Find the Domain and Range y = square root of 1/(x-2)

$y = \sqrt{\frac{1}{x - 2}}$

Set the radicand in $\sqrt{\frac{1}{x - 2}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{1}{x - 2} \geq 0$

Solve for $x$ .

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

Set the denominator in $\frac{1}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$(- \infty, 2) \cup (2, \infty)$

The solution consists of all of the true intervals.

$x > 2$

Set the denominator in $\frac{1}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

Interval Notation:

$(2, \infty)$

Set-Builder Notation:

${x | x > 2}$

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

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${y | y > 0}$

Determine the domain and range.

Domain: $(2, \infty), {x | x > 2}$

Range: $(0, \infty), {y | y > 0}$