Calculus Examples

$f (x) = 10 x \ln (| x |)$

Step 1

Find the first derivative of the function.

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Step 1.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x \ln (| x |)$ with respect to $x$ is $10 \frac{d}{d x} [x \ln (| x |)]$ .

$10 \frac{d}{d x} [x \ln (| x |)]$

Step 1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (| x |)$ .

$10 (x \frac{d}{d x} [\ln (| x |)] + \ln (| x |) \frac{d}{d x} [x])$

Step 1.3

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

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Step 1.3.1

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

$10 (x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [| x |]) + \ln (| x |) \frac{d}{d x} [x])$

Step 1.3.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$10 (x (\frac{1}{u} \frac{d}{d x} [| x |]) + \ln (| x |) \frac{d}{d x} [x])$

Step 1.3.3

Replace all occurrences of $u$ with $| x |$ .

$10 (x (\frac{1}{| x |} \frac{d}{d x} [| x |]) + \ln (| x |) \frac{d}{d x} [x])$

Step 1.4

Combine $\frac{1}{| x |}$ and $x$ .

$10 (\frac{x}{| x |} \frac{d}{d x} [| x |] + \ln (| x |) \frac{d}{d x} [x])$

Step 1.5

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

$10 (\frac{x}{| x |} \cdot \frac{x}{| x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.6

Multiply $\frac{x}{| x |}$ by $\frac{x}{| x |}$ .

$10 (\frac{x \cdot x}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.7

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

$10 (\frac{x^{1} x}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.8

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

$10 (\frac{x^{1} x^{1}}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.9

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

$10 (\frac{x^{1 + 1}}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.10

Add $1$ and $1$ .

$10 (\frac{x^{2}}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.11

To multiply absolute values, multiply the terms inside each absolute value.

$10 (\frac{x^{2}}{| x \cdot x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.12

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

$10 (\frac{x^{2}}{| x^{1} x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.13

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

$10 (\frac{x^{2}}{| x^{1} x^{1} |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.14

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

$10 (\frac{x^{2}}{| x^{1 + 1} |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.15

Add $1$ and $1$ .

$10 (\frac{x^{2}}{| x^{2} |} + \ln (| x |) \frac{d}{d x} [x])$

Step 1.16

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

$10 (\frac{x^{2}}{| x^{2} |} + \ln (| x |) \cdot 1)$

Step 1.17

Multiply $\ln (| x |)$ by $1$ .

$10 (\frac{x^{2}}{| x^{2} |} + \ln (| x |))$

Step 1.18

Simplify.

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Step 1.18.1

Apply the distributive property.

$10 \frac{x^{2}}{| x^{2} |} + 10 \ln (| x |)$

Step 1.18.2

Combine $10$ and $\frac{x^{2}}{| x^{2} |}$ .

$\frac{10 x^{2}}{| x^{2} |} + 10 \ln (| x |)$

Step 1.18.3

Simplify each term.

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Step 1.18.3.1

Remove non-negative terms from the absolute value.

$\frac{10 x^{2}}{x^{2}} + 10 \ln (| x |)$

Step 1.18.3.2

Cancel the common factor of $x^{2}$ .

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Step 1.18.3.2.1

Cancel the common factor.

$\frac{10 x^{2}}{x^{2}} + 10 \ln (| x |)$

Step 1.18.3.2.2

Divide $10$ by $1$ .

$10 + 10 \ln (| x |)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $10 + 10 \ln (| x |)$ with respect to $x$ is $\frac{d}{d x} [10] + \frac{d}{d x} [10 \ln (| x |)]$ .

$f'' (x) = \frac{d}{d x} (10) + \frac{d}{d x} (10 \ln (| x |))$

Step 2.1.2

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (10 \ln (| x |))$

Step 2.2

Evaluate $\frac{d}{d x} [10 \ln (| x |)]$ .

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Step 2.2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 \ln (| x |)$ with respect to $x$ is $10 \frac{d}{d x} [\ln (| x |)]$ .

$f'' (x) = 0 + 10 \frac{d}{d x} (\ln (| x |))$

Step 2.2.2

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

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Step 2.2.2.1

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

$f'' (x) = 0 + 10 (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (| x |))$

Step 2.2.2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$f'' (x) = 0 + 10 (\frac{1}{u} \cdot \frac{d}{d x} (| x |))$

Step 2.2.2.3

Replace all occurrences of $u$ with $| x |$ .

$f'' (x) = 0 + 10 (\frac{1}{| x |} \cdot \frac{d}{d x} (| x |))$

Step 2.2.3

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

$f'' (x) = 0 + 10 (\frac{1}{| x |} \cdot \frac{x}{| x |})$

Step 2.2.4

Multiply $\frac{1}{| x |}$ by $\frac{x}{| x |}$ .

$f'' (x) = 0 + 10 (\frac{x}{| x | | x |})$

Step 2.2.5

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = 0 + 10 (\frac{x}{| x \cdot x |})$

Step 2.2.6

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

$f'' (x) = 0 + 10 (\frac{x}{| x \cdot x |})$

Step 2.2.7

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

$f'' (x) = 0 + 10 (\frac{x}{| x \cdot x |})$

Step 2.2.8

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

$f'' (x) = 0 + 10 (\frac{x}{| x^{1 + 1} |})$

Step 2.2.9

Add $1$ and $1$ .

$f'' (x) = 0 + 10 (\frac{x}{| x^{2} |})$

Step 2.2.10

Combine $10$ and $\frac{x}{| x^{2} |}$ .

$f'' (x) = 0 + \frac{10 x}{| x^{2} |}$

Step 2.3

Simplify.

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Step 2.3.1

Add $0$ and $\frac{10 x}{| x^{2} |}$ .

$f'' (x) = \frac{10 x}{| x^{2} |}$

Step 2.3.2

Remove non-negative terms from the absolute value.

$f'' (x) = \frac{10 x}{x^{2}}$

Step 2.3.3

Cancel the common factor of $x$ and $x^{2}$ .

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Step 2.3.3.1

Factor $x$ out of $10 x$ .

$f'' (x) = \frac{x \cdot 10}{x^{2}}$

Step 2.3.3.2

Cancel the common factors.

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Step 2.3.3.2.1

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

$f'' (x) = \frac{x \cdot 10}{x \cdot x}$

Step 2.3.3.2.2

Cancel the common factor.

$f'' (x) = \frac{x \cdot 10}{x \cdot x}$

Step 2.3.3.2.3

Rewrite the expression.

$f'' (x) = \frac{10}{x}$

Step 3

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

$10 + 10 \ln (| x |) = 0$

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

Since $10$ is constant with respect to $x$ , the derivative of $10 x \ln (| x |)$ with respect to $x$ is $10 \frac{d}{d x} [x \ln (| x |)]$ .

$10 \frac{d}{d x} [x \ln (| x |)]$

Step 4.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (| x |)$ .

$10 (x \frac{d}{d x} [\ln (| x |)] + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.3

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

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Step 4.1.3.1

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

$10 (x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [| x |]) + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.3.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$10 (x (\frac{1}{u} \frac{d}{d x} [| x |]) + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.3.3

Replace all occurrences of $u$ with $| x |$ .

$10 (x (\frac{1}{| x |} \frac{d}{d x} [| x |]) + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.4

Combine $\frac{1}{| x |}$ and $x$ .

$10 (\frac{x}{| x |} \frac{d}{d x} [| x |] + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.5

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

$10 (\frac{x}{| x |} \cdot \frac{x}{| x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.6

Multiply $\frac{x}{| x |}$ by $\frac{x}{| x |}$ .

$10 (\frac{x \cdot x}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.7

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

$10 (\frac{x^{1} x}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.8

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

$10 (\frac{x^{1} x^{1}}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.9

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

$10 (\frac{x^{1 + 1}}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.10

Add $1$ and $1$ .

$10 (\frac{x^{2}}{| x | | x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.11

To multiply absolute values, multiply the terms inside each absolute value.

$10 (\frac{x^{2}}{| x \cdot x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.12

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

$10 (\frac{x^{2}}{| x^{1} x |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.13

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

$10 (\frac{x^{2}}{| x^{1} x^{1} |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.14

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

$10 (\frac{x^{2}}{| x^{1 + 1} |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.15

Add $1$ and $1$ .

$10 (\frac{x^{2}}{| x^{2} |} + \ln (| x |) \frac{d}{d x} [x])$

Step 4.1.16

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

$10 (\frac{x^{2}}{| x^{2} |} + \ln (| x |) \cdot 1)$

Step 4.1.17

Multiply $\ln (| x |)$ by $1$ .

$10 (\frac{x^{2}}{| x^{2} |} + \ln (| x |))$

Step 4.1.18

Simplify.

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Step 4.1.18.1

Apply the distributive property.

$10 \frac{x^{2}}{| x^{2} |} + 10 \ln (| x |)$

Step 4.1.18.2

Combine $10$ and $\frac{x^{2}}{| x^{2} |}$ .

$\frac{10 x^{2}}{| x^{2} |} + 10 \ln (| x |)$

Step 4.1.18.3

Simplify each term.

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Step 4.1.18.3.1

Remove non-negative terms from the absolute value.

$\frac{10 x^{2}}{x^{2}} + 10 \ln (| x |)$

Step 4.1.18.3.2

Cancel the common factor of $x^{2}$ .

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Step 4.1.18.3.2.1

Cancel the common factor.

$\frac{10 x^{2}}{x^{2}} + 10 \ln (| x |)$

Step 4.1.18.3.2.2

Divide $10$ by $1$ .

$f' (x) = 10 + 10 \ln (| x |)$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $10 + 10 \ln (| x |)$ .

$10 + 10 \ln (| x |)$

Step 5

Set the first derivative equal to $0$ then solve the equation $10 + 10 \ln (| x |) = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$10 + 10 \ln (| x |) = 0$

Step 5.2

Subtract $10$ from both sides of the equation.

$10 \ln (| x |) = - 10$

Step 5.3

Divide each term in $10 \ln (| x |) = - 10$ by $10$ and simplify.

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Step 5.3.1

Divide each term in $10 \ln (| x |) = - 10$ by $10$ .

$\frac{10 \ln (| x |)}{10} = \frac{- 10}{10}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $10$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{10 \ln (| x |)}{10} = \frac{- 10}{10}$

Step 5.3.2.1.2

Divide $\ln (| x |)$ by $1$ .

$\ln (| x |) = \frac{- 10}{10}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $- 10$ by $10$ .

$\ln (| x |) = - 1$

Step 5.4

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (| x |)} = e^{- 1}$

Step 5.5

Rewrite $\ln (| x |) = - 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- 1} = | x |$

Step 5.6

Solve for $x$ .

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Step 5.6.1

Rewrite the equation as $| x | = e^{- 1}$ .

$| x | = e^{- 1}$

Step 5.6.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$| x | = \frac{1}{e}$

Step 5.6.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

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

Step 5.6.4

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

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Step 5.6.4.1

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

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

Step 5.6.4.2

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

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

Step 5.6.4.3

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

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

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the argument in $\ln (| x |)$ less than or equal to $0$ to find where the expression is undefined.

$| x | \leq 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Write $| x | \leq 0$ as a piecewise.

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Step 6.2.1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 6.2.1.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \leq 0$

Step 6.2.1.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 6.2.1.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \leq 0$

Step 6.2.1.5

Write as a piecewise.

${\begin{cases} x \leq 0 & x \geq 0 \\ - x \leq 0 & x < 0 \end{cases}$

Step 6.2.2

Find the intersection of $x \leq 0$ and $x \geq 0$ .

$x = 0$

Step 6.2.3

Solve $- x \leq 0$ when $x < 0$ .

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Step 6.2.3.1

Divide each term in $- x \leq 0$ by $- 1$ and simplify.

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Step 6.2.3.1.1

Divide each term in $- x \leq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{0}{- 1}$

Step 6.2.3.1.2

Simplify the left side.

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Step 6.2.3.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{0}{- 1}$

Step 6.2.3.1.2.2

Divide $x$ by $1$ .

$x \geq \frac{0}{- 1}$

Step 6.2.3.1.3

Simplify the right side.

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Step 6.2.3.1.3.1

Divide $0$ by $- 1$ .

$x \geq 0$

Step 6.2.3.2

Find the intersection of $x \geq 0$ and $x < 0$ .

No solution

Step 6.2.4

Find the union of the solutions.

$x = 0$

Step 7

Critical points to evaluate.

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

Step 8

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

$\frac{10}{\frac{1}{e}}$

Step 9

Multiply the numerator by the reciprocal of the denominator.

$10 e$

Step 10

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

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

Step 11

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

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Step 11.1

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

$f (\frac{1}{e}) = 10 (\frac{1}{e}) \ln (| \frac{1}{e} |)$

Step 11.2

Simplify the result.

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Step 11.2.1

Combine $10$ and $\frac{1}{e}$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot \ln (| \frac{1}{e} |)$

Step 11.2.2

$\frac{1}{e}$ is approximately $0.36787944$ which is positive so remove the absolute value

$f (\frac{1}{e}) = \frac{10}{e} \cdot \ln (\frac{1}{e})$

Step 11.2.3

Rewrite $\frac{1}{e}$ as $1 e^{- 1}$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot \ln (1 e^{- 1})$

Step 11.2.4

Rewrite $\ln (1 e^{- 1})$ as $\ln (1) + \ln (e^{- 1})$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot (\ln (1) + \ln (e^{- 1}))$

Step 11.2.5

Use logarithm rules to move $- 1$ out of the exponent.

$f (\frac{1}{e}) = \frac{10}{e} \cdot (\ln (1) - \ln (e))$

Step 11.2.6

The natural logarithm of $e$ is $1$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot (\ln (1) - 1 \cdot 1)$

Step 11.2.7

Multiply $- 1$ by $1$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot (\ln (1) - 1)$

Step 11.2.8

The natural logarithm of $1$ is $0$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot (0 - 1)$

Step 11.2.9

Subtract $1$ from $0$ .

$f (\frac{1}{e}) = \frac{10}{e} \cdot - 1$

Step 11.2.10

Multiply $\frac{10}{e} \cdot - 1$ .

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Step 11.2.10.1

Combine $\frac{10}{e}$ and $- 1$ .

$f (\frac{1}{e}) = \frac{10 \cdot - 1}{e}$

Step 11.2.10.2

Multiply $10$ by $- 1$ .

$f (\frac{1}{e}) = \frac{- 10}{e}$

Step 11.2.11

Move the negative in front of the fraction.

$f (\frac{1}{e}) = - \frac{10}{e}$

Step 11.2.12

The final answer is $- \frac{10}{e}$ .

$y = - \frac{10}{e}$

Step 12

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

$\frac{10}{- \frac{1}{e}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Multiply the numerator by the reciprocal of the denominator.

$10 (- e)$

Step 13.2

Multiply $- 1$ by $10$ .

$- 10 e$

Step 14

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

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

Step 15

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

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Step 15.1

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

$f (- \frac{1}{e}) = 10 (- \frac{1}{e}) \ln (| - \frac{1}{e} |)$

Step 15.2

Simplify the result.

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Step 15.2.1

Multiply $10 (- \frac{1}{e})$ .

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Step 15.2.1.1

Multiply $- 1$ by $10$ .

$f (- \frac{1}{e}) = - 10 \frac{1}{e} \cdot \ln (| - \frac{1}{e} |)$

Step 15.2.1.2

Combine $- 10$ and $\frac{1}{e}$ .

$f (- \frac{1}{e}) = \frac{- 10}{e} \cdot \ln (| - \frac{1}{e} |)$

Step 15.2.2

Move the negative in front of the fraction.

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot \ln (| - \frac{1}{e} |)$

Step 15.2.3

$- \frac{1}{e}$ is approximately $- 0.36787944$ which is negative so negate $- \frac{1}{e}$ and remove the absolute value

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot \ln (\frac{1}{e})$

Step 15.2.4

Rewrite $\frac{1}{e}$ as $1 e^{- 1}$ .

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot \ln (1 e^{- 1})$

Step 15.2.5

Rewrite $\ln (1 e^{- 1})$ as $\ln (1) + \ln (e^{- 1})$ .

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot (\ln (1) + \ln (e^{- 1}))$

Step 15.2.6

Use logarithm rules to move $- 1$ out of the exponent.

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot (\ln (1) - \ln (e))$

Step 15.2.7

The natural logarithm of $e$ is $1$ .

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot (\ln (1) - 1 \cdot 1)$

Step 15.2.8

Multiply $- 1$ by $1$ .

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot (\ln (1) - 1)$

Step 15.2.9

The natural logarithm of $1$ is $0$ .

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot (0 - 1)$

Step 15.2.10

Subtract $1$ from $0$ .

$f (- \frac{1}{e}) = - \frac{10}{e} \cdot - 1$

Step 15.2.11

Multiply $- \frac{10}{e} \cdot - 1$ .

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Step 15.2.11.1

Multiply $- 1$ by $- 1$ .

$f (- \frac{1}{e}) = 1 (\frac{10}{e})$

Step 15.2.11.2

Multiply $\frac{10}{e}$ by $1$ .

$f (- \frac{1}{e}) = \frac{10}{e}$

Step 15.2.12

The final answer is $\frac{10}{e}$ .

$y = \frac{10}{e}$

Step 16

These are the local extrema for $f (x) = 10 x \ln (| x |)$ .

$(\frac{1}{e}, - \frac{10}{e})$ is a local minima

$(- \frac{1}{e}, \frac{10}{e})$ is a local maxima

Step 17