Solve y = (1/2)x² - 2: Finding Positive Function Values

Find the positive and negative domains of the function below:

$y=\frac{1}{2}x^2-2$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

To solve for the positive and negative domains of the function $y = \frac{1}{2}x^2 - 2$, we will go through the following steps:

• Step 1: Find the roots of the equation by setting the function equal to zero: $\frac{1}{2}x^2 - 2 = 0$.
• Step 2: Solve $\frac{1}{2}x^2 = 2$, which simplifies to $x^2 = 4$. Taking the square root of both sides gives $x = 2$ or $x = -2$.
• Step 3: Test intervals between the roots $(- \infty, -2)$, $(-2, 2)$, and $(2, \infty)$ to determine where the function is positive.

Now, let's determine which intervals yield positive values:

For $x < -2$, pick $x = -3$ and plug into the function: $y = \frac{1}{2}(-3)^2 - 2 = \frac{1}{2}(9) - 2 = 4.5 - 2 = 2.5$, which is positive.

For $-2 < x < 2$, pick $x = 0$: $y = \frac{1}{2}(0)^2 - 2 = -2$, which is negative.

For $x > 2$, pick $x = 3$: $y = \frac{1}{2}(3)^2 - 2 = \frac{1}{2}(9) - 2 = 4.5 - 2 = 2.5$, which is positive.

Thus, the function $y = \frac{1}{2}x^2 - 2$ is positive for $x > 2$ or $x < -2$.

Therefore, the solution is $x > 2$ or $x < -2$.