Find the Domain and Range g(x)=(x+7)/(x^2-4)

$g (x) = \frac{x + 7}{x^{2} - 4}$

Step 1

Set the denominator in $\frac{x + 7}{x^{2} - 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 4 = 0$

Step 2

Solve for $x$ .

Add $4$ to both sides of the equation.

$x^{2} = 4$

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$x = \pm 2$

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.

$x = 2$

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

$x = - 2$

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

$x = 2, - 2$

Step 3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 2, - 2}$

Step 4

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

Interval Notation:

$(- \infty, - \frac{7 + 3 \sqrt{5}}{8}] \cup [- \frac{7 - 3 \sqrt{5}}{8}, \infty)$

Set-Builder Notation:

${y | y \leq - \frac{7 + 3 \sqrt{5}}{8}, y \geq - \frac{7 - 3 \sqrt{5}}{8}}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty), {x | x \neq 2, - 2}$

Range: $(- \infty, - \frac{7 + 3 \sqrt{5}}{8}] \cup [- \frac{7 - 3 \sqrt{5}}{8}, \infty), {y | y \leq - \frac{7 + 3 \sqrt{5}}{8}, y \geq - \frac{7 - 3 \sqrt{5}}{8}}$

Step 6

Find the Domain and Range g(x)=(x+7)/(x^2-4)

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