Precalculus Examples

$g (x) = \ln (| x + 1 |)$

Step 1

Set the argument in $\ln (| x + 1 |)$ greater than $0$ to find where the expression is defined.

$| x + 1 | > 0$

Step 2

Solve for $x$ .

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Step 2.1

Write $| x + 1 | > 0$ as a piecewise.

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Step 2.1.1

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

$x + 1 \geq 0$

Step 2.1.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 2.1.3

In the piece where $x + 1$ is non-negative, remove the absolute value.

$x + 1 > 0$

Step 2.1.4

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

$x + 1 < 0$

Step 2.1.5

Subtract $1$ from both sides of the inequality.

$x < - 1$

Step 2.1.6

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

$- (x + 1) > 0$

Step 2.1.7

Write as a piecewise.

${\begin{cases} x + 1 > 0 & x \geq - 1 \\ - (x + 1) > 0 & x < - 1 \end{cases}$

Step 2.1.8

Simplify $- (x + 1) > 0$ .

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Step 2.1.8.1

Apply the distributive property.

${\begin{cases} x + 1 > 0 & x \geq - 1 \\ - x - 1 \cdot 1 > 0 & x < - 1 \end{cases}$

Step 2.1.8.2

Multiply $- 1$ by $1$ .

${\begin{cases} x + 1 > 0 & x \geq - 1 \\ - x - 1 > 0 & x < - 1 \end{cases}$

Step 2.2

Subtract $1$ from both sides of the inequality.

$x > - 1$

Step 2.3

Solve $- x - 1 > 0$ for $x$ .

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Step 2.3.1

Add $1$ to both sides of the inequality.

$- x > 1$

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

Simplify the left side.

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Step 2.3.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.3.2.2.2

Divide $x$ by $1$ .

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

Step 2.3.2.3

Simplify the right side.

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Step 2.3.2.3.1

Divide $1$ by $- 1$ .

$x < - 1$

Step 2.4

Find the union of the solutions.

$x < - 1$ or $x > - 1$

Step 3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq - 1}$

Step 4