Find Intervals of Increase and Decrease: y = -1/6x² + 11/3x

To determine the intervals of increase and decrease for the quadratic function $y = -\frac{1}{6}x^2 + 3\frac{2}{3}x$, follow these steps:

• Step 1: Find the derivative of the function:

The function is given by $y = -\frac{1}{6}x^2 + \frac{11}{3}x$.
The derivative, using power rules, is $\frac{dy}{dx} = -\frac{1}{3}x + \frac{11}{3}$.

• Step 2: Set the derivative equal to zero and solve for $x$ (Critical points):

Set $-\frac{1}{3}x + \frac{11}{3} = 0$.
Solving for $x$, we get $\frac{1}{3}x = \frac{11}{3}$,
Thus, $x = 11$.

• Step 3: Determine the intervals by testing values around the critical point:

For $x < 11$, choose any point like $x = 0$:
$\frac{dy}{dx} = -\frac{1}{3}(0) + \frac{11}{3} = \frac{11}{3} > 0$. So, the function is increasing on $x < 11$.
For $x > 11$, choose any point like $x = 12$:
$\frac{dy}{dx} = -\frac{1}{3}(12) + \frac{11}{3} = -4 + \frac{11}{3} = -\frac{1}{3} < 0$. So, the function is decreasing on $x > 11$.

Thus, the function is increasing on the interval $(-\infty, 11)$ and decreasing on the interval $(11, \infty)$.

Therefore, the intervals of increase and decrease for the function are:
$\nearrow$ for $x < 11$; $\searrow$ for $x > 11$.