Calculus Examples

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

Step 1

Find the critical points.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1.1

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

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

Step 1.1.1.2

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

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

Step 1.1.1.3

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

Tap for more steps...

Step 1.1.1.3.1

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

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

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 $g (x) = 1 - x$ .

Tap for more steps...

Step 1.1.1.3.2.1

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

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

Step 1.1.1.3.2.2

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

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

Step 1.1.1.3.2.3

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

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

Step 1.1.1.3.3

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

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

Step 1.1.1.3.4

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

$\frac{x}{| x |} + 2 (\frac{1 - x}{| 1 - x |} (0 + \frac{d}{d x} [- x]))$

Step 1.1.1.3.5

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

$\frac{x}{| x |} + 2 (\frac{1 - x}{| 1 - x |} (0 - \frac{d}{d x} [x]))$

Step 1.1.1.3.6

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

$\frac{x}{| x |} + 2 (\frac{1 - x}{| 1 - x |} (0 - 1 \cdot 1))$

Step 1.1.1.3.7

Multiply $- 1$ by $1$ .

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

Step 1.1.1.3.8

Subtract $1$ from $0$ .

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

Step 1.1.1.3.9

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

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

Step 1.1.1.3.10

Move $- 1$ to the left of $1 - x$ .

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

Step 1.1.1.3.11

Rewrite $- 1 (1 - x)$ as $- (1 - x)$ .

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

Step 1.1.1.3.12

Move the negative in front of the fraction.

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

Step 1.1.1.3.13

Multiply $- 1$ by $2$ .

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

Step 1.1.1.3.14

Combine $- 2$ and $\frac{1 - x}{| 1 - x |}$ .

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

Step 1.1.1.3.15

Move the negative in front of the fraction.

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

Step 1.1.2

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

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

Step 1.2

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

Tap for more steps...

Step 1.2.1

Set the first derivative equal to $0$ .

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

Step 1.2.2

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

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

Step 1.2.3

Find the LCD of the terms in the equation.

Tap for more steps...

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.

$| 1 - x |, | x |$

Step 1.2.3.2

Since $| 1 - x |, | x |$ 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 ${| 1 - x |}^{1}, {| x |}^{1}$ .

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 ${| 1 - x |}^{1}$ is $| 1 - x |$ itself.

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

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

$| 1 - x | | x |$

Step 1.2.4

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

Tap for more steps...

Step 1.2.4.1

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

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

Step 1.2.4.2

Simplify the left side.

Tap for more steps...

Step 1.2.4.2.1

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

Tap for more steps...

Step 1.2.4.2.1.1

Move the leading negative in $- \frac{2 (1 - x)}{| 1 - x |}$ into the numerator.

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

Step 1.2.4.2.1.2

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

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

Step 1.2.4.2.1.3

Cancel the common factor.

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

Step 1.2.4.2.1.4

Rewrite the expression.

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

Step 1.2.4.2.2

Apply the distributive property.

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

Step 1.2.4.2.3

Multiply.

Tap for more steps...

Step 1.2.4.2.3.1

Multiply $- 2$ by $1$ .

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

Step 1.2.4.2.3.2

Multiply $- 1$ by $- 2$ .

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

Step 1.2.4.2.4

Apply the distributive property.

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

Step 1.2.4.3

Simplify the right side.

Tap for more steps...

Step 1.2.4.3.1

Cancel the common factor of $| x |$ .

Tap for more steps...

Step 1.2.4.3.1.1

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

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

Step 1.2.4.3.1.2

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

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

Step 1.2.4.3.1.3

Cancel the common factor.

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

Step 1.2.4.3.1.4

Rewrite the expression.

$- 2 | x | + 2 x | x | = - x | 1 - x |$

Step 1.2.5

Solve the equation.

Tap for more steps...

Step 1.2.5.1

Rewrite the equation as $- x | 1 - x | = - 2 | x | + 2 x | x |$ .

$- x | 1 - x | = - 2 | x | + 2 x | x |$

Step 1.2.5.2

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

Tap for more steps...

Step 1.2.5.2.1

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

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

Step 1.2.5.2.2

Simplify the left side.

Tap for more steps...

Step 1.2.5.2.2.1

Dividing two negative values results in a positive value.

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

Step 1.2.5.2.2.2

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.2.5.2.2.2.1

Cancel the common factor.

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

Step 1.2.5.2.2.2.2

Divide $| 1 - x |$ by $1$ .

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

Step 1.2.5.2.3

Simplify the right side.

Tap for more steps...

Step 1.2.5.2.3.1

Simplify each term.

Tap for more steps...

Step 1.2.5.2.3.1.1

Dividing two negative values results in a positive value.

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

Step 1.2.5.2.3.1.2

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.2.5.2.3.1.2.1

Cancel the common factor.

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

Step 1.2.5.2.3.1.2.2

Rewrite the expression.

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

Step 1.2.5.2.3.1.2.3

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

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

Step 1.2.5.2.3.1.3

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

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

Step 1.2.5.2.3.1.4

Multiply $2$ by $- 1$ .

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

Step 1.2.6

Solve for $| x |$ .

Tap for more steps...

Step 1.2.6.1

Rewrite the equation as $\frac{2 | x |}{x} - 2 | x | = | 1 - x |$ .

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

Step 1.2.6.2

Find the LCD of the terms in the equation.

Tap for more steps...

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.

$x, 1, 1$

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

Tap for more steps...

Step 1.2.6.3.1

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

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

Step 1.2.6.3.2

Simplify the left side.

Tap for more steps...

Step 1.2.6.3.2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.2.6.3.2.1.1

Cancel the common factor.

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

Step 1.2.6.3.2.1.2

Rewrite the expression.

$2 | x | - 2 | x | x = | 1 - x | x$

Step 1.2.6.4

Solve the equation.

Tap for more steps...

Step 1.2.6.4.1

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

Tap for more steps...

Step 1.2.6.4.1.1

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

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

Step 1.2.6.4.1.2

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

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

Step 1.2.6.4.1.3

Factor $2 | x |$ out of $2 | x | \cdot 1 + 2 | x | (- 1 x)$ .

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

Step 1.2.6.4.2

Rewrite $- 1 x$ as $- x$ .

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

Step 1.2.6.4.3

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

Tap for more steps...

Step 1.2.6.4.3.1

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

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

Step 1.2.6.4.3.2

Simplify the left side.

Tap for more steps...

Step 1.2.6.4.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.2.6.4.3.2.1.1

Cancel the common factor.

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

Step 1.2.6.4.3.2.1.2

Rewrite the expression.

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

Step 1.2.6.4.3.2.2

Cancel the common factor of $1 - x$ .

Tap for more steps...

Step 1.2.6.4.3.2.2.1

Cancel the common factor.

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

Step 1.2.6.4.3.2.2.2

Divide $| x |$ by $1$ .

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

Step 1.2.7

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

$x = \pm (\frac{| 1 - x | x}{2 (1 - x)})$

Step 1.2.8

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

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

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

Step 1.2.9

Solve $x = \frac{| 1 - x | x}{2 (1 - x)}$ for $x$ .

Tap for more steps...

Step 1.2.9.1

Solve for $| 1 - x |$ .

Tap for more steps...

Step 1.2.9.1.1

Rewrite the equation as $\frac{| 1 - x | x}{2 (1 - x)} = x$ .

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

Step 1.2.9.1.2

Multiply both sides by $2 (1 - x)$ .

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

Step 1.2.9.1.3

Simplify.

Tap for more steps...

Step 1.2.9.1.3.1

Simplify the left side.

Tap for more steps...

Step 1.2.9.1.3.1.1

Simplify $\frac{| 1 - x | x}{2 (1 - x)} (2 (1 - x))$ .

Tap for more steps...

Step 1.2.9.1.3.1.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.2.9.1.3.1.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.2.9.1.3.1.1.2.1

Cancel the common factor.

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

Step 1.2.9.1.3.1.1.2.2

Rewrite the expression.

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

Step 1.2.9.1.3.1.1.3

Cancel the common factor of $1 - x$ .

Tap for more steps...

Step 1.2.9.1.3.1.1.3.1

Cancel the common factor.

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

Step 1.2.9.1.3.1.1.3.2

Rewrite the expression.

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

Step 1.2.9.1.3.2

Simplify the right side.

Tap for more steps...

Step 1.2.9.1.3.2.1

Simplify $x (2 (1 - x))$ .

Tap for more steps...

Step 1.2.9.1.3.2.1.1

Simplify by multiplying through.

Tap for more steps...

Step 1.2.9.1.3.2.1.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.2.9.1.3.2.1.1.2

Apply the distributive property.

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

Step 1.2.9.1.3.2.1.1.3

Simplify the expression.

Tap for more steps...

Step 1.2.9.1.3.2.1.1.3.1

Multiply $2$ by $1$ .

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

Step 1.2.9.1.3.2.1.1.3.2

Rewrite using the commutative property of multiplication.

$| 1 - x | x = 2 x + 2 \cdot - 1 x \cdot x$

Step 1.2.9.1.3.2.1.2

Simplify each term.

Tap for more steps...

Step 1.2.9.1.3.2.1.2.1

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

Tap for more steps...

Step 1.2.9.1.3.2.1.2.1.1

Move $x$ .

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

Step 1.2.9.1.3.2.1.2.1.2

Multiply $x$ by $x$ .

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

Step 1.2.9.1.3.2.1.2.2

Multiply $2$ by $- 1$ .

$| 1 - x | x = 2 x - 2 x^{2}$

Step 1.2.9.1.3.2.1.3

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

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

Step 1.2.9.1.4

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

Tap for more steps...

Step 1.2.9.1.4.1

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

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

Step 1.2.9.1.4.2

Simplify the left side.

Tap for more steps...

Step 1.2.9.1.4.2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.2.9.1.4.2.1.1

Cancel the common factor.

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

Step 1.2.9.1.4.2.1.2

Divide $| 1 - x |$ by $1$ .

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

Step 1.2.9.1.4.3

Simplify the right side.

Tap for more steps...

Step 1.2.9.1.4.3.1

Simplify each term.

Tap for more steps...

Step 1.2.9.1.4.3.1.1

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

Tap for more steps...

Step 1.2.9.1.4.3.1.1.1

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

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

Step 1.2.9.1.4.3.1.1.2

Cancel the common factors.

Tap for more steps...

Step 1.2.9.1.4.3.1.1.2.1

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

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

Step 1.2.9.1.4.3.1.1.2.2

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

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

Step 1.2.9.1.4.3.1.1.2.3

Cancel the common factor.

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

Step 1.2.9.1.4.3.1.1.2.4

Rewrite the expression.

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

Step 1.2.9.1.4.3.1.1.2.5

Divide $- 2 x$ by $1$ .

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

Step 1.2.9.1.4.3.1.2

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.2.9.1.4.3.1.2.1

Cancel the common factor.

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

Step 1.2.9.1.4.3.1.2.2

Divide $2$ by $1$ .

$| 1 - x | = - 2 x + 2$

Step 1.2.9.2

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

$1 - x = \pm (- 2 x + 2)$

Step 1.2.9.3

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

$1 - x = - 2 x + 2$

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

Step 1.2.9.4

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

Tap for more steps...

Step 1.2.9.4.1

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

Tap for more steps...

Step 1.2.9.4.1.1

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

$1 - x + 2 x = 2$

Step 1.2.9.4.1.2

Add $- x$ and $2 x$ .

$1 + x = 2$

Step 1.2.9.4.2

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

Tap for more steps...

Step 1.2.9.4.2.1

Subtract $1$ from both sides of the equation.

$x = 2 - 1$

Step 1.2.9.4.2.2

Subtract $1$ from $2$ .

$x = 1$

Step 1.2.9.5

Consolidate the solutions.

$x = 1$

Step 1.2.10

Find the domain of $\frac{x}{| x |} - \frac{2 (1 - x)}{| 1 - x |}$ .

Tap for more steps...

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 $x$ .

Tap for more steps...

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 $\frac{2 (1 - x)}{| 1 - x |}$ equal to $0$ to find where the expression is undefined.

$| 1 - x | = 0$

Step 1.2.10.4

Solve for $x$ .

Tap for more steps...

Step 1.2.10.4.1

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

$1 - x = \pm 0$

Step 1.2.10.4.2

Plus or minus $0$ is $0$ .

$1 - x = 0$

Step 1.2.10.4.3

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 1.2.10.4.4

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

Tap for more steps...

Step 1.2.10.4.4.1

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

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

Step 1.2.10.4.4.2

Simplify the left side.

Tap for more steps...

Step 1.2.10.4.4.2.1

Dividing two negative values results in a positive value.

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

Step 1.2.10.4.4.2.2

Divide $x$ by $1$ .

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

Step 1.2.10.4.4.3

Simplify the right side.

Tap for more steps...

Step 1.2.10.4.4.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 1.2.10.5

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

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

Step 1.2.11

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

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.

Tap for more steps...

Step 1.2.12.1

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

Tap for more steps...

Step 1.2.12.1.1

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

$x = - 2$

Step 1.2.12.1.2

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

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

Step 1.2.12.1.3

The left side $- 3$ 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 $0 < x < 1$ to see if it makes the inequality true.

Tap for more steps...

Step 1.2.12.2.1

Choose a value on the interval $0 < x < 1$ 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.

$\frac{0.5}{| 0.5 |} - \frac{2 (1 - (0.5))}{| 1 - (0.5) |} = 0$

Step 1.2.12.2.3

The left side $- 1$ 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 $x > 1$ to see if it makes the inequality true.

Tap for more steps...

Step 1.2.12.3.1

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

$x = 4$

Step 1.2.12.3.2

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

$\frac{4}{| 4 |} - \frac{2 (1 - (4))}{| 1 - (4) |} = 0$

Step 1.2.12.3.3

The left side $3$ 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.

$x < 0$ False

$0 < x < 1$ False

$x > 1$ False

$x < 0$ False

$0 < x < 1$ False

$x > 1$ 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.

Tap for more steps...

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 $x$ .

Tap for more steps...

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 $\frac{2 (1 - x)}{| 1 - x |}$ equal to $0$ to find where the expression is undefined.

$| 1 - x | = 0$

Step 1.3.4

Solve for $x$ .

Tap for more steps...

Step 1.3.4.1

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

$1 - x = \pm 0$

Step 1.3.4.2

Plus or minus $0$ is $0$ .

$1 - x = 0$

Step 1.3.4.3

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 1.3.4.4

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

Tap for more steps...

Step 1.3.4.4.1

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

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

Step 1.3.4.4.2

Simplify the left side.

Tap for more steps...

Step 1.3.4.4.2.1

Dividing two negative values results in a positive value.

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

Step 1.3.4.4.2.2

Divide $x$ by $1$ .

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

Step 1.3.4.4.3

Simplify the right side.

Tap for more steps...

Step 1.3.4.4.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

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 $0$ .

$x = 0, x = 1$

Step 1.4

Evaluate $| x | + 2 | 1 - x |$ at each $x$ value where the derivative is $0$ or undefined.

Tap for more steps...

Step 1.4.1

Evaluate at $x = 0$ .

Tap for more steps...

Step 1.4.1.1

Substitute $0$ for $x$ .

$| 0 | + 2 | 1 - (0) |$

Step 1.4.1.2

Simplify.

Tap for more steps...

Step 1.4.1.2.1

Simplify each term.

Tap for more steps...

Step 1.4.1.2.1.1

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

$0 + 2 | 1 - (0) |$

Step 1.4.1.2.1.2

Subtract $0$ from $1$ .

$0 + 2 | 1 |$

Step 1.4.1.2.1.3

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

$0 + 2 \cdot 1$

Step 1.4.1.2.1.4

Multiply $2$ by $1$ .

$0 + 2$

Step 1.4.1.2.2

Add $0$ and $2$ .

$2$

Step 1.4.2

Evaluate at $x = 1$ .

Tap for more steps...

Step 1.4.2.1

Substitute $1$ for $x$ .

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

Step 1.4.2.2

Simplify.

Tap for more steps...

Step 1.4.2.2.1

Simplify each term.

Tap for more steps...

Step 1.4.2.2.1.1

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

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

Step 1.4.2.2.1.2

Multiply $- 1$ by $1$ .

$1 + 2 | 1 - 1 |$

Step 1.4.2.2.1.3

Subtract $1$ from $1$ .

$1 + 2 | 0 |$

Step 1.4.2.2.1.4

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

$1 + 2 \cdot 0$

Step 1.4.2.2.1.5

Multiply $2$ by $0$ .

$1 + 0$

Step 1.4.2.2.2

Add $1$ and $0$ .

$1$

Step 1.4.3

List all of the points.

$(0, 2), (1, 1)$

Step 2

Evaluate at the included endpoints.

Tap for more steps...

Step 2.1

Evaluate at $x = - 2$ .

Tap for more steps...

Step 2.1.1

Substitute $- 2$ for $x$ .

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

Step 2.1.2

Simplify.

Tap for more steps...

Step 2.1.2.1

Simplify each term.

Tap for more steps...

Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

Multiply $- 1$ by $- 2$ .

$2 + 2 | 1 + 2 |$

Step 2.1.2.1.3

Add $1$ and $2$ .

$2 + 2 | 3 |$

Step 2.1.2.1.4

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

$2 + 2 \cdot 3$

Step 2.1.2.1.5

Multiply $2$ by $3$ .

$2 + 6$

Step 2.1.2.2

Add $2$ and $6$ .

$8$

Step 2.2

Evaluate at $x = 2$ .

Tap for more steps...

Step 2.2.1

Substitute $2$ for $x$ .

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

Step 2.2.2

Simplify.

Tap for more steps...

Step 2.2.2.1

Simplify each term.

Tap for more steps...

Step 2.2.2.1.1

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

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

Step 2.2.2.1.2

Multiply $- 1$ by $2$ .

$2 + 2 | 1 - 2 |$

Step 2.2.2.1.3

Subtract $2$ from $1$ .

$2 + 2 | - 1 |$

Step 2.2.2.1.4

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

$2 + 2 \cdot 1$

Step 2.2.2.1.5

Multiply $2$ by $1$ .

$2 + 2$

Step 2.2.2.2

Add $2$ and $2$ .

$4$

Step 2.3

List all of the points.

$(- 2, 8), (2, 4)$

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: $(- 2, 8)$

Absolute Minimum: $(1, 1)$

Step 4