Determine the Domain: Analyzing 8x/√(2x²-2)

To find the domain of the function $\frac{8x}{\sqrt{2x^2-2}}$, we need to ensure the denominator is not zero and is defined.

Since the denominator is $\sqrt{2x^2 - 2}$, we have the condition:

• $2x^2 - 2 > 0$

Let's solve the inequality $2x^2 - 2 > 0$.

First, set the equation to zero to find critical points:

$2x^2 - 2 = 0$

Simplify and solve for $x$:

$2x^2 = 2$

$x^2 = 1$

$x = \pm 1$

The critical points divide the number line into three intervals: $x < -1$, $-1 < x < 1$, and $x > 1$.

We need to test these intervals to see where $2x^2 - 2 > 0$.

• For $x < -1$, choose $x = -2$:
• $2(-2)^2 - 2 = 8 - 2 = 6$ (positive)

• For $-1 < x < 1$, choose $x = 0$:
• $2(0)^2 - 2 = -2$ (negative)

• For $x > 1$, choose $x = 2$:
• $2(2)^2 - 2 = 8 - 2 = 6$ (positive)

Therefore, the intervals where $2x^2 - 2 > 0$ are $x < -1$ or $x > 1$.

Thus, the domain of the function is $x > 1$ or $x < -1$, in interval notation this is $(-\infty, -1) \cup (1, \infty)$.

So, the correct choice is $x > 1, x < -1$, corresponding to choice 4.

Therefore, the domain of the function is $x > 1, x < -1$.