Solve y = (1/2)x² - 2: Finding Negative 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:

$f(x) < 0$

The function $y = \frac{1}{2}x^2 - 2$ is quadratic. To find where $f(x) < 0$, identify where $f(x) = 0$ by setting the equation equal to zero and solving for $x$.

The equation is $\frac{1}{2}x^2 - 2 = 0$.

Multiply through by 2 to eliminate the fraction:
$x^2 - 4 = 0$.

Set the equation as $(x - 2)(x + 2) = 0$ to find the roots.

Solving gives roots $x = 2$ and $x = -2$.

The function $f(x)$ will change signs at these roots $x = -2$ and $x = 2$.

Check intervals determined by the roots to find where $f(x) < 0$:

• For $-2 < x < 2$: Evaluate one test point like $x = 0$ in $f(x)$:
  $f(0) = \frac{1}{2}(0)^2 - 2 = -2$, indicating $f(x) < 0$.
• For intervals $x < -2$ or $x > 2$: Select $x = -3$ and $x = 3$ respectively, both yield positive results, therefore $f(x) > 0$.

Ultimately, the function $f(x)$ is negative only on the interval $-2 < x < 2$.

Therefore, the values of $x$ for which $f(x) < 0$ are such that $-2 < x < 2$.