Find Intervals of Increase and Decrease: y = (1/3)x² + 2(1/3)x

To determine the intervals of increase and decrease for the function $y = \frac{1}{3}x^2 + 2\frac{1}{3}x$, we will perform the following steps:

• Step 1: Differentiate the function with respect to $x$.
• Step 2: Set the derivative equal to zero to find the critical points.
• Step 3: Use sign analysis on intervals determined by the critical points to identify where the function is increasing or decreasing.

Let's proceed with the solution:

Step 1: Differentiate $y = \frac{1}{3}x^2 + 2\frac{1}{3}x$.

The derivative $f'(x)$ is given by:

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

This simplifies to:

$f'(x) = \frac{2}{3}x + 2\frac{1}{3}$.

Converting $2\frac{1}{3}$ to an improper fraction gives $\frac{7}{3}$, hence:

$f'(x) = \frac{2}{3}x + \frac{7}{3}$.

Step 2: Solve $f'(x) = 0$ to find critical points.

Set $\frac{2}{3}x + \frac{7}{3} = 0$.

Multiply through by 3 to eliminate fractions:

$2x + 7 = 0$.

This simplifies to:

$2x = -7$ $\Rightarrow x = -\frac{7}{2}$ or $x = -3.5$.

Step 3: Perform sign analysis around the critical point $x = -3.5$.

• For $x < -3.5$, choose a test point like $x = -4$.

$f'(-4) = \frac{2}{3}(-4) + \frac{7}{3} = -\frac{8}{3} + \frac{7}{3} = -\frac{1}{3}$.

This is negative, indicating the function is decreasing on this interval.

• For $x > -3.5$, choose a test point like $x = 0$.

$f'(0) = \frac{2}{3}(0) + \frac{7}{3} = \frac{7}{3}$.

This is positive, indicating the function is increasing on this interval.

Thus, the function is decreasing for $x < -3.5$ and increasing for $x > -3.5$.

Therefore, the correct answer is: $\searrow: x > -3.5$; $\nearrow: x < -3.5$.