Find Positive and Negative Domains of y = (1/3)x² - 2: Domain Analysis

To solve the problem, we will determine the positive and negative domains of the quadratic function $y = \frac{1}{3}x^2 - 2$ by following these steps:

• Step 1: Find the roots of the quadratic function to identify the intervals.

• Step 2: Analyze the sign of the function in each interval determined by the roots.

Step 1: Finding the Roots
To find the roots, set $y = 0$ in the equation:
$\frac{1}{3}x^2 - 2 = 0$.
Multiply through by 3 to eliminate the fraction:
$x^2 - 6 = 0$.
Rearranging gives:
$x^2 = 6$.
The solutions to this equation are the roots $x = \pm \sqrt{6}$.

Step 2: Analyze Sign in Each Interval
The roots $x = -\sqrt{6}$ and $x = \sqrt{6}$ split the real number line into three intervals: $(-\infty, -\sqrt{6})$, $(-\sqrt{6}, \sqrt{6})$, $(\sqrt{6}, \infty)$.
For each interval, choose a test point and determine the sign of $y$.

- Interval $(-\infty, -\sqrt{6})$: Choose $x = -3$ (since $-3 \lt -\sqrt{6} \approx -2.45$).
$y = \frac{1}{3}(-3)^2 - 2 = \frac{9}{3} - 2 = 3 - 2 = 1 > 0$.
Thus, $y > 0$ in the interval $(-\infty, -\sqrt{6})$.

- Interval $(-\sqrt{6}, \sqrt{6})$: Choose $x = 0$.
$y = \frac{1}{3}(0)^2 - 2 = -2 < 0$.
Thus, $y < 0$ in the interval $(-\sqrt{6}, \sqrt{6})$.

- Interval $(\sqrt{6}, \infty)$: Choose $x = 3$ (since $3 \gt \sqrt{6} \approx 2.45$).
$y = \frac{1}{3}(3)^2 - 2 = \frac{9}{3} - 2 = 3 - 2 = 1 > 0$.
Thus, $y > 0$ in the interval $(\sqrt{6}, \infty)$.

Therefore, the positive domain is given by the intervals $x < -\sqrt{6}$ and $x > \sqrt{6}$, while the negative domain is $-\sqrt{6} < x < \sqrt{6}$.

Thus, the positive and negative domains of the function are:

- Positive: $x < -\sqrt{6}$ or $x > \sqrt{6}$

- Negative: $-\sqrt{6} < x < \sqrt{6}$

These match the correct answer choice as follows:

$x < 0 : -\sqrt{6} < x < \sqrt{6}$

$x > \sqrt{6}$ or $x> 0 : x <-\sqrt{6}$