

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Step 18.5.1

Subtract $9$ from $4$ .

$\frac{- 5 - 12 + 8}{4}$

Step 18.5.2

Subtract $12$ from $- 5$ .

$\frac{- 17 + 8}{4}$

Step 18.5.3

Add $- 17$ and $8$ .

$\frac{- 9}{4}$

Step 18.5.4

Move the negative in front of the fraction.

$- \frac{9}{4}$

Step 19

$x = \frac{1}{2}$ 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 20

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

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Step 20.1

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

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5}}{5} - \frac{3 {(\frac{1}{2})}^{4}}{4} - {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2}$

Step 20.2

Simplify the result.

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Step 20.2.1

Find the common denominator.

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Step 20.2.1.1

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

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5}}{5} \cdot \frac{4}{4} - \frac{3 {(\frac{1}{2})}^{4}}{4} - {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2}$

Step 20.2.1.2

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

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4}}{4} - {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2}$

Step 20.2.1.3

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

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - (\frac{3 {(\frac{1}{2})}^{4}}{4} \cdot \frac{5}{5}) - {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2}$

Step 20.2.1.4

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

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} - {(\frac{1}{2})}^{3} + {(\frac{1}{2})}^{2}$

Step 20.2.1.5

Write $- {(\frac{1}{2})}^{3}$ as a fraction with denominator $1$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3}}{1} + {(\frac{1}{2})}^{2}$

Step 20.2.1.6

Multiply $\frac{- {(\frac{1}{2})}^{3}}{1}$ by $\frac{20}{20}$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3}}{1} \cdot \frac{20}{20} + {(\frac{1}{2})}^{2}$

Step 20.2.1.7

Multiply $\frac{- {(\frac{1}{2})}^{3}}{1}$ by $\frac{20}{20}$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3} \cdot 20}{20} + {(\frac{1}{2})}^{2}$

Step 20.2.1.8

Write ${(\frac{1}{2})}^{2}$ as a fraction with denominator $1$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3} \cdot 20}{20} + \frac{{(\frac{1}{2})}^{2}}{1}$

Step 20.2.1.9

Multiply $\frac{{(\frac{1}{2})}^{2}}{1}$ by $\frac{20}{20}$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3} \cdot 20}{20} + \frac{{(\frac{1}{2})}^{2}}{1} \cdot \frac{20}{20}$

Step 20.2.1.10

Multiply $\frac{{(\frac{1}{2})}^{2}}{1}$ by $\frac{20}{20}$ .

Step 20.2.1.11

Reorder the factors of $5 \cdot 4$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{4 \cdot 5} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3} \cdot 20}{20} + \frac{{(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.1.12

Multiply $4$ by $5$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{20} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(\frac{1}{2})}^{3} \cdot 20}{20} + \frac{{(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.1.13

Multiply $4$ by $5$ .

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4}{20} - \frac{3 {(\frac{1}{2})}^{4} \cdot 5}{20} + \frac{- {(\frac{1}{2})}^{3} \cdot 20}{20} + \frac{{(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.2

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{2 {(\frac{1}{2})}^{5} \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3

Simplify each term.

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Step 20.2.3.1

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

$f (\frac{1}{2}) = \frac{2 (\frac{1^{5}}{2^{5}}) \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.2

One to any power is one.

$f (\frac{1}{2}) = \frac{2 (\frac{1}{2^{5}}) \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.3

Raise $2$ to the power of $5$ .

$f (\frac{1}{2}) = \frac{2 (\frac{1}{32}) \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.4

Cancel the common factor of $2$ .

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Step 20.2.3.4.1

Factor $2$ out of $32$ .

$f (\frac{1}{2}) = \frac{2 (\frac{1}{2 (16)}) \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.4.2

Cancel the common factor.

$f (\frac{1}{2}) = \frac{2 (\frac{1}{2 \cdot 16}) \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.4.3

Rewrite the expression.

$f (\frac{1}{2}) = \frac{\frac{1}{16} \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.5

Cancel the common factor of $4$ .

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Step 20.2.3.5.1

Factor $4$ out of $16$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4 (4)} \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.5.2

Cancel the common factor.

$f (\frac{1}{2}) = \frac{\frac{1}{4 \cdot 4} \cdot 4 - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.5.3

Rewrite the expression.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - 3 {(\frac{1}{2})}^{4} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.6

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

$f (\frac{1}{2}) = \frac{\frac{1}{4} - 3 \frac{1^{4}}{2^{4}} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.7

One to any power is one.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - 3 \frac{1}{2^{4}} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.8

Raise $2$ to the power of $4$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - 3 (\frac{1}{16}) \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.9

Combine $- 3$ and $\frac{1}{16}$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} + \frac{- 3}{16} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.10

Move the negative in front of the fraction.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{3}{16} \cdot 5 - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.11

Multiply $- \frac{3}{16} \cdot 5$ .

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Step 20.2.3.11.1

Multiply $5$ by $- 1$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - 5 (\frac{3}{16}) - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.11.2

Combine $- 5$ and $\frac{3}{16}$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} + \frac{- 5 \cdot 3}{16} - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.11.3

Multiply $- 5$ by $3$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} + \frac{- 15}{16} - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.12

Move the negative in front of the fraction.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - {(\frac{1}{2})}^{3} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.13

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

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{1^{3}}{2^{3}} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.14

One to any power is one.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{1}{2^{3}} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.15

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

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{1}{8} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.16

Cancel the common factor of $4$ .

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Step 20.2.3.16.1

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

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 1}{8} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.16.2

Factor $4$ out of $8$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 1}{4 (2)} \cdot 20 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.16.3

Factor $4$ out of $20$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 1}{4 \cdot 2} \cdot (4 \cdot 5) + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.16.4

Cancel the common factor.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 1}{4 \cdot 2} \cdot (4 \cdot 5) + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.16.5

Rewrite the expression.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 1}{2} \cdot 5 + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.17

Combine $\frac{- 1}{2}$ and $5$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 1 \cdot 5}{2} + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.18

Multiply $- 1$ by $5$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} + \frac{- 5}{2} + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.19

Move the negative in front of the fraction.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + {(\frac{1}{2})}^{2} \cdot 20}{20}$

Step 20.2.3.20

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

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + \frac{1^{2}}{2^{2}} \cdot 20}{20}$

Step 20.2.3.21

One to any power is one.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + \frac{1}{2^{2}} \cdot 20}{20}$

Step 20.2.3.22

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

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + \frac{1}{4} \cdot 20}{20}$

Step 20.2.3.23

Cancel the common factor of $4$ .

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Step 20.2.3.23.1

Factor $4$ out of $20$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + \frac{1}{4} \cdot (4 (5))}{20}$

Step 20.2.3.23.2

Cancel the common factor.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + \frac{1}{4} \cdot (4 \cdot 5)}{20}$

Step 20.2.3.23.3

Rewrite the expression.

$f (\frac{1}{2}) = \frac{\frac{1}{4} - \frac{15}{16} - \frac{5}{2} + 5}{20}$

Step 20.2.4

Find the common denominator.

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Step 20.2.4.1

Multiply $\frac{1}{4}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{\frac{1}{4} \cdot \frac{4}{4} - \frac{15}{16} - \frac{5}{2} + 5}{20}$

Step 20.2.4.2

Multiply $\frac{1}{4}$ by $\frac{4}{4}$ .

$f (\frac{1}{2}) = \frac{\frac{4}{4 \cdot 4} - \frac{15}{16} - \frac{5}{2} + 5}{20}$

Step 20.2.4.3

Multiply $\frac{5}{2}$ by $\frac{8}{8}$ .

$f (\frac{1}{2}) = \frac{\frac{4}{4 \cdot 4} - \frac{15}{16} - (\frac{5}{2} \cdot \frac{8}{8}) + 5}{20}$

Step 20.2.4.4

Multiply $\frac{5}{2}$ by $\frac{8}{8}$ .

$f (\frac{1}{2}) = \frac{\frac{4}{4 \cdot 4} - \frac{15}{16} - \frac{5 \cdot 8}{2 \cdot 8} + 5}{20}$

Step 20.2.4.5

Write $5$ as a fraction with denominator $1$ .

$f (\frac{1}{2}) = \frac{\frac{4}{4 \cdot 4} - \frac{15}{16} - \frac{5 \cdot 8}{2 \cdot 8} + \frac{5}{1}}{20}$

Step 20.2.4.6

Multiply $\frac{5}{1}$ by $\frac{16}{16}$ .

$f (\frac{1}{2}) = \frac{\frac{4}{4 \cdot 4} - \frac{15}{16} - \frac{5 \cdot 8}{2 \cdot 8} + \frac{5}{1} \cdot \frac{16}{16}}{20}$

Step 20.2.4.7

Multiply $\frac{5}{1}$ by $\frac{16}{16}$ .

$f (\frac{1}{2}) = \frac{\frac{4}{4 \cdot 4} - \frac{15}{16} - \frac{5 \cdot 8}{2 \cdot 8} + \frac{5 \cdot 16}{16}}{20}$

Step 20.2.4.8

Multiply $4$ by $4$ .

$f (\frac{1}{2}) = \frac{\frac{4}{16} - \frac{15}{16} - \frac{5 \cdot 8}{2 \cdot 8} + \frac{5 \cdot 16}{16}}{20}$

Step 20.2.4.9

Multiply $2$ by $8$ .

$f (\frac{1}{2}) = \frac{\frac{4}{16} - \frac{15}{16} - \frac{5 \cdot 8}{16} + \frac{5 \cdot 16}{16}}{20}$

Step 20.2.5

Combine the numerators over the common denominator.

$f (\frac{1}{2}) = \frac{\frac{4 - 15 - 5 \cdot 8 + 5 \cdot 16}{16}}{20}$

Step 20.2.6

Simplify each term.

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Step 20.2.6.1

Multiply $- 5$ by $8$ .

$f (\frac{1}{2}) = \frac{\frac{4 - 15 - 40 + 5 \cdot 16}{16}}{20}$

Step 20.2.6.2

Multiply $5$ by $16$ .

$f (\frac{1}{2}) = \frac{\frac{4 - 15 - 40 + 80}{16}}{20}$

Step 20.2.7

Simplify by adding and subtracting.

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Step 20.2.7.1

Subtract $15$ from $4$ .

$f (\frac{1}{2}) = \frac{\frac{- 11 - 40 + 80}{16}}{20}$

Step 20.2.7.2

Subtract $40$ from $- 11$ .

$f (\frac{1}{2}) = \frac{\frac{- 51 + 80}{16}}{20}$

Step 20.2.7.3

Add $- 51$ and $80$ .

$f (\frac{1}{2}) = \frac{\frac{29}{16}}{20}$

Step 20.2.8

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{1}{2}) = \frac{29}{16} \cdot \frac{1}{20}$

Step 20.2.9

Multiply $\frac{29}{16} \cdot \frac{1}{20}$ .

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Step 20.2.9.1

Multiply $\frac{29}{16}$ by $\frac{1}{20}$ .

$f (\frac{1}{2}) = \frac{29}{16 \cdot 20}$

Step 20.2.9.2

Multiply $16$ by $20$ .

$f (\frac{1}{2}) = \frac{29}{320}$

Step 20.2.10

The final answer is $\frac{29}{320}$ .

$y = \frac{29}{320}$

Step 21

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.

$8 {(2)}^{3} - 9 {(2)}^{2} - 6 \cdot 2 + 2$

Step 22

Evaluate the second derivative.

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Step 22.1

Simplify each term.

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Step 22.1.1

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

$8 \cdot 8 - 9 {(2)}^{2} - 6 \cdot 2 + 2$

Step 22.1.2

Multiply $8$ by $8$ .

$64 - 9 {(2)}^{2} - 6 \cdot 2 + 2$

Step 22.1.3

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

$64 - 9 \cdot 4 - 6 \cdot 2 + 2$

Step 22.1.4

Multiply $- 9$ by $4$ .

$64 - 36 - 6 \cdot 2 + 2$

Step 22.1.5

Multiply $- 6$ by $2$ .

$64 - 36 - 12 + 2$

Step 22.2

Simplify by adding and subtracting.

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Step 22.2.1

Subtract $36$ from $64$ .

$28 - 12 + 2$

Step 22.2.2

Subtract $12$ from $28$ .

$16 + 2$

Step 22.2.3

Add $16$ and $2$ .

$18$

Step 23

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

$x = 2$ is a local minimum

Step 24

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

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Step 24.1

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

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

Step 24.2

Simplify the result.

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Step 24.2.1

Find the common denominator.

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Step 24.2.1.1

Multiply $\frac{2 {(2)}^{5}}{5}$ by $\frac{4}{4}$ .

$f (2) = \frac{2 {(2)}^{5}}{5} \cdot \frac{4}{4} - \frac{3 {(2)}^{4}}{4} - {(2)}^{3} + {(2)}^{2}$

Step 24.2.1.2

Multiply $\frac{2 {(2)}^{5}}{5}$ by $\frac{4}{4}$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4}}{4} - {(2)}^{3} + {(2)}^{2}$

Step 24.2.1.3

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

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - (\frac{3 {(2)}^{4}}{4} \cdot \frac{5}{5}) - {(2)}^{3} + {(2)}^{2}$

Step 24.2.1.4

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

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} - {(2)}^{3} + {(2)}^{2}$

Step 24.2.1.5

Write $- {(2)}^{3}$ as a fraction with denominator $1$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3}}{1} + {(2)}^{2}$

Step 24.2.1.6

Multiply $\frac{- {(2)}^{3}}{1}$ by $\frac{20}{20}$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3}}{1} \cdot \frac{20}{20} + {(2)}^{2}$

Step 24.2.1.7

Multiply $\frac{- {(2)}^{3}}{1}$ by $\frac{20}{20}$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3} \cdot 20}{20} + {(2)}^{2}$

Step 24.2.1.8

Write ${(2)}^{2}$ as a fraction with denominator $1$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3} \cdot 20}{20} + \frac{{(2)}^{2}}{1}$

Step 24.2.1.9

Multiply $\frac{{(2)}^{2}}{1}$ by $\frac{20}{20}$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3} \cdot 20}{20} + \frac{{(2)}^{2}}{1} \cdot \frac{20}{20}$

Step 24.2.1.10

Multiply $\frac{{(2)}^{2}}{1}$ by $\frac{20}{20}$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{5 \cdot 4} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3} \cdot 20}{20} + \frac{{(2)}^{2} \cdot 20}{20}$

Step 24.2.1.11

Reorder the factors of $5 \cdot 4$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{4 \cdot 5} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3} \cdot 20}{20} + \frac{{(2)}^{2} \cdot 20}{20}$

Step 24.2.1.12

Multiply $4$ by $5$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{20} - \frac{3 {(2)}^{4} \cdot 5}{4 \cdot 5} + \frac{- {(2)}^{3} \cdot 20}{20} + \frac{{(2)}^{2} \cdot 20}{20}$

Step 24.2.1.13

Multiply $4$ by $5$ .

$f (2) = \frac{2 {(2)}^{5} \cdot 4}{20} - \frac{3 {(2)}^{4} \cdot 5}{20} + \frac{- {(2)}^{3} \cdot 20}{20} + \frac{{(2)}^{2} \cdot 20}{20}$

Step 24.2.2

Combine the numerators over the common denominator.

$f (2) = \frac{2 {(2)}^{5} \cdot 4 - 3 {(2)}^{4} \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3

Simplify each term.

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Step 24.2.3.1

Multiply $2$ by ${(2)}^{5}$ by adding the exponents.

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Step 24.2.3.1.1

Multiply $2$ by ${(2)}^{5}$ .

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Step 24.2.3.1.1.1

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

$f (2) = \frac{2 {(2)}^{5} \cdot 4 - 3 {(2)}^{4} \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.1.1.2

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

$f (2) = \frac{2^{1 + 5} \cdot 4 - 3 {(2)}^{4} \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.1.2

Add $1$ and $5$ .

$f (2) = \frac{2^{6} \cdot 4 - 3 {(2)}^{4} \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.2

Raise $2$ to the power of $6$ .

$f (2) = \frac{64 \cdot 4 - 3 {(2)}^{4} \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.3

Multiply $64$ by $4$ .

$f (2) = \frac{256 - 3 {(2)}^{4} \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.4

Raise $2$ to the power of $4$ .

$f (2) = \frac{256 - 3 \cdot 16 \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.5

Multiply $- 3 \cdot 16 \cdot 5$ .

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Step 24.2.3.5.1

Multiply $- 3$ by $16$ .

$f (2) = \frac{256 - 48 \cdot 5 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.5.2

Multiply $- 48$ by $5$ .

$f (2) = \frac{256 - 240 - {(2)}^{3} \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.6

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

$f (2) = \frac{256 - 240 - 1 \cdot 8 \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.7

Multiply $- 1 \cdot 8 \cdot 20$ .

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Step 24.2.3.7.1

Multiply $- 1$ by $8$ .

$f (2) = \frac{256 - 240 - 8 \cdot 20 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.7.2

Multiply $- 8$ by $20$ .

$f (2) = \frac{256 - 240 - 160 + {(2)}^{2} \cdot 20}{20}$

Step 24.2.3.8

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

$f (2) = \frac{256 - 240 - 160 + 4 \cdot 20}{20}$

Step 24.2.3.9

Multiply $4$ by $20$ .

$f (2) = \frac{256 - 240 - 160 + 80}{20}$

Step 24.2.4

Reduce the expression by cancelling the common factors.

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Step 24.2.4.1

Subtract $240$ from $256$ .

$f (2) = \frac{16 - 160 + 80}{20}$

Step 24.2.4.2

Simplify by adding and subtracting.

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Step 24.2.4.2.1

Subtract $160$ from $16$ .

$f (2) = \frac{- 144 + 80}{20}$

Step 24.2.4.2.2

Add $- 144$ and $80$ .

$f (2) = \frac{- 64}{20}$

Step 24.2.4.3

Cancel the common factor of $- 64$ and $20$ .

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Step 24.2.4.3.1

Factor $4$ out of $- 64$ .

$f (2) = \frac{4 (- 16)}{20}$

Step 24.2.4.3.2

Cancel the common factors.

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Step 24.2.4.3.2.1

Factor $4$ out of $20$ .

$f (2) = \frac{4 \cdot - 16}{4 \cdot 5}$

Step 24.2.4.3.2.2

Cancel the common factor.

$f (2) = \frac{4 \cdot - 16}{4 \cdot 5}$

Step 24.2.4.3.2.3

Rewrite the expression.

$f (2) = \frac{- 16}{5}$

Step 24.2.4.4

Move the negative in front of the fraction.

$f (2) = - \frac{16}{5}$

Step 24.2.5

The final answer is $- \frac{16}{5}$ .

$y = - \frac{16}{5}$

Step 25

These are the local extrema for $f (x) = \frac{2 x^{5}}{5} - \frac{3 x^{4}}{4} - x^{3} + x^{2}$ .

$(0, 0)$ is a local minima

$(- 1, \frac{17}{20})$ is a local maxima

$(\frac{1}{2}, \frac{29}{320})$ is a local maxima

$(2, - \frac{16}{5})$ is a local minima

Step 26