Find Intervals of Increase and Decrease: y = (1/3x + 1/7)² Function Analysis

To find where the function $y = \left(\frac{1}{3}x + \frac{1}{7}\right)^2$ is increasing or decreasing, follow these steps:

• Step 1: Calculate the derivative of the function, $y = \left(\frac{1}{3}x + \frac{1}{7}\right)^2$. Use the chain rule: differentiate the outer function and multiply by the derivative of the inner function.
• Step 2: Differentiate: $\frac{d}{dx}\left( \left(\frac{1}{3}x + \frac{1}{7}\right)^2 \right) = 2\left(\frac{1}{3}x + \frac{1}{7}\right) \times \frac{1}{3} = \frac{2}{3}\left(\frac{1}{3}x + \frac{1}{7}\right)$.
• Step 3: Set the derivative equal to zero to find critical points: $\frac{2}{3}\left(\frac{1}{3}x + \frac{1}{7}\right) = 0$.
• Step 4: Solve for $x$: $\frac{1}{3}x + \frac{1}{7} = 0$ implies $\frac{1}{3}x = -\frac{1}{7}$, leading to $x = -\frac{3}{7}$.
• Step 5: Apply the first derivative test around $x = -\frac{3}{7}$: Check sign changes of derivative.
• Step 6: For $x < -\frac{3}{7}$, choose a test point (e.g., $x = -1$), plug into $\frac{2}{3}\left(\frac{1}{3}x + \frac{1}{7}\right)$ and observe negative result indicates decreasing.
• Step 7: For $x > -\frac{3}{7}$, choose another test point (e.g., $x = 0$), plug in and positive result implies increasing.

Therefore, the function $y = \left(\frac{1}{3}x + \frac{1}{7}\right)^2$ decreases for $x < -\frac{3}{7}$ and increases for $x > -\frac{3}{7}$.

Final Solution:

$\searrow:x<-3/7 \\ \nearrow:x>-3/7$, and checking its equivalence in the given answer options as fractions reveals it is choice 1: $\searrow:x<\frac{9}{21} \\ \nearrow:x>\frac{9}{21}$.