Finding Intervals of Increase and Decrease: y = (1/2)x² + 4(3/5)x

To determine where the given function increases or decreases, we follow these steps:

• Step 1: Differentiate the given function. Given the function $y = \frac{1}{2}x^2 + 4\frac{3}{5}x$, we first express $4\frac{3}{5}$ as an improper fraction, which is $\frac{23}{5}$. This makes the function $y = \frac{1}{2}x^2 + \frac{23}{5}x$.
• Step 2: Calculate the derivative of the function. The derivative $y'$ is computed as follows: $y' = \frac{d}{dx}\left(\frac{1}{2}x^2 + \frac{23}{5}x\right) = x + \frac{23}{5}.$
• Step 3: Set the derivative to zero to find critical points. Solve the equation $x + \frac{23}{5} = 0$: $x = -\frac{23}{5}.$
• Step 4: Identify intervals around the critical point. The critical point is $x = -\frac{23}{5}$, corresponding to a point of potential change from increasing to decreasing or vice versa.
• Step 5: Test the intervals determined by the critical points to find the sign of the derivative, determining whether the function is increasing or decreasing in those intervals.

- For $x < -\frac{23}{5}$, choose a test point such as $x = -5$:
$y'(-5) = -5 + \frac{23}{5} = -5 + 4.6 = -0.4$ which is negative, so the function is decreasing ($\searrow$).

- For $x > -\frac{23}{5}$, choose a test point such as $x = 0$:
$y'(0) = 0 + \frac{23}{5} = 4.6$ which is positive, indicating the function is increasing ($\nearrow$).

Therefore, the function decreases for $x < -\frac{23}{5}$ and increases for $x > -\frac{23}{5}$.

Thus, the intervals of increase and decrease for the function are:

$\searrow:x > -4\frac{3}{5}~~\\ \nearrow:x < -4\frac{3}{5}$