Precalculus Examples

$| 8 x + 8 |$

Step 1

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

$8 x + 8 \geq 0$

Step 2

Solve the inequality.

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Step 2.1

Subtract $8$ from both sides of the inequality.

$8 x \geq - 8$

Step 2.2

Divide each term in $8 x \geq - 8$ by $8$ and simplify.

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Step 2.2.1

Divide each term in $8 x \geq - 8$ by $8$ .

$\frac{8 x}{8} \geq \frac{- 8}{8}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $8$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{8 x}{8} \geq \frac{- 8}{8}$

Step 2.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{- 8}{8}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 8$ by $8$ .

$x \geq - 1$

Step 3

In the piece where $8 x + 8$ is non-negative, remove the absolute value.

$8 x + 8$

Step 4

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

$8 x + 8 < 0$

Step 5

Solve the inequality.

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Step 5.1

Subtract $8$ from both sides of the inequality.

$8 x < - 8$

Step 5.2

Divide each term in $8 x < - 8$ by $8$ and simplify.

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Step 5.2.1

Divide each term in $8 x < - 8$ by $8$ .

$\frac{8 x}{8} < \frac{- 8}{8}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $8$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{8 x}{8} < \frac{- 8}{8}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x < \frac{- 8}{8}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide $- 8$ by $8$ .

$x < - 1$

Step 6

In the piece where $8 x + 8$ is negative, remove the absolute value and multiply by $- 1$ .

$- (8 x + 8)$

Step 7

Write as a piecewise.

${\begin{cases} 8 x + 8 & x \geq - 1 \\ - (8 x + 8) & x < - 1 \end{cases}$

Step 8

Simplify $- (8 x + 8)$ .

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Step 8.1

Apply the distributive property.

${\begin{cases} 8 x + 8 & x \geq - 1 \\ - (8 x) - 1 \cdot 8 & x < - 1 \end{cases}$

Step 8.2

Multiply $8$ by $- 1$ .

${\begin{cases} 8 x + 8 & x \geq - 1 \\ - 8 x - 1 \cdot 8 & x < - 1 \end{cases}$

Step 8.3

Multiply $- 1$ by $8$ .

${\begin{cases} 8 x + 8 & x \geq - 1 \\ - 8 x - 8 & x < - 1 \end{cases}$

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