Find the Domain and Range (2x^2)/(x^2-1)

$\frac{2 x^{2}}{x^{2} - 1}$

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

$x^{2} - 1 = 0$

Solve for $x$ .

Add $1$ to both sides of the equation.

$x^{2} = 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 1, - 1}$

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

Interval Notation:

$(- \infty, 0] \cup (2, \infty)$

Set-Builder Notation:

${y | y \leq 0, y > 2}$

Determine the domain and range.

Domain: $(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty), {x | x \neq 1, - 1}$

Range: $(- \infty, 0] \cup (2, \infty), {y | y \leq 0, y > 2}$