Find Intervals of Increase and Decrease: y = (1/5)x² + 1⅓x

To find the intervals of increase and decrease for the function $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$, we first calculate the derivative to analyze the behavior of the function.

Step 1: Finding the Derivative
The function is $y = \frac{1}{5}x^2 + \frac{4}{3}x$. To find the derivative, we use the power rule:

$y' = \frac{d}{dx}\left(\frac{1}{5}x^2 + \frac{4}{3}x\right) = \frac{2}{5}x + \frac{4}{3}$.

Step 2: Set the Derivative to Zero
To find critical points, set the derivative equal to zero:

$\frac{2}{5}x + \frac{4}{3} = 0$.

Solving for $x$, we multiply the equation by 15 (to eliminate fractions):

$6x + 20 = 0$.
$6x = -20$.
$x = -\frac{20}{6} = -\frac{10}{3}$.

So, the critical point is at $x = -\frac{10}{3}$, or $x = -3\frac{1}{3}$.

Step 3: Determine the Sign of the Derivative
Test the sign of the derivative on intervals around the critical point $x = -3\frac{1}{3}$:

• For $x < -3\frac{1}{3}$, choose $x = -4$. Then $y' = \frac{2}{5}(-4) + \frac{4}{3} = -\frac{8}{5} + \frac{4}{3}$. Compute to check if negative.
• For $x > -3\frac{1}{3}$, choose $x = 0$. Then $y' = \frac{2}{5}(0) + \frac{4}{3} = \frac{4}{3}$, which is positive.

Conclusion:
The function is decreasing ($\searrow$) for $x > -3\frac{1}{3}$ and increasing ($\nearrow$) for $x < -3\frac{1}{3}$.

Therefore, the correct intervals are:

$\searrow~:x > -3\frac{1}{3}$
$\nearrow~:x < -3\frac{1}{3}$

This matches with choice 4 of the provided options.