Find the Domain and Range f(x) = log of (x+7)/(x-1)

$f (x) = \log (\frac{x + 7}{x - 1})$

Step 1

Set the argument in $\log (\frac{x + 7}{x - 1})$ greater than $0$ to find where the expression is defined.

$\frac{x + 7}{x - 1} > 0$

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.

$x + 7 = 0$

$x - 1 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Add $1$ to 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.

$x = - 7$

$x = 1$

Consolidate the solutions.

$x = - 7, 1$

Find the domain of $\frac{x + 7}{x - 1}$ .

Set the denominator in $\frac{x + 7}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Add $1$ to 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.

$x < - 7$

$- 7 < x < 1$

$x > 1$

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$\frac{(- 10) + 7}{(- 10) - 1} > 0$

The left side $0.\overline{27}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 7 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 7}{(0) - 1} > 0$

The left side $- 7$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

$\frac{(4) + 7}{(4) - 1} > 0$

The left side $3.\overline{6}$ 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.

$x < - 7$ True

$- 7 < x < 1$ False

$x > 1$ True

$x < - 7$ True

$- 7 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$x < - 7$ or $x > 1$

Step 3

Set the denominator in $\frac{x + 7}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

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

Interval Notation:

$(- \infty, - 7) \cup (1, \infty)$

Set-Builder Notation:

${x | x < - 7, x > 1}$

Step 6

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

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

Set-Builder Notation:

${y | y \neq 0}$

Step 7

Determine the domain and range.

Domain: $(- \infty, - 7) \cup (1, \infty), {x | x < - 7, x > 1}$

Range: $(- \infty, 0) \cup (0, \infty), {y | y \neq 0}$

Step 8

Find the Domain and Range f(x) = log of (x+7)/(x-1)

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