Find Intervals of Increase and Decrease: y = (1/√3)x² - 1

To determine the intervals of increase and decrease for the function $y = \frac{1}{\sqrt{3}}x^2 - 1$, we follow these steps:

• Step 1: Calculate the Derivative

The derivative of the function $y = \frac{1}{\sqrt{3}}x^2 - 1$ is found using basic differentiation rules. Differentiating with respect to $x$, we have:

$y' = \frac{d}{dx}\left(\frac{1}{\sqrt{3}}x^2 - 1\right) = \frac{2}{\sqrt{3}}x$.

• Step 2: Find the Critical Point

To find the critical points, set the derivative $y'$ equal to zero:

$\frac{2}{\sqrt{3}}x = 0$

This gives $x = 0$ as the critical point.

• Step 3: Analyze the Sign of the Derivative

Now, we determine the sign of the derivative on either side of the critical point $x = 0$.

• For $x < 0$:

The derivative $y' = \frac{2}{\sqrt{3}}x$ is negative, thus the function is decreasing.

• For $x > 0$:

The derivative $y' = \frac{2}{\sqrt{3}}x$ is positive, thus the function is increasing.

Conclusion:

The function decreases for $x < 0$ and increases for $x > 0$.

Therefore, the intervals of increase and decrease are:

$\nearrow: x > 0$

$\searrow: x < 0$

The correct answer is:

$\searrow: x > 0 \; | \; \nearrow: x < 0$

This matches the provided answer in choice option 4.