Calculus Examples

$f (x) = | x | - 3 | x + 1 |$ , $[- 2, 2]$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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

$\frac{d}{d x} [| x |] + \frac{d}{d x} [- 3 | x + 1 |]$

Step 1.1.1.2

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

$\frac{x}{| x |} + \frac{d}{d x} [- 3 | x + 1 |]$

Step 1.1.1.3

Evaluate $\frac{d}{d x} [- 3 | x + 1 |]$ .

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Step 1.1.1.3.1

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

$\frac{x}{| x |} - 3 \frac{d}{d x} [| x + 1 |]$

Step 1.1.1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = x + 1$ .

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Step 1.1.1.3.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$\frac{x}{| x |} - 3 (\frac{d}{d u} [| u |] \frac{d}{d x} [x + 1])$

Step 1.1.1.3.2.2

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

$\frac{x}{| x |} - 3 (\frac{u}{| u |} \frac{d}{d x} [x + 1])$

Step 1.1.1.3.2.3

Replace all occurrences of $u$ with $x + 1$ .

$\frac{x}{| x |} - 3 (\frac{x + 1}{| x + 1 |} \frac{d}{d x} [x + 1])$

Step 1.1.1.3.3

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

$\frac{x}{| x |} - 3 (\frac{x + 1}{| x + 1 |} (\frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 1.1.1.3.4

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

$\frac{x}{| x |} - 3 (\frac{x + 1}{| x + 1 |} (1 + \frac{d}{d x} [1]))$

Step 1.1.1.3.5

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

$\frac{x}{| x |} - 3 (\frac{x + 1}{| x + 1 |} (1 + 0))$

Step 1.1.1.3.6

Add $1$ and $0$ .

$\frac{x}{| x |} - 3 (\frac{x + 1}{| x + 1 |} \cdot 1)$

Step 1.1.1.3.7

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

$\frac{x}{| x |} - 3 \frac{x + 1}{| x + 1 |}$

Step 1.1.1.3.8

Combine $- 3$ and $\frac{x + 1}{| x + 1 |}$ .

$\frac{x}{| x |} + \frac{- 3 (x + 1)}{| x + 1 |}$

Step 1.1.1.3.9

Move the negative in front of the fraction.

$f' (x) = \frac{x}{| x |} - \frac{3 (x + 1)}{| x + 1 |}$

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x}{| x |} - \frac{3 (x + 1)}{| x + 1 |}$ .

$\frac{x}{| x |} - \frac{3 (x + 1)}{| x + 1 |}$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $\frac{x}{| x |} - \frac{3 (x + 1)}{| x + 1 |} = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$\frac{x}{| x |} - \frac{3 (x + 1)}{| x + 1 |} = 0$

Step 1.2.2

Subtract $\frac{x}{| x |}$ from both sides of the equation.

$- \frac{3 (x + 1)}{| x + 1 |} = - \frac{x}{| x |}$

Step 1.2.3

Find the LCD of the terms in the equation.

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Step 1.2.3.1

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

$| x + 1 |, | x |$

Step 1.2.3.2

Since $| x + 1 |, | x |$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1$ then find LCM for the variable part ${| x + 1 |}^{1}, {| x |}^{1}$ .

Step 1.2.3.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 1.2.3.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 1.2.3.5

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 1.2.3.6

The factor for ${| x + 1 |}^{1}$ is $| x + 1 |$ itself.

${| x + 1 |}^{1} = | x + 1 |$

$| x + 1 |$ occurs $1$ time.

Step 1.2.3.7

The factor for ${| x |}^{1}$ is $| x |$ itself.

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

$| x |$ occurs $1$ time.

Step 1.2.3.8

The LCM of ${| x + 1 |}^{1}, {| x |}^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$| x + 1 | | x |$

Step 1.2.4

Multiply each term in $- \frac{3 (x + 1)}{| x + 1 |} = - \frac{x}{| x |}$ by $| x + 1 | | x |$ to eliminate the fractions.

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Step 1.2.4.1

Multiply each term in $- \frac{3 (x + 1)}{| x + 1 |} = - \frac{x}{| x |}$ by $| x + 1 | | x |$ .

$- \frac{3 (x + 1)}{| x + 1 |} (| x + 1 | | x |) = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2

Simplify the left side.

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Step 1.2.4.2.1

Cancel the common factor of $| x + 1 |$ .

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Step 1.2.4.2.1.1

Move the leading negative in $- \frac{3 (x + 1)}{| x + 1 |}$ into the numerator.

$\frac{- 3 (x + 1)}{| x + 1 |} (| x + 1 | | x |) = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2.1.2

Factor $| x + 1 |$ out of $| x + 1 | | x |$ .

$\frac{- 3 (x + 1)}{| x + 1 |} (| x + 1 | | x |) = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2.1.3

Cancel the common factor.

$\frac{- 3 (x + 1)}{| x + 1 |} (| x + 1 | | x |) = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2.1.4

Rewrite the expression.

$- 3 (x + 1) | x | = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2.2

Apply the distributive property.

$(- 3 x - 3 \cdot 1) | x | = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2.3

Multiply $- 3$ by $1$ .

$(- 3 x - 3) | x | = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.2.4

Apply the distributive property.

$- 3 x | x | - 3 | x | = - \frac{x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.3

Simplify the right side.

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Step 1.2.4.3.1

Cancel the common factor of $| x |$ .

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Step 1.2.4.3.1.1

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

$- 3 x | x | - 3 | x | = \frac{- x}{| x |} (| x + 1 | | x |)$

Step 1.2.4.3.1.2

Factor $| x |$ out of $| x + 1 | | x |$ .

$- 3 x | x | - 3 | x | = \frac{- x}{| x |} (| x | | x + 1 |)$

Step 1.2.4.3.1.3

Cancel the common factor.

$- 3 x | x | - 3 | x | = \frac{- x}{| x |} (| x | | x + 1 |)$

Step 1.2.4.3.1.4

Rewrite the expression.

$- 3 x | x | - 3 | x | = - x | x + 1 |$

Step 1.2.5

Solve the equation.

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Step 1.2.5.1

Rewrite the equation as $- x | x + 1 | = - 3 x | x | - 3 | x |$ .

$- x | x + 1 | = - 3 x | x | - 3 | x |$

Step 1.2.5.2

Divide each term in $- x | x + 1 | = - 3 x | x | - 3 | x |$ by $- x$ and simplify.

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Step 1.2.5.2.1

Divide each term in $- x | x + 1 | = - 3 x | x | - 3 | x |$ by $- x$ .

$\frac{- x | x + 1 |}{- x} = \frac{- 3 x | x |}{- x} + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.2

Simplify the left side.

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Step 1.2.5.2.2.1

Dividing two negative values results in a positive value.

$\frac{x | x + 1 |}{x} = \frac{- 3 x | x |}{- x} + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.2.2

Cancel the common factor of $x$ .

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Step 1.2.5.2.2.2.1

Cancel the common factor.

$\frac{x | x + 1 |}{x} = \frac{- 3 x | x |}{- x} + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.2.2.2

Divide $| x + 1 |$ by $1$ .

$| x + 1 | = \frac{- 3 x | x |}{- x} + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.3

Simplify the right side.

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Step 1.2.5.2.3.1

Simplify each term.

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Step 1.2.5.2.3.1.1

Cancel the common factor of $x$ .

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Step 1.2.5.2.3.1.1.1

Cancel the common factor.

$| x + 1 | = \frac{- 3 x | x |}{- x} + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.3.1.1.2

Rewrite the expression.

$| x + 1 | = \frac{- 3 | x |}{- 1} + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.3.1.1.3

Move the negative one from the denominator of $\frac{- 3 | x |}{- 1}$ .

$| x + 1 | = - 1 \cdot (- 3 | x |) + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.3.1.2

Rewrite $- 1 \cdot (- 3 | x |)$ as $- (- 3 | x |)$ .

$| x + 1 | = - (- 3 | x |) + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.3.1.3

Multiply $- 3$ by $- 1$ .

$| x + 1 | = 3 | x | + \frac{- 3 | x |}{- x}$

Step 1.2.5.2.3.1.4

Dividing two negative values results in a positive value.

$| x + 1 | = 3 | x | + \frac{3 | x |}{x}$

Step 1.2.6

Solve for $| x |$ .

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Step 1.2.6.1

Rewrite the equation as $3 | x | + \frac{3 | x |}{x} = | x + 1 |$ .

$3 | x | + \frac{3 | x |}{x} = | x + 1 |$

Step 1.2.6.2

Find the LCD of the terms in the equation.

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Step 1.2.6.2.1

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

$1, x, 1$

Step 1.2.6.2.2

The LCM of one and any expression is the expression.

$x$

Step 1.2.6.3

Multiply each term in $3 | x | + \frac{3 | x |}{x} = | x + 1 |$ by $x$ to eliminate the fractions.

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Step 1.2.6.3.1

Multiply each term in $3 | x | + \frac{3 | x |}{x} = | x + 1 |$ by $x$ .

$3 | x | x + \frac{3 | x |}{x} x = | x + 1 | x$

Step 1.2.6.3.2

Simplify the left side.

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Step 1.2.6.3.2.1

Cancel the common factor of $x$ .

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Step 1.2.6.3.2.1.1

Cancel the common factor.

$3 | x | x + \frac{3 | x |}{x} x = | x + 1 | x$

Step 1.2.6.3.2.1.2

Rewrite the expression.

$3 | x | x + 3 | x | = | x + 1 | x$

Step 1.2.6.4

Solve the equation.

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Step 1.2.6.4.1

Factor $3 | x |$ out of $3 | x | x + 3 | x |$ .

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Step 1.2.6.4.1.1

Factor $3 | x |$ out of $3 | x | x$ .

$3 | x | x + 3 | x | = | x + 1 | x$

Step 1.2.6.4.1.2

Factor $3 | x |$ out of $3 | x |$ .

$3 | x | x + 3 | x | \cdot 1 = | x + 1 | x$

Step 1.2.6.4.1.3

Factor $3 | x |$ out of $3 | x | x + 3 | x | \cdot 1$ .

$3 | x | (x + 1) = | x + 1 | x$

Step 1.2.6.4.2

Divide each term in $3 | x | (x + 1) = | x + 1 | x$ by $3 (x + 1)$ and simplify.

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Step 1.2.6.4.2.1

Divide each term in $3 | x | (x + 1) = | x + 1 | x$ by $3 (x + 1)$ .

$\frac{3 | x | (x + 1)}{3 (x + 1)} = \frac{| x + 1 | x}{3 (x + 1)}$

Step 1.2.6.4.2.2

Simplify the left side.

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Step 1.2.6.4.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.6.4.2.2.1.1

Cancel the common factor.

$\frac{3 | x | (x + 1)}{3 (x + 1)} = \frac{| x + 1 | x}{3 (x + 1)}$

Step 1.2.6.4.2.2.1.2

Rewrite the expression.

$\frac{| x | (x + 1)}{x + 1} = \frac{| x + 1 | x}{3 (x + 1)}$

Step 1.2.6.4.2.2.2

Cancel the common factor of $x + 1$ .

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Step 1.2.6.4.2.2.2.1

Cancel the common factor.

$\frac{| x | (x + 1)}{x + 1} = \frac{| x + 1 | x}{3 (x + 1)}$

Step 1.2.6.4.2.2.2.2

Divide $| x |$ by $1$ .

$| x | = \frac{| x + 1 | x}{3 (x + 1)}$

Step 1.2.7

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

$x = \pm (\frac{| x + 1 | x}{3 (x + 1)})$

Step 1.2.8

The result consists of both the positive and negative portions of the $\pm$ .

$x = \frac{| x + 1 | x}{3 (x + 1)}$

$x = - (\frac{| x + 1 | x}{3 (x + 1)})$

Step 1.2.9

Solve $x = \frac{| x + 1 | x}{3 (x + 1)}$ for $x$ .

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Step 1.2.9.1

Solve for $| x + 1 |$ .

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Step 1.2.9.1.1

Rewrite the equation as $\frac{| x + 1 | x}{3 (x + 1)} = x$ .

$\frac{| x + 1 | x}{3 (x + 1)} = x$

Step 1.2.9.1.2

Multiply both sides by $3 (x + 1)$ .

$\frac{| x + 1 | x}{3 (x + 1)} (3 (x + 1)) = x (3 (x + 1))$

Step 1.2.9.1.3

Simplify.

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Step 1.2.9.1.3.1

Simplify the left side.

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Step 1.2.9.1.3.1.1

Simplify $\frac{| x + 1 | x}{3 (x + 1)} (3 (x + 1))$ .

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Step 1.2.9.1.3.1.1.1

Rewrite using the commutative property of multiplication.

$3 \frac{| x + 1 | x}{3 (x + 1)} (x + 1) = x (3 (x + 1))$

Step 1.2.9.1.3.1.1.2

Cancel the common factor of $3$ .

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Step 1.2.9.1.3.1.1.2.1

Cancel the common factor.

$3 \frac{| x + 1 | x}{3 (x + 1)} (x + 1) = x (3 (x + 1))$

Step 1.2.9.1.3.1.1.2.2

Rewrite the expression.

$\frac{| x + 1 | x}{x + 1} (x + 1) = x (3 (x + 1))$

Step 1.2.9.1.3.1.1.3

Cancel the common factor of $x + 1$ .

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Step 1.2.9.1.3.1.1.3.1

Cancel the common factor.

$\frac{| x + 1 | x}{x + 1} (x + 1) = x (3 (x + 1))$

Step 1.2.9.1.3.1.1.3.2

Rewrite the expression.

$| x + 1 | x = x (3 (x + 1))$

Step 1.2.9.1.3.2

Simplify the right side.

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Step 1.2.9.1.3.2.1

Simplify $x (3 (x + 1))$ .

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Step 1.2.9.1.3.2.1.1

Rewrite using the commutative property of multiplication.

$| x + 1 | x = 3 x (x + 1)$

Step 1.2.9.1.3.2.1.2

Apply the distributive property.

$| x + 1 | x = 3 x \cdot x + 3 x \cdot 1$

Step 1.2.9.1.3.2.1.3

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.9.1.3.2.1.3.1

Move $x$ .

$| x + 1 | x = 3 (x \cdot x) + 3 x \cdot 1$

Step 1.2.9.1.3.2.1.3.2

Multiply $x$ by $x$ .

$| x + 1 | x = 3 x^{2} + 3 x \cdot 1$

Step 1.2.9.1.3.2.1.4

Multiply $3$ by $1$ .

$| x + 1 | x = 3 x^{2} + 3 x$

Step 1.2.9.1.4

Divide each term in $| x + 1 | x = 3 x^{2} + 3 x$ by $x$ and simplify.

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Step 1.2.9.1.4.1

Divide each term in $| x + 1 | x = 3 x^{2} + 3 x$ by $x$ .

$\frac{| x + 1 | x}{x} = \frac{3 x^{2}}{x} + \frac{3 x}{x}$

Step 1.2.9.1.4.2

Simplify the left side.

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Step 1.2.9.1.4.2.1

Cancel the common factor of $x$ .

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Step 1.2.9.1.4.2.1.1

Cancel the common factor.

$\frac{| x + 1 | x}{x} = \frac{3 x^{2}}{x} + \frac{3 x}{x}$

Step 1.2.9.1.4.2.1.2

Divide $| x + 1 |$ by $1$ .

$| x + 1 | = \frac{3 x^{2}}{x} + \frac{3 x}{x}$

Step 1.2.9.1.4.3

Simplify the right side.

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Step 1.2.9.1.4.3.1

Simplify each term.

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Step 1.2.9.1.4.3.1.1

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

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Step 1.2.9.1.4.3.1.1.1

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

$| x + 1 | = \frac{x (3 x)}{x} + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.1.2

Cancel the common factors.

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Step 1.2.9.1.4.3.1.1.2.1

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

$| x + 1 | = \frac{x (3 x)}{x^{1}} + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.1.2.2

Factor $x$ out of $x^{1}$ .

$| x + 1 | = \frac{x (3 x)}{x \cdot 1} + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.1.2.3

Cancel the common factor.

$| x + 1 | = \frac{x (3 x)}{x \cdot 1} + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.1.2.4

Rewrite the expression.

$| x + 1 | = \frac{3 x}{1} + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.1.2.5

Divide $3 x$ by $1$ .

$| x + 1 | = 3 x + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.2

Cancel the common factor of $x$ .

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Step 1.2.9.1.4.3.1.2.1

Cancel the common factor.

$| x + 1 | = 3 x + \frac{3 x}{x}$

Step 1.2.9.1.4.3.1.2.2

Divide $3$ by $1$ .

$| x + 1 | = 3 x + 3$

Step 1.2.9.2

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

$x + 1 = \pm (3 x + 3)$

Step 1.2.9.3

The result consists of both the positive and negative portions of the $\pm$ .

$x + 1 = 3 x + 3$

$x + 1 = - (3 x + 3)$

Step 1.2.9.4

Solve $x + 1 = 3 x + 3$ for $x$ .

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Step 1.2.9.4.1

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

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Step 1.2.9.4.1.1

Subtract $3 x$ from both sides of the equation.

$x + 1 - 3 x = 3$

Step 1.2.9.4.1.2

Subtract $3 x$ from $x$ .

$- 2 x + 1 = 3$

Step 1.2.9.4.2

Move all terms not containing $x$ to the right side of the equation.

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Step 1.2.9.4.2.1

Subtract $1$ from both sides of the equation.

$- 2 x = 3 - 1$

Step 1.2.9.4.2.2

Subtract $1$ from $3$ .

$- 2 x = 2$

Step 1.2.9.4.3

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

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Step 1.2.9.4.3.1

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

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

Step 1.2.9.4.3.2

Simplify the left side.

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Step 1.2.9.4.3.2.1

Cancel the common factor of $- 2$ .

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Step 1.2.9.4.3.2.1.1

Cancel the common factor.

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

Step 1.2.9.4.3.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.9.4.3.3

Simplify the right side.

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Step 1.2.9.4.3.3.1

Divide $2$ by $- 2$ .

$x = - 1$

Step 1.2.9.5

Consolidate the solutions.

$x = - 1$

Step 1.2.10

Find the domain of $\frac{x}{| x |} - \frac{3 (x + 1)}{| x + 1 |}$ .

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Step 1.2.10.1

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

$| x | = 0$

Step 1.2.10.2

Solve for $x$ .

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Step 1.2.10.2.1

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

$x = \pm 0$

Step 1.2.10.2.2

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.2.10.3

Set the denominator in $\frac{3 (x + 1)}{| x + 1 |}$ equal to $0$ to find where the expression is undefined.

$| x + 1 | = 0$

Step 1.2.10.4

Solve for $x$ .

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Step 1.2.10.4.1

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

$x + 1 = \pm 0$

Step 1.2.10.4.2

Plus or minus $0$ is $0$ .

$x + 1 = 0$

Step 1.2.10.4.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.10.5

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 1.2.11

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

Step 1.2.12

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.12.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

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Step 1.2.12.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 1.2.12.1.2

Replace $x$ with $- 4$ in the original inequality.

$\frac{- 4}{| - 4 |} - \frac{3 ((- 4) + 1)}{| (- 4) + 1 |} = 0$

Step 1.2.12.1.3

The left side $2$ does not equal to the right side $0$ , which means that the given statement is false.

False

Step 1.2.12.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 1.2.12.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 1.2.12.2.2

Replace $x$ with $- 0.5$ in the original inequality.

$\frac{- 0.5}{| - 0.5 |} - \frac{3 ((- 0.5) + 1)}{| (- 0.5) + 1 |} = 0$

Step 1.2.12.2.3

The left side $- 4$ does not equal to the right side $0$ , which means that the given statement is false.

False

Step 1.2.12.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 1.2.12.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 1.2.12.3.2

Replace $x$ with $2$ in the original inequality.

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

Step 1.2.12.3.3

The left side $- 2$ does not equal to the right side $0$ , which means that the given statement is false.

False

Step 1.2.12.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 1$ False

$- 1 < x < 0$ False

$x > 0$ False

$x < - 1$ False

$- 1 < x < 0$ False

$x > 0$ False

Step 1.2.13

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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

$| x | = 0$

Step 1.3.2

Solve for $x$ .

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Step 1.3.2.1

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

$x = \pm 0$

Step 1.3.2.2

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.3.3

Set the denominator in $\frac{3 (x + 1)}{| x + 1 |}$ equal to $0$ to find where the expression is undefined.

$| x + 1 | = 0$

Step 1.3.4

Solve for $x$ .

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Step 1.3.4.1

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

$x + 1 = \pm 0$

Step 1.3.4.2

Plus or minus $0$ is $0$ .

$x + 1 = 0$

Step 1.3.4.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.3.5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 1, x = 0$

Step 1.4

Evaluate $| x | - 3 | x + 1 |$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = - 1$ .

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Step 1.4.1.1

Substitute $- 1$ for $x$ .

$| - 1 | - 3 | (- 1) + 1 |$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$1 - 3 | (- 1) + 1 |$

Step 1.4.1.2.1.2

Add $- 1$ and $1$ .

$1 - 3 | 0 |$

Step 1.4.1.2.1.3

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

$1 - 3 \cdot 0$

Step 1.4.1.2.1.4

Multiply $- 3$ by $0$ .

$1 + 0$

Step 1.4.1.2.2

Add $1$ and $0$ .

$1$

Step 1.4.2

Evaluate at $x = 0$ .

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Step 1.4.2.1

Substitute $0$ for $x$ .

$| 0 | - 3 | (0) + 1 |$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

Simplify each term.

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Step 1.4.2.2.1.1

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

$0 - 3 | (0) + 1 |$

Step 1.4.2.2.1.2

Add $0$ and $1$ .

$0 - 3 | 1 |$

Step 1.4.2.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$0 - 3 \cdot 1$

Step 1.4.2.2.1.4

Multiply $- 3$ by $1$ .

$0 - 3$

Step 1.4.2.2.2

Subtract $3$ from $0$ .

$- 3$

Step 1.4.3

List all of the points.

$(- 1, 1), (0, - 3)$

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at $x = - 2$ .

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Step 2.1.1

Substitute $- 2$ for $x$ .

$| - 2 | - 3 | (- 2) + 1 |$

Step 2.1.2

Simplify.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

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

$2 - 3 | (- 2) + 1 |$

Step 2.1.2.1.2

Add $- 2$ and $1$ .

$2 - 3 | - 1 |$

Step 2.1.2.1.3

The absolute value is the distance between a number and zero. The distance between $- 1$ and $0$ is $1$ .

$2 - 3 \cdot 1$

Step 2.1.2.1.4

Multiply $- 3$ by $1$ .

$2 - 3$

Step 2.1.2.2

Subtract $3$ from $2$ .

$- 1$

Step 2.2

Evaluate at $x = 2$ .

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Step 2.2.1

Substitute $2$ for $x$ .

$| 2 | - 3 | (2) + 1 |$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

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

$2 - 3 | (2) + 1 |$

Step 2.2.2.1.2

Add $2$ and $1$ .

$2 - 3 | 3 |$

Step 2.2.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$2 - 3 \cdot 3$

Step 2.2.2.1.4

Multiply $- 3$ by $3$ .

$2 - 9$

Step 2.2.2.2

Subtract $9$ from $2$ .

$- 7$

Step 2.3

List all of the points.

$(- 2, - 1), (2, - 7)$

Step 3

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

Absolute Maximum: $(- 1, 1)$

Absolute Minimum: $(2, - 7)$

Step 4