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

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

• Step 1: Find the roots of the quadratic equation $\frac{1}{2}x^2 - 1 = 0$ using the quadratic formula.
• Step 2: Determine intervals based on the roots and examine the sign of the function within each interval.
• Step 3: Identify where the function takes positive and negative values.

Step 1: The equation $\frac{1}{2}x^2 - 1 = 0$ can be rewritten as $x^2 = 2$. Solving for $x$ gives $x = \pm \sqrt{2}$.

Step 2: The roots $x = -\sqrt{2}$ and $x = \sqrt{2}$ divide the number line into three intervals:
a) $x < -\sqrt{2}$
b) $-\sqrt{2} < x < \sqrt{2}$
c) $x > \sqrt{2}$

Step 3: Analyze the sign of the function in each interval:

• Interval $x < -\sqrt{2}$: Pick $x = -2$ (any point in the interval). The function $y = \frac{1}{2} x^2 - 1$ becomes $y = 2 - 1 = 1$, which is positive.
• Interval $-\sqrt{2} < x < \sqrt{2}$: Pick $x = 0$. Then $y = \frac{1}{2} \times 0 - 1 = -1$, which is negative.
• Interval $x > \sqrt{2}$: Pick $x = 2$. The function $y = \frac{1}{2} \times 4 - 1 = 1$, which is positive.

Therefore, the positive domain of the function is $x < 0 : x < -\sqrt{2}$ and $x > \sqrt{2}$. The negative domain is $x < 0 : -\sqrt{2} < x < \sqrt{2}$.