Analyzing y = (1/3)x² - 3: Domain and Positive Values

Find the positive and negative domains of the function below:

$y=\frac{1}{3}x^2-3$

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

$f(x) > 0$

We will solve the quadratic inequality $\frac{1}{3}x^2 - 3 > 0$.

Step 1: Find roots of the equation $\frac{1}{3}x^2 - 3 = 0$.

Rewriting, we have:

$\frac{1}{3}x^2 = 3$

Multiply both sides by 3 to clear the fraction:

$x^2 = 9$

Then take the square root of both sides:

$x = \pm 3$

Step 2: Determine the sign of $f(x) = \frac{1}{3}x^2 - 3$ on the intervals determined by the roots:

• For $x < -3$, let $x = -4$: $f(-4) = \frac{1}{3}(-4)^2 - 3 = \frac{16}{3} - 3 = \frac{7}{3} > 0$.
• For $-3 < x < 3$, let $x = 0$: $f(0) = \frac{1}{3}(0)^2 - 3 = -3 < 0$.
• For $x > 3$, let $x = 4$: $f(4) = \frac{1}{3}(4)^2 - 3 = \frac{16}{3} - 3 = \frac{7}{3} > 0$.

Step 3: From the interval test, we find that $f(x) > 0$ when $x < -3$ or $x > 3$.

Therefore, the solution to the problem is $x > 3$ or $x < -3$.