Find Intervals of Increase and Decrease: Analyzing y = (x+1)(x+1/6)

To solve this problem, we'll follow these steps:

• Step 1: Expand the original function
• Step 2: Differentiate the function
• Step 3: Find the critical points by setting the derivative to zero
• Step 4: Determine intervals of increase and decrease based on the derivative's sign

Let's explore each step in detail:
Step 1: Expand the function:

The function $y = (x+1)(x+\frac{1}{6})$ expands to:

$y = x^2 + \frac{1}{6}x + x + \frac{1}{6} = x^2 + \frac{7}{6}x + \frac{1}{6}$.

Step 2: Differentiate the function:
The derivative of the quadratic function $y = x^2 + \frac{7}{6}x + \frac{1}{6}$ is:

$y' = 2x + \frac{7}{6}$.

Step 3: Set the derivative equal to zero to find critical points:
Solve $2x + \frac{7}{6} = 0$.

Subtract $\frac{7}{6}$ from both sides: $2x = -\frac{7}{6}$.

Divide both sides by 2 to solve for $x$:

$x = -\frac{7}{12}$.

Step 4: Determine the sign of the derivative on either side of the critical point:

- For $x < -\frac{7}{12}$, the derivative $y' = 2x + \frac{7}{6}$ is negative, indicating the function is decreasing.

- For $x > -\frac{7}{12}$, the derivative $y'$ is positive, indicating the function is increasing.

Thus, the function is decreasing on $x < -\frac{7}{12}$ and increasing on $x > -\frac{7}{12}$.

Therefore, the final solution is:

$\searrow:x < -\frac{7}{12} \\ \nearrow: x > -\frac{7}{12}$.