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

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

Step 1

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

$x^{2} - 16 = 0$

Step 2

Solve for $x$ .

Add $16$ to both sides of the equation.

$x^{2} = 16$

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.

$x = 4, - 4$

Step 3

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:

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

Step 4

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

Interval Notation:

$(- \infty, \frac{9 - 3 \sqrt{5}}{16}] \cup [\frac{9 + 3 \sqrt{5}}{16}, \infty)$

Set-Builder Notation:

${y | y \leq \frac{9 - 3 \sqrt{5}}{16}, y \geq \frac{9 + 3 \sqrt{5}}{16}}$

Step 5

Determine the domain and range.

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

Range: $(- \infty, \frac{9 - 3 \sqrt{5}}{16}] \cup [\frac{9 + 3 \sqrt{5}}{16}, \infty), {y | y \leq \frac{9 - 3 \sqrt{5}}{16}, y \geq \frac{9 + 3 \sqrt{5}}{16}}$

Step 6

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

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