Find the Domain f(x) = natural log of (x-3)/(x+1)

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

Step 1

Set the argument in $\ln (\frac{x - 3}{x + 1})$ greater than $0$ to find where the expression is defined.

$\frac{x - 3}{x + 1} > 0$

Step 2

Solve for $x$ .

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 3 = 0$

$x + 1 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Subtract $1$ from both sides of the equation.

$x = - 1$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 3$

$x = - 1$

Consolidate the solutions.

$x = 3, - 1$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

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

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

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

$x = - 4$

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

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

The left side $2.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

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

The left side $- 3$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{(6) - 3}{(6) + 1} > 0$

The left side $0.\overline{428571}$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

$x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x < - 1$ or $x > 3$

Step 3

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

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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

Interval Notation:

$(- \infty, - 1) \cup (3, \infty)$

Set-Builder Notation:

${x | x < - 1, x > 3}$

Step 6

Find the Domain f(x) = natural log of (x-3)/(x+1)

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