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

To determine where the function $y=\left(\frac{1}{3}x+\frac{1}{2}\right)^2$ is increasing or decreasing, we need to first find its derivative.

Let's compute the derivative $y'$ of the function:

$y = \left(\frac{1}{3}x + \frac{1}{2}\right)^2$.

Using the chain rule, let $u = \frac{1}{3}x + \frac{1}{2}$, then $y = u^2$.

The derivative of $u^2$ with respect to $x$ is $2u \cdot \frac{du}{dx}$.

Now, find $\frac{du}{dx}$:

$\frac{du}{dx} = \frac{1}{3}$.

Thus, the derivative of the function is:

$y' = 2 \left( \frac{1}{3}x + \frac{1}{2} \right) \cdot \frac{1}{3} = \frac{2}{3} \left( \frac{1}{3}x + \frac{1}{2} \right)$.

Set this derivative to zero to find critical points:

$\frac{2}{3} \left( \frac{1}{3}x + \frac{1}{2} \right) = 0$.

Simplify to find $x$:

$\frac{1}{3}x + \frac{1}{2} = 0$.

$\frac{1}{3}x = -\frac{1}{2}$.

$x = -\frac{1}{2} \times 3$.

$x = -\frac{3}{2}$.

The critical point is at $x = -\frac{3}{2}$.

To determine the nature of intervals around this critical point, test $y'$ on intervals around $x = -\frac{3}{2}$.

- For $x < -\frac{3}{2}$: Choose $x = -2$.
$y' = \frac{2}{3} \left( \frac{1}{3}(-2) + \frac{1}{2} \right) = \frac{2}{3}(-\frac{2}{3} + \frac{1}{2}) < 0$.
$y'$ is negative, so $y$ is decreasing.

- For $x > -\frac{3}{2}$: Choose $x = 0$.
$y' = \frac{2}{3} \left( \frac{1}{3}(0) + \frac{1}{2} \right) = \frac{2}{3} \times \frac{1}{2} > 0$.
$y'$ is positive, so $y$ is increasing.

Thus, the function decreases on $x < -\frac{3}{2}$ and increases on $x > -\frac{3}{2}$.

The intervals of increase and decrease are $\searrow: x > -1\frac{1}{2}$ and $\nearrow: x < -1\frac{1}{2}$.

Analyzing the multiple-choice answers, the correct one matches choice 3.