Find the Domain (6b-3)/(3b^2-12)

$\frac{6 b - 3}{3 b^{2} - 12}$

Step 1

Set the denominator in $\frac{6 b - 3}{3 b^{2} - 12}$ equal to $0$ to find where the expression is undefined.

$3 b^{2} - 12 = 0$

Step 2

Solve for $b$ .

Add $12$ to both sides of the equation.

$3 b^{2} = 12$

Divide each term in $3 b^{2} = 12$ by $3$ and simplify.

Divide each term in $3 b^{2} = 12$ by $3$ .

$\frac{3 b^{2}}{3} = \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 b^{2}}{3} = \frac{12}{3}$

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

$b^{2} = \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$b^{2} = 4$

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

$b = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$b = \pm 2$

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.

$b = 2$

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

$b = - 2$

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

$b = 2, - 2$

Step 3

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

Interval Notation:

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

Set-Builder Notation:

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

Step 4