Find Intervals of Increase and Decrease: y = -2/3x² + 1/4x - 1/5

To find the intervals of increase and decrease for the function $y = -\frac{2}{3}x^2 + \frac{1}{4}x - \frac{1}{5}$, we begin by finding its first derivative.

• Step 1: Compute the Derivative
  The derivative of the function $y$ is $y' = \frac{d}{dx} \left(-\frac{2}{3}x^2 + \frac{1}{4}x - \frac{1}{5} \right)$.
  Using the power rule, this yields $y' = -\frac{4}{3}x + \frac{1}{4}$.
• Step 2: Find Critical Points
  Set the derivative equal to zero to find critical points: $-\frac{4}{3}x + \frac{1}{4} = 0$.
  Solving for $x$, we get:
  $-\frac{4}{3}x = -\frac{1}{4}$
  $x = \frac{-\frac{1}{4}}{-\frac{4}{3}} = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} = 0.1875$.
• Step 3: Determine Intervals of Increase and Decrease
  The critical point divides the number line into two intervals: $x < 0.1875$ and $x > 0.1875$.
  Evaluate the sign of the derivative $y'$ in these intervals:
• For $x < 0.1875$: Choose a test point like $x = 0$. Evaluating $y'$ gives $y' = -\frac{4}{3}(0) + \frac{1}{4} = \frac{1}{4} > 0$. So, $y$ is increasing.
• For $x > 0.1875$: Choose a test point like $x = 1$. Evaluating $y'$ gives $y' = -\frac{4}{3}(1) + \frac{1}{4} = -\frac{4}{3} + \frac{1}{4} = -\frac{13}{12} < 0$. So, $y$ is decreasing.

Therefore, the function is increasing for $x < 0.1875$ and decreasing for $x > 0.1875$.

The correct answer is: $\searrow: x > 0.1875 \\\nearrow: x < 0.1875$