Find Intervals of Increase and Decrease for y = √(2x²)

To solve this problem, we'll follow these steps:

• Simplify the function expression
• Find the derivative of the function
• Determine where the derivative is positive, zero, or negative
• Draw conclusions about the function's increase or decrease intervals

Let's go through each step:

Simplify the Function:
The function given is $y = \sqrt{2x^2}$. We can simplify this to $y = \sqrt{2}|x|$ because $\sqrt{2x^2} = \sqrt{2} \cdot |x|$.

Calculate the Derivative:
Since $y = \sqrt{2}|x|$, let's consider the derivative of $y = \sqrt{2}|x|$. Note that when differentiating absolute value functions: - For $x > 0$, $|x| = x$, so $y = \sqrt{2}x$ and its derivative is $y' = \sqrt{2}$. - For $x < 0$, $|x| = -x$, so $y = -\sqrt{2}x$ and its derivative is $y' = -\sqrt{2}$.

Determine Sign of the Derivative:
Analyzing $y'$: - For $x > 0$, the derivative $y' = \sqrt{2} > 0$, implying the function is increasing on this interval. - For $x < 0$, the derivative $y' = -\sqrt{2} < 0$, implying the function is decreasing on this interval.

Conclusion:
The function decreases on the interval $x < 0$ and increases on the interval $x > 0$.

Thus, the intervals of increase and decrease of the function are $\searrow:x < 0$ (decreasing), and $\nearrow:x > 0$ (increasing).