Calculus Examples

$f (x) = \frac{1}{25} x^{3} - \frac{1}{5} x^{2} - x + 5$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $\frac{1}{25} x^{3} - \frac{1}{5} x^{2} - x + 5$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{25} x^{3}] + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [\frac{1}{25} x^{3}] + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.2

Evaluate $\frac{d}{d x} [\frac{1}{25} x^{3}]$ .

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Step 1.2.1

Since $\frac{1}{25}$ is constant with respect to $x$ , the derivative of $\frac{1}{25} x^{3}$ with respect to $x$ is $\frac{1}{25} \frac{d}{d x} [x^{3}]$ .

$\frac{1}{25} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.2.2

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

$\frac{1}{25} (3 x^{2}) + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.2.3

Combine $3$ and $\frac{1}{25}$ .

$\frac{3}{25} x^{2} + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.2.4

Combine $\frac{3}{25}$ and $x^{2}$ .

$\frac{3 x^{2}}{25} + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.3

Evaluate $\frac{d}{d x} [- \frac{1}{5} x^{2}]$ .

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Step 1.3.1

Since $- \frac{1}{5}$ is constant with respect to $x$ , the derivative of $- \frac{1}{5} x^{2}$ with respect to $x$ is $- \frac{1}{5} \frac{d}{d x} [x^{2}]$ .

$\frac{3 x^{2}}{25} - \frac{1}{5} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.3.2

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

$\frac{3 x^{2}}{25} - \frac{1}{5} (2 x) + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.3.3

Multiply $2$ by $- 1$ .

$\frac{3 x^{2}}{25} - 2 (\frac{1}{5}) x + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.3.4

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

$\frac{3 x^{2}}{25} + \frac{- 2}{5} x + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.3.5

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

$\frac{3 x^{2}}{25} + \frac{- 2 x}{5} + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.3.6

Move the negative in front of the fraction.

$\frac{3 x^{2}}{25} - \frac{2 x}{5} + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 1.4

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

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Step 1.4.1

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

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - \frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 1.4.2

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

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 \cdot 1 + \frac{d}{d x} [5]$

Step 1.4.3

Multiply $- 1$ by $1$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 + \frac{d}{d x} [5]$

Step 1.5

Differentiate using the Constant Rule.

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Step 1.5.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 + 0$

Step 1.5.2

Add $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$ and $0$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$ with respect to $x$ is $\frac{d}{d x} [\frac{3 x^{2}}{25}] + \frac{d}{d x} [- \frac{2 x}{5}] + \frac{d}{d x} [- 1]$ .

$f'' (x) = \frac{d}{d x} (\frac{3 x^{2}}{25}) + \frac{d}{d x} (- \frac{2 x}{5}) + \frac{d}{d x} (- 1)$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{3 x^{2}}{25}]$ .

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Step 2.2.1

Since $\frac{3}{25}$ is constant with respect to $x$ , the derivative of $\frac{3 x^{2}}{25}$ with respect to $x$ is $\frac{3}{25} \frac{d}{d x} [x^{2}]$ .

$f'' (x) = \frac{3}{25} \cdot \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- \frac{2 x}{5}) + \frac{d}{d x} (- 1)$

Step 2.2.2

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

$f'' (x) = \frac{3}{25} \cdot (2 x) + \frac{d}{d x} (- \frac{2 x}{5}) + \frac{d}{d x} (- 1)$

Step 2.2.3

Combine $2$ and $\frac{3}{25}$ .

$f'' (x) = \frac{2 \cdot 3}{25} x + \frac{d}{d x} (- \frac{2 x}{5}) + \frac{d}{d x} (- 1)$

Step 2.2.4

Multiply $2$ by $3$ .

$f'' (x) = \frac{6}{25} x + \frac{d}{d x} (- \frac{2 x}{5}) + \frac{d}{d x} (- 1)$

Step 2.2.5

Combine $\frac{6}{25}$ and $x$ .

$f'' (x) = \frac{6 x}{25} + \frac{d}{d x} (- \frac{2 x}{5}) + \frac{d}{d x} (- 1)$

Step 2.3

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

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Step 2.3.1

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

$f'' (x) = \frac{6 x}{25} - \frac{2}{5} \cdot \frac{d}{d x} x + \frac{d}{d x} (- 1)$

Step 2.3.2

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

$f'' (x) = \frac{6 x}{25} - \frac{2}{5} \cdot 1 + \frac{d}{d x} (- 1)$

Step 2.3.3

Multiply $- 1$ by $1$ .

$f'' (x) = \frac{6 x}{25} - \frac{2}{5} + \frac{d}{d x} (- 1)$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$f'' (x) = \frac{6 x}{25} - \frac{2}{5} + 0$

Step 2.4.2

Add $\frac{6 x}{25} - \frac{2}{5}$ and $0$ .

$f'' (x) = \frac{6 x}{25} - \frac{2}{5}$

Step 3

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

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 = 0$

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

$\frac{d}{d x} [\frac{1}{25} x^{3}] + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.2

Evaluate $\frac{d}{d x} [\frac{1}{25} x^{3}]$ .

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Step 4.1.2.1

Since $\frac{1}{25}$ is constant with respect to $x$ , the derivative of $\frac{1}{25} x^{3}$ with respect to $x$ is $\frac{1}{25} \frac{d}{d x} [x^{3}]$ .

$\frac{1}{25} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

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 $n = 3$ .

$\frac{1}{25} (3 x^{2}) + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.2.3

Combine $3$ and $\frac{1}{25}$ .

$\frac{3}{25} x^{2} + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.2.4

Combine $\frac{3}{25}$ and $x^{2}$ .

$\frac{3 x^{2}}{25} + \frac{d}{d x} [- \frac{1}{5} x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.3

Evaluate $\frac{d}{d x} [- \frac{1}{5} x^{2}]$ .

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Step 4.1.3.1

Since $- \frac{1}{5}$ is constant with respect to $x$ , the derivative of $- \frac{1}{5} x^{2}$ with respect to $x$ is $- \frac{1}{5} \frac{d}{d x} [x^{2}]$ .

$\frac{3 x^{2}}{25} - \frac{1}{5} \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

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 $n = 2$ .

$\frac{3 x^{2}}{25} - \frac{1}{5} (2 x) + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.3.3

Multiply $2$ by $- 1$ .

$\frac{3 x^{2}}{25} - 2 (\frac{1}{5}) x + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.3.4

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

$\frac{3 x^{2}}{25} + \frac{- 2}{5} x + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.3.5

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

$\frac{3 x^{2}}{25} + \frac{- 2 x}{5} + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.3.6

Move the negative in front of the fraction.

$\frac{3 x^{2}}{25} - \frac{2 x}{5} + \frac{d}{d x} [- x] + \frac{d}{d x} [5]$

Step 4.1.4

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

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Step 4.1.4.1

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

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - \frac{d}{d x} [x] + \frac{d}{d x} [5]$

Step 4.1.4.2

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

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 \cdot 1 + \frac{d}{d x} [5]$

Step 4.1.4.3

Multiply $- 1$ by $1$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 + \frac{d}{d x} [5]$

Step 4.1.5

Differentiate using the Constant Rule.

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Step 4.1.5.1

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 + 0$

Step 4.1.5.2

Add $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$ and $0$ .

$f' (x) = \frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 = 0$

Step 5.2

Multiply each term in $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 = 0$ by $25$ to eliminate the fractions.

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Step 5.2.1

Multiply each term in $\frac{3 x^{2}}{25} - \frac{2 x}{5} - 1 = 0$ by $25$ .

$\frac{3 x^{2}}{25} \cdot 25 - \frac{2 x}{5} \cdot 25 - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Cancel the common factor of $25$ .

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Step 5.2.2.1.1.1

Cancel the common factor.

$\frac{3 x^{2}}{25} \cdot 25 - \frac{2 x}{5} \cdot 25 - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.1.2

Rewrite the expression.

$3 x^{2} - \frac{2 x}{5} \cdot 25 - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.2

Cancel the common factor of $5$ .

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Step 5.2.2.1.2.1

Move the leading negative in $- \frac{2 x}{5}$ into the numerator.

$3 x^{2} + \frac{- 2 x}{5} \cdot 25 - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.2.2

Factor $5$ out of $25$ .

$3 x^{2} + \frac{- 2 x}{5} \cdot (5 (5)) - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.2.3

Cancel the common factor.

$3 x^{2} + \frac{- 2 x}{5} \cdot (5 \cdot 5) - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.2.4

Rewrite the expression.

$3 x^{2} - 2 x \cdot 5 - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.3

Multiply $5$ by $- 2$ .

$3 x^{2} - 10 x - 1 \cdot 25 = 0 \cdot 25$

Step 5.2.2.1.4

Multiply $- 1$ by $25$ .

$3 x^{2} - 10 x - 25 = 0 \cdot 25$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Multiply $0$ by $25$ .

$3 x^{2} - 10 x - 25 = 0$

Step 5.3

Factor by grouping.

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Step 5.3.1

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 $a \cdot c = 3 \cdot - 25 = - 75$ and whose sum is $b = - 10$ .

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Step 5.3.1.1

Factor $- 10$ out of $- 10 x$ .

$3 x^{2} - 10 x - 25 = 0$

Step 5.3.1.2

Rewrite $- 10$ as $5$ plus $- 15$

$3 x^{2} + (5 - 15) x - 25 = 0$

Step 5.3.1.3

Apply the distributive property.

$3 x^{2} + 5 x - 15 x - 25 = 0$

Step 5.3.2

Factor out the greatest common factor from each group.

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Step 5.3.2.1

Group the first two terms and the last two terms.

$(3 x^{2} + 5 x) - 15 x - 25 = 0$

Step 5.3.2.2

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

$x (3 x + 5) - 5 (3 x + 5) = 0$

Step 5.3.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 5$ .

$(3 x + 5) (x - 5) = 0$

Step 5.4

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

$3 x + 5 = 0$

$x - 5 = 0$

Step 5.5

Set $3 x + 5$ equal to $0$ and solve for $x$ .

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Step 5.5.1

Set $3 x + 5$ equal to $0$ .

$3 x + 5 = 0$

Step 5.5.2

Solve $3 x + 5 = 0$ for $x$ .

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Step 5.5.2.1

Subtract $5$ from both sides of the equation.

$3 x = - 5$

Step 5.5.2.2

Divide each term in $3 x = - 5$ by $3$ and simplify.

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Step 5.5.2.2.1

Divide each term in $3 x = - 5$ by $3$ .

$\frac{3 x}{3} = \frac{- 5}{3}$

Step 5.5.2.2.2

Simplify the left side.

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Step 5.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 5.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 5}{3}$

Step 5.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{3}$

Step 5.5.2.2.3

Simplify the right side.

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Step 5.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{3}$

Step 5.6

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 5.6.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 5.6.2

Add $5$ to both sides of the equation.

$x = 5$

Step 5.7

The final solution is all the values that make $(3 x + 5) (x - 5) = 0$ true.

$x = - \frac{5}{3}, 5$

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.

$x = - \frac{5}{3}, 5$

Step 8

Evaluate the second derivative at $x = - \frac{5}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{6 (- \frac{5}{3})}{25} - \frac{2}{5}$

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 numerator.

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Step 9.1.1.1

Multiply $- 1$ by $6$ .

$\frac{- 6 (\frac{5}{3})}{25} - \frac{2}{5}$

Step 9.1.1.2

Combine $- 6$ and $\frac{5}{3}$ .

$\frac{\frac{- 6 \cdot 5}{3}}{25} - \frac{2}{5}$

Step 9.1.2

Multiply $- 6$ by $5$ .

$\frac{\frac{- 30}{3}}{25} - \frac{2}{5}$

Step 9.1.3

Divide $- 30$ by $3$ .

$\frac{- 10}{25} - \frac{2}{5}$

Step 9.1.4

Cancel the common factor of $- 10$ and $25$ .

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Step 9.1.4.1

Factor $5$ out of $- 10$ .

$\frac{5 (- 2)}{25} - \frac{2}{5}$

Step 9.1.4.2

Cancel the common factors.

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Step 9.1.4.2.1

Factor $5$ out of $25$ .

$\frac{5 \cdot - 2}{5 \cdot 5} - \frac{2}{5}$

Step 9.1.4.2.2

Cancel the common factor.

$\frac{5 \cdot - 2}{5 \cdot 5} - \frac{2}{5}$

Step 9.1.4.2.3

Rewrite the expression.

$\frac{- 2}{5} - \frac{2}{5}$

Step 9.1.5

Move the negative in front of the fraction.

$- \frac{2}{5} - \frac{2}{5}$

Step 9.2

Combine fractions.

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Step 9.2.1

Combine the numerators over the common denominator.

$\frac{- 2 - 2}{5}$

Step 9.2.2

Simplify the expression.

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Step 9.2.2.1

Subtract $2$ from $- 2$ .

$\frac{- 4}{5}$

Step 9.2.2.2

Move the negative in front of the fraction.

$- \frac{4}{5}$

Step 10

$x = - \frac{5}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \frac{5}{3}$ is a local maximum

Step 11

Find the y-value when $x = - \frac{5}{3}$ .

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Step 11.1

Replace the variable $x$ with $- \frac{5}{3}$ in the expression.

$f (- \frac{5}{3}) = \frac{1}{25} \cdot {(- \frac{5}{3})}^{3} - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

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

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

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Step 11.2.1.1.1

Apply the product rule to $- \frac{5}{3}$ .

$f (- \frac{5}{3}) = \frac{1}{25} \cdot ({(- 1)}^{3} {(\frac{5}{3})}^{3}) - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.1.2

Apply the product rule to $\frac{5}{3}$ .

$f (- \frac{5}{3}) = \frac{1}{25} \cdot ({(- 1)}^{3} (\frac{5^{3}}{3^{3}})) - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.2

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

$f (- \frac{5}{3}) = \frac{1}{25} \cdot (- \frac{5^{3}}{3^{3}}) - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.3

Raise $5$ to the power of $3$ .

$f (- \frac{5}{3}) = \frac{1}{25} \cdot (- \frac{125}{3^{3}}) - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.4

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

$f (- \frac{5}{3}) = \frac{1}{25} \cdot (- \frac{125}{27}) - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.5

Cancel the common factor of $25$ .

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Step 11.2.1.5.1

Move the leading negative in $- \frac{125}{27}$ into the numerator.

$f (- \frac{5}{3}) = \frac{1}{25} \cdot \frac{- 125}{27} - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.5.2

Factor $25$ out of $- 125$ .

$f (- \frac{5}{3}) = \frac{1}{25} \cdot \frac{25 (- 5)}{27} - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.5.3

Cancel the common factor.

$f (- \frac{5}{3}) = \frac{1}{25} \cdot \frac{25 \cdot - 5}{27} - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.5.4

Rewrite the expression.

$f (- \frac{5}{3}) = \frac{- 5}{27} - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.6

Move the negative in front of the fraction.

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{1}{5} \cdot {(- \frac{5}{3})}^{2} - (- \frac{5}{3}) + 5$

Step 11.2.1.7

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

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Step 11.2.1.7.1

Apply the product rule to $- \frac{5}{3}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{1}{5} \cdot ({(- 1)}^{2} {(\frac{5}{3})}^{2}) - (- \frac{5}{3}) + 5$

Step 11.2.1.7.2

Apply the product rule to $\frac{5}{3}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{1}{5} \cdot ({(- 1)}^{2} (\frac{5^{2}}{3^{2}})) - (- \frac{5}{3}) + 5$

Step 11.2.1.8

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 11.2.1.8.1

Move ${(- 1)}^{2}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} + {(- 1)}^{2} \cdot (- 1 (\frac{1}{5})) \cdot \frac{5^{2}}{3^{2}} - (- \frac{5}{3}) + 5$

Step 11.2.1.8.2

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

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Step 11.2.1.8.2.1

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

$f (- \frac{5}{3}) = - \frac{5}{27} + {(- 1)}^{2} \cdot ((- 1) (\frac{1}{5})) \cdot \frac{5^{2}}{3^{2}} - (- \frac{5}{3}) + 5$

Step 11.2.1.8.2.2

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

$f (- \frac{5}{3}) = - \frac{5}{27} + {(- 1)}^{2 + 1} (\frac{1}{5}) \cdot \frac{5^{2}}{3^{2}} - (- \frac{5}{3}) + 5$

Step 11.2.1.8.3

Add $2$ and $1$ .

$f (- \frac{5}{3}) = - \frac{5}{27} + {(- 1)}^{3} (\frac{1}{5}) \cdot \frac{5^{2}}{3^{2}} - (- \frac{5}{3}) + 5$

Step 11.2.1.9

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

$f (- \frac{5}{3}) = - \frac{5}{27} + {(- 1)}^{3} (\frac{1}{5}) \cdot \frac{25}{3^{2}} - (- \frac{5}{3}) + 5$

Step 11.2.1.10

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

$f (- \frac{5}{3}) = - \frac{5}{27} + {(- 1)}^{3} (\frac{1}{5}) \cdot \frac{25}{9} - (- \frac{5}{3}) + 5$

Step 11.2.1.11

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

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{1}{5} \cdot \frac{25}{9} - (- \frac{5}{3}) + 5$

Step 11.2.1.12

Cancel the common factor of $5$ .

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Step 11.2.1.12.1

Move the leading negative in $- \frac{1}{5}$ into the numerator.

$f (- \frac{5}{3}) = - \frac{5}{27} + \frac{- 1}{5} \cdot \frac{25}{9} - (- \frac{5}{3}) + 5$

Step 11.2.1.12.2

Factor $5$ out of $25$ .

$f (- \frac{5}{3}) = - \frac{5}{27} + \frac{- 1}{5} \cdot \frac{5 (5)}{9} - (- \frac{5}{3}) + 5$

Step 11.2.1.12.3

Cancel the common factor.

$f (- \frac{5}{3}) = - \frac{5}{27} + \frac{- 1}{5} \cdot \frac{5 \cdot 5}{9} - (- \frac{5}{3}) + 5$

Step 11.2.1.12.4

Rewrite the expression.

$f (- \frac{5}{3}) = - \frac{5}{27} - 1 \cdot \frac{5}{9} - (- \frac{5}{3}) + 5$

Step 11.2.1.13

Rewrite $- 1 (\frac{5}{9})$ as $- (\frac{5}{9})$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - (\frac{5}{9}) - (- \frac{5}{3}) + 5$

Step 11.2.1.14

Multiply $- (- \frac{5}{3})$ .

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Step 11.2.1.14.1

Multiply $- 1$ by $- 1$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5}{9} + 1 (\frac{5}{3}) + 5$

Step 11.2.1.14.2

Multiply $\frac{5}{3}$ by $1$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5}{9} + \frac{5}{3} + 5$

Step 11.2.2

Find the common denominator.

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Step 11.2.2.1

Multiply $\frac{5}{9}$ by $\frac{3}{3}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - (\frac{5}{9} \cdot \frac{3}{3}) + \frac{5}{3} + 5$

Step 11.2.2.2

Multiply $\frac{5}{9}$ by $\frac{3}{3}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{5}{3} + 5$

Step 11.2.2.3

Multiply $\frac{5}{3}$ by $\frac{9}{9}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{5}{3} \cdot \frac{9}{9} + 5$

Step 11.2.2.4

Multiply $\frac{5}{3}$ by $\frac{9}{9}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{5 \cdot 9}{3 \cdot 9} + 5$

Step 11.2.2.5

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

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{5 \cdot 9}{3 \cdot 9} + \frac{5}{1}$

Step 11.2.2.6

Multiply $\frac{5}{1}$ by $\frac{27}{27}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{5 \cdot 9}{3 \cdot 9} + \frac{5}{1} \cdot \frac{27}{27}$

Step 11.2.2.7

Multiply $\frac{5}{1}$ by $\frac{27}{27}$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{9 \cdot 3} + \frac{5 \cdot 9}{3 \cdot 9} + \frac{5 \cdot 27}{27}$

Step 11.2.2.8

Reorder the factors of $9 \cdot 3$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{3 \cdot 9} + \frac{5 \cdot 9}{3 \cdot 9} + \frac{5 \cdot 27}{27}$

Step 11.2.2.9

Multiply $3$ by $9$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{27} + \frac{5 \cdot 9}{3 \cdot 9} + \frac{5 \cdot 27}{27}$

Step 11.2.2.10

Multiply $3$ by $9$ .

$f (- \frac{5}{3}) = - \frac{5}{27} - \frac{5 \cdot 3}{27} + \frac{5 \cdot 9}{27} + \frac{5 \cdot 27}{27}$

Step 11.2.3

Combine the numerators over the common denominator.

$f (- \frac{5}{3}) = \frac{- 5 - 5 \cdot 3 + 5 \cdot 9 + 5 \cdot 27}{27}$

Step 11.2.4

Simplify each term.

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Step 11.2.4.1

Multiply $- 5$ by $3$ .

$f (- \frac{5}{3}) = \frac{- 5 - 15 + 5 \cdot 9 + 5 \cdot 27}{27}$

Step 11.2.4.2

Multiply $5$ by $9$ .

$f (- \frac{5}{3}) = \frac{- 5 - 15 + 45 + 5 \cdot 27}{27}$

Step 11.2.4.3

Multiply $5$ by $27$ .

$f (- \frac{5}{3}) = \frac{- 5 - 15 + 45 + 135}{27}$

Step 11.2.5

Simplify by adding and subtracting.

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Step 11.2.5.1

Subtract $15$ from $- 5$ .

$f (- \frac{5}{3}) = \frac{- 20 + 45 + 135}{27}$

Step 11.2.5.2

Add $- 20$ and $45$ .

$f (- \frac{5}{3}) = \frac{25 + 135}{27}$

Step 11.2.5.3

Add $25$ and $135$ .

$f (- \frac{5}{3}) = \frac{160}{27}$

Step 11.2.6

The final answer is $\frac{160}{27}$ .

$y = \frac{160}{27}$

Step 12

Evaluate the second derivative at $x = 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{6 (5)}{25} - \frac{2}{5}$

Step 13

Evaluate the second derivative.

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Step 13.1

Multiply $6$ by $5$ .

$\frac{30}{25} - \frac{2}{5}$

Step 13.2

Cancel the common factor of $30$ and $25$ .

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Step 13.2.1

Factor $5$ out of $30$ .

$\frac{5 (6)}{25} - \frac{2}{5}$

Step 13.2.2

Cancel the common factors.

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Step 13.2.2.1

Factor $5$ out of $25$ .

$\frac{5 \cdot 6}{5 \cdot 5} - \frac{2}{5}$

Step 13.2.2.2

Cancel the common factor.

$\frac{5 \cdot 6}{5 \cdot 5} - \frac{2}{5}$

Step 13.2.2.3

Rewrite the expression.

$\frac{6}{5} - \frac{2}{5}$

Step 13.3

Combine the numerators over the common denominator.

$\frac{6 - 2}{5}$

Step 13.4

Subtract $2$ from $6$ .

$\frac{4}{5}$

Step 14

$x = 5$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 5$ is a local minimum

Step 15

Find the y-value when $x = 5$ .

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Step 15.1

Replace the variable $x$ with $5$ in the expression.

$f (5) = \frac{1}{25} \cdot {(5)}^{3} - \frac{1}{5} \cdot {(5)}^{2} - (5) + 5$

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

Raise $5$ to the power of $3$ .

$f (5) = \frac{1}{25} \cdot 125 - \frac{1}{5} \cdot {(5)}^{2} - (5) + 5$

Step 15.2.1.2

Cancel the common factor of $25$ .

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Step 15.2.1.2.1

Factor $25$ out of $125$ .

$f (5) = \frac{1}{25} \cdot (25 (5)) - \frac{1}{5} \cdot {(5)}^{2} - (5) + 5$

Step 15.2.1.2.2

Cancel the common factor.

$f (5) = \frac{1}{25} \cdot (25 \cdot 5) - \frac{1}{5} \cdot {(5)}^{2} - (5) + 5$

Step 15.2.1.2.3

Rewrite the expression.

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

Step 15.2.1.3

Cancel the common factor of $5$ .

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Step 15.2.1.3.1

Move the leading negative in $- \frac{1}{5}$ into the numerator.

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

Step 15.2.1.3.2

Factor $5$ out of ${(5)}^{2}$ .

$f (5) = 5 + \frac{- 1}{5} \cdot (5 \cdot 5) - (5) + 5$

Step 15.2.1.3.3

Cancel the common factor.

$f (5) = 5 + \frac{- 1}{5} \cdot (5 \cdot 5) - (5) + 5$

Step 15.2.1.3.4

Rewrite the expression.

$f (5) = 5 - 1 \cdot 5 - (5) + 5$

Step 15.2.1.4

Multiply $- 1$ by $5$ .

$f (5) = 5 - 5 - (5) + 5$

Step 15.2.1.5

Multiply $- 1$ by $5$ .

$f (5) = 5 - 5 - 5 + 5$

Step 15.2.2

Simplify by adding and subtracting.

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Step 15.2.2.1

Subtract $5$ from $5$ .

$f (5) = 0 - 5 + 5$

Step 15.2.2.2

Subtract $5$ from $0$ .

$f (5) = - 5 + 5$

Step 15.2.2.3

Add $- 5$ and $5$ .

$f (5) = 0$

Step 15.2.3

The final answer is $0$ .

$y = 0$

Step 16

These are the local extrema for $f (x) = \frac{1}{25} x^{3} - \frac{1}{5} x^{2} - x + 5$ .

$(- \frac{5}{3}, \frac{160}{27})$ is a local maxima

$(5, 0)$ is a local minima

Step 17