Determining the Valid Domain for (x+2)/√(3x²-9)

To solve this problem, we must determine the domain of the function $\frac{x+2}{\sqrt{3x^2-9}}$.

To begin, the expression inside the square root, $3x^2 - 9$, must be greater than 0 for the square root to be real and the function to be defined. Thus, we set up the inequality:

$3x^2 - 9 > 0$

Next, solve this inequality for $x$:

• Start by factoring: $3x^2 - 9 = 3(x^2 - 3)$
• Set the factor equal to zero to find the critical points: $x^2 - 3 = 0$
• Solve for $x$:
• $x^2 = 3$
• $x = \pm \sqrt{3}$

Now, determine the intervals where $3x^2 - 9$ is positive. Consider the intervals defined by the critical points $x = \sqrt{3}$ and $x = -\sqrt{3}$:

• Interval 1: $(-\infty, -\sqrt{3})$
• Interval 2: $(- \sqrt{3}, \sqrt{3})$
• Interval 3: $(\sqrt{3}, \infty)$

Test a value from each interval in the inequality $3x^2 - 9 > 0$:

• For interval 1, test $x = -2$ :
• $3(-2)^2 - 9 = 12 - 9 = 3 > 0$ (True)
• For interval 2, test $x = 0$ :
• $3(0)^2 - 9 = -9 < 0$ (False)
• For interval 3, test $x = 2$ :
• $3(2)^2 - 9 = 12 - 9 = 3 > 0$ (True)

Thus, the function is defined for $x$ in the intervals $(-\infty, -\sqrt{3})$ and $(\sqrt{3}, \infty)$.

This means that the domain of the function is:

$x>\sqrt{3},x<-\sqrt{3}$

Therefore, the solution to the problem is $x>\sqrt{3},x<-\sqrt{3}$.