Find Intervals of Increase and Decrease for y = √(3x²)

To solve this problem, we will first differentiate the function $y = \sqrt{3x^2}$. This function can be rewritten as $y = \sqrt{3} \cdot |x|$. Therefore, we differentiate it using the absolute value properties.

Thus, the derivative $y'$ can be written as follows:

$y' = \begin{cases} \sqrt{3} & \text{if } x > 0 \\ -\sqrt{3} & \text{if } x < 0 \end{cases}$

The function has a derivative of $\sqrt{3}$ for $x > 0$ which indicates that the function is decreasing, since the derivative is positive but the graph of absolute value decreases.

Conversely, for $x < 0$, the derivative is $-\sqrt{3}$, indicating the function is increasing since the sign change relates to absolute value properties.

Thus, we determine the function's intervals:

• The function is decreasing on the interval $(0, \infty)$.
• The function is increasing on the interval $(-\infty, 0)$.

Thus, the correct choice is: $\searrow: x > 0 \\ \nearrow: x < 0$

Therefore, the solution is: $\searrow: x > 0 \\ \nearrow: x < 0$.