Find the Domain and Range f(x)=(x^2+16)/(2x^2-32)

$f (x) = \frac{x^{2} + 16}{2 x^{2} - 32}$

Set the denominator in $\frac{x^{2} + 16}{2 x^{2} - 32}$ equal to $0$ to find where the expression is undefined.

$2 x^{2} - 32 = 0$

Solve for $x$ .

Add $32$ to both sides of the equation.

$2 x^{2} = 32$

Divide each term in $2 x^{2} = 32$ by $2$ and simplify.

Divide each term in $2 x^{2} = 32$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{32}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{32}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{32}{2}$

Simplify the right side.

Divide $32$ by $2$ .

$x^{2} = 16$

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

$x = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 4$

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.

$x = 4$

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

$x = - 4$

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

$x = 4, - 4$

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, 4) \cup (4, \infty)$

Set-Builder Notation:

${x | x \neq 4, - 4}$

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

Interval Notation:

$(- \infty, - \frac{1}{2}] \cup (\frac{1}{2}, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{1}{2}, y > \frac{1}{2}}$

Determine the domain and range.

Domain: $(- \infty, - 4) \cup (- 4, 4) \cup (4, \infty), {x | x \neq 4, - 4}$

Range: $(- \infty, - \frac{1}{2}] \cup (\frac{1}{2}, \infty), {y | y \leq - \frac{1}{2}, y > \frac{1}{2}}$