Solving y = (1/3)x² - 3: Finding Negative Function Values and Domain

To solve this problem, we'll follow a systematic approach:

• Step 1: Determine the roots of the function by setting $f(x) = 0$.
• Step 2: Solve the equation $\frac{1}{3}x^2 - 3 = 0$.
• Step 3: Rearrange it to $\frac{1}{3}x^2 = 3$.
• Step 4: Multiply through by 3 to clear the fraction, resulting in $x^2 = 9$.
• Step 5: Solve for $x$ to find the roots $x = 3$ and $x = -3$ (since $\pm\sqrt{9} = \pm 3$).
• Step 6: Build a number line and test intervals: $x < -3$, $-3 < x < 3$, and $x > 3$.
• Step 7: Test a value within each interval to see where the inequality $\frac{1}{3}x^2 - 3 < 0$ holds.

Interval testing:

• For $x < -3$, try $x = -4$: $\frac{1}{3}(-4)^2 - 3 = \frac{16}{3} - 3 \approx 5.33 - 3 = 2.33 > 0$.
• For $-3 < x < 3$, try $x = 0$: $\frac{1}{3}(0)^2 - 3 = -3 < 0$.
• For $x > 3$, try $x = 4$: $\frac{1}{3}(4)^2 - 3 = \frac{16}{3} - 3 \approx 2.33 > 0$.

Conclusion:

The value of $x$ for which $f(x) < 0$ is the interval $-3 < x < 3$.

Therefore, the correct answer is $-3 < x < 3$.