Calculus Examples

$f (x) = x + | 2 x |$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

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

$\frac{d}{d x} [x] + \frac{d}{d x} [| 2 x |]$

Step 1.1.2

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

$1 + \frac{d}{d x} [| 2 x |]$

Step 1.2

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

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Step 1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = 2 x$ .

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Step 1.2.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$1 + \frac{d}{d u} [| u |] \frac{d}{d x} [2 x]$

Step 1.2.1.2

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

$1 + \frac{u}{| u |} \frac{d}{d x} [2 x]$

Step 1.2.1.3

Replace all occurrences of $u$ with $2 x$ .

$1 + \frac{2 x}{| 2 x |} \frac{d}{d x} [2 x]$

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Multiply $2$ by $1$ .

$1 + \frac{2 x}{| 2 x |} \cdot 2$

Step 1.2.5

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

$1 + \frac{2 x \cdot 2}{| 2 x |}$

Step 1.2.6

Multiply $2$ by $2$ .

$1 + \frac{4 x}{| 2 x |}$

Step 1.3

Simplify.

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Step 1.3.1

Reorder terms.

$\frac{4 x}{| 2 x |} + 1$

Step 1.3.2

Simplify each term.

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Step 1.3.2.1

Remove non-negative terms from the absolute value.

$\frac{4 x}{2 | x |} + 1$

Step 1.3.2.2

Cancel the common factor of $4$ and $2$ .

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Step 1.3.2.2.1

Factor $2$ out of $4 x$ .

$\frac{2 (2 x)}{2 | x |} + 1$

Step 1.3.2.2.2

Cancel the common factors.

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Step 1.3.2.2.2.1

Factor $2$ out of $2 | x |$ .

$\frac{2 (2 x)}{2 (| x |)} + 1$

Step 1.3.2.2.2.2

Cancel the common factor.

$\frac{2 (2 x)}{2 | x |} + 1$

Step 1.3.2.2.2.3

Rewrite the expression.

$\frac{2 x}{| x |} + 1$

Step 2

Find the second derivative of the function.

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Step 2.1

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

$f'' (x) = \frac{d}{d x} (\frac{2 x}{| x |}) + \frac{d}{d x} (1)$

Step 2.2

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

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Step 2.2.1

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

$f'' (x) = 2 \frac{d}{d x} (\frac{x}{| x |}) + \frac{d}{d x} (1)$

Step 2.2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = | x |$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $| x |$ by $1$ .

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

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

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

Step 2.2.10

Add $1$ and $1$ .

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

Step 2.2.11

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

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

Step 2.3

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

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $- 1$ by $2$ .

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

Step 2.4.2.2

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

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

Step 2.4.2.3

Move the negative in front of the fraction.

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

Step 2.4.2.4

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

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

Step 2.4.3

Reorder terms.

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

Step 2.4.4

Simplify the numerator.

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Step 2.4.4.1

Factor $2$ out of $- \frac{2 x^{2}}{| x |} + 2 | x |$ .

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Step 2.4.4.1.1

Factor $2$ out of $- \frac{2 x^{2}}{| x |}$ .

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

Step 2.4.4.1.2

Factor $2$ out of $2 | x |$ .

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

Step 2.4.4.1.3

Factor $2$ out of $2 (- \frac{x^{2}}{| x |}) + 2 (| x |)$ .

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

Step 2.4.4.2

To write $| x |$ as a fraction with a common denominator, multiply by $\frac{| x |}{| x |}$ .

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

Step 2.4.4.3

Combine the numerators over the common denominator.

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

Step 2.4.4.4

Simplify the numerator.

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Step 2.4.4.4.1

Multiply $| x | | x |$ .

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Step 2.4.4.4.1.1

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

$f'' (x) = \frac{2 (\frac{- x^{2} + | x \cdot x |}{| x |})}{{| x |}^{2}}$

Step 2.4.4.4.1.2

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

$f'' (x) = \frac{2 (\frac{- x^{2} + | x \cdot x |}{| x |})}{{| x |}^{2}}$

Step 2.4.4.4.1.3

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

$f'' (x) = \frac{2 (\frac{- x^{2} + | x \cdot x |}{| x |})}{{| x |}^{2}}$

Step 2.4.4.4.1.4

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

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

Step 2.4.4.4.1.5

Add $1$ and $1$ .

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

Step 2.4.4.4.2

Remove non-negative terms from the absolute value.

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

Step 2.4.4.4.3

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

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

Step 2.4.4.5

Divide $0$ by $| x |$ .

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

Step 2.4.5

Remove the absolute value in ${| x |}^{2}$ because exponentiations with even powers are always positive.

$f'' (x) = \frac{2 \cdot 0}{x^{2}}$

Step 2.4.6

Multiply $2$ by $0$ .

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

Step 2.4.7

Divide $0$ by $x^{2}$ .

$f'' (x) = 0$

Step 3

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

$\frac{2 x}{| x |} + 1 = 0$

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

Differentiate.

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Step 4.1.1.1

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

$\frac{d}{d x} [x] + \frac{d}{d x} [| 2 x |]$

Step 4.1.1.2

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

$1 + \frac{d}{d x} [| 2 x |]$

Step 4.1.2

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

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Step 4.1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = 2 x$ .

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Step 4.1.2.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$1 + \frac{d}{d u} [| u |] \frac{d}{d x} [2 x]$

Step 4.1.2.1.2

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

$1 + \frac{u}{| u |} \frac{d}{d x} [2 x]$

Step 4.1.2.1.3

Replace all occurrences of $u$ with $2 x$ .

$1 + \frac{2 x}{| 2 x |} \frac{d}{d x} [2 x]$

Step 4.1.2.2

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

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

Step 4.1.2.3

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

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

Step 4.1.2.4

Multiply $2$ by $1$ .

$1 + \frac{2 x}{| 2 x |} \cdot 2$

Step 4.1.2.5

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

$1 + \frac{2 x \cdot 2}{| 2 x |}$

Step 4.1.2.6

Multiply $2$ by $2$ .

$1 + \frac{4 x}{| 2 x |}$

Step 4.1.3

Simplify.

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Step 4.1.3.1

Reorder terms.

$\frac{4 x}{| 2 x |} + 1$

Step 4.1.3.2

Simplify each term.

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Step 4.1.3.2.1

Remove non-negative terms from the absolute value.

$\frac{4 x}{2 | x |} + 1$

Step 4.1.3.2.2

Cancel the common factor of $4$ and $2$ .

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Step 4.1.3.2.2.1

Factor $2$ out of $4 x$ .

$\frac{2 (2 x)}{2 | x |} + 1$

Step 4.1.3.2.2.2

Cancel the common factors.

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Step 4.1.3.2.2.2.1

Factor $2$ out of $2 | x |$ .

$\frac{2 (2 x)}{2 (| x |)} + 1$

Step 4.1.3.2.2.2.2

Cancel the common factor.

$\frac{2 (2 x)}{2 | x |} + 1$

Step 4.1.3.2.2.2.3

Rewrite the expression.

$f' (x) = \frac{2 x}{| x |} + 1$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{2 x}{| x |} + 1$ .

$\frac{2 x}{| x |} + 1$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{2 x}{| x |} + 1 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{2 x}{| x |} + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

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

Step 5.3

Find the LCD of the terms in the equation.

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Step 5.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$| x |, 1$

Step 5.3.2

The LCM of one and any expression is the expression.

$| x |$

Step 5.4

Multiply each term in $\frac{2 x}{| x |} = - 1$ by $| x |$ to eliminate the fractions.

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Step 5.4.1

Multiply each term in $\frac{2 x}{| x |} = - 1$ by $| x |$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $| x |$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

$2 x = - | x |$

Step 5.5

Solve the equation.

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Step 5.5.1

Rewrite the equation as $- | x | = 2 x$ .

$- | x | = 2 x$

Step 5.5.2

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

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Step 5.5.2.1

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

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

Step 5.5.2.2

Simplify the left side.

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Step 5.5.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.5.2.2.2

Divide $| x |$ by $1$ .

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

Step 5.5.2.3

Simplify the right side.

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Step 5.5.2.3.1

Move the negative one from the denominator of $\frac{2 x}{- 1}$ .

$| x | = - 1 \cdot (2 x)$

Step 5.5.2.3.2

Rewrite $- 1 \cdot (2 x)$ as $- (2 x)$ .

$| x | = - (2 x)$

Step 5.5.2.3.3

Multiply $2$ by $- 1$ .

$| x | = - 2 x$

Step 5.6

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

$x = \pm (- 2 x)$

Step 5.7

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

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Step 5.7.1

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

$x = - 2 x$

Step 5.7.2

Move all terms containing $x$ to the left side of the equation.

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Step 5.7.2.1

Add $2 x$ to both sides of the equation.

$x + 2 x = 0$

Step 5.7.2.2

Add $x$ and $2 x$ .

$3 x = 0$

Step 5.7.3

Divide each term in $3 x = 0$ by $3$ and simplify.

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Step 5.7.3.1

Divide each term in $3 x = 0$ by $3$ .

$\frac{3 x}{3} = \frac{0}{3}$

Step 5.7.3.2

Simplify the left side.

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Step 5.7.3.2.1

Cancel the common factor of $3$ .

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Step 5.7.3.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{0}{3}$

Step 5.7.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{3}$

Step 5.7.3.3

Simplify the right side.

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Step 5.7.3.3.1

Divide $0$ by $3$ .

$x = 0$

Step 5.7.4

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

$x = 2 x$

Step 5.7.5

Move all terms containing $x$ to the left side of the equation.

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Step 5.7.5.1

Subtract $2 x$ from both sides of the equation.

$x - 2 x = 0$

Step 5.7.5.2

Subtract $2 x$ from $x$ .

$- x = 0$

Step 5.7.6

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

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Step 5.7.6.1

Divide each term in $- x = 0$ by $- 1$ .

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

Step 5.7.6.2

Simplify the left side.

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Step 5.7.6.2.1

Dividing two negative values results in a positive value.

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

Step 5.7.6.2.2

Divide $x$ by $1$ .

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

Step 5.7.6.3

Simplify the right side.

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Step 5.7.6.3.1

Divide $0$ by $- 1$ .

$x = 0$

Step 5.8

Exclude the solutions that do not make $\frac{2 x}{| x |} + 1 = 0$ true.

$No solution$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{2 x}{| x |}$ equal to $0$ to find where the expression is undefined.

$| x | = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

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

$x = \pm 0$

Step 6.2.2

Plus or minus $0$ is $0$ .

$x = 0$

Step 7

Critical points to evaluate.

$x = 0$

Step 8

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

$0$

Step 9

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

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Step 9.1

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

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

Step 9.2

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

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Step 9.2.1

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

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

Step 9.2.2

Simplify the result.

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Step 9.2.2.1

Multiply $2$ by $- 2$ .

$f' (- 2) = \frac{- 4}{| - 2 |} + 1$

Step 9.2.2.2

Simplify each term.

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Step 9.2.2.2.1

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

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

Step 9.2.2.2.2

Divide $- 4$ by $2$ .

$f' (- 2) = - 2 + 1$

Step 9.2.2.3

Add $- 2$ and $1$ .

$f' (- 2) = - 1$

Step 9.2.2.4

The final answer is $- 1$ .

$- 1$

Step 9.3

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

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Step 9.3.1

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

$f' (2) = \frac{2 (2)}{| 2 |} + 1$

Step 9.3.2

Simplify the result.

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Step 9.3.2.1

Multiply $2$ by $2$ .

$f' (2) = \frac{4}{| 2 |} + 1$

Step 9.3.2.2

Simplify each term.

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Step 9.3.2.2.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f' (2) = \frac{4}{2} + 1$

Step 9.3.2.2.2

Divide $4$ by $2$ .

$f' (2) = 2 + 1$

Step 9.3.2.3

Add $2$ and $1$ .

$f' (2) = 3$

Step 9.3.2.4

The final answer is $3$ .

$3$

Step 9.4

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

$x = 0$ is a local minimum

Step 10