Find Intervals of Increase and Decrease for y = (3x+1)(4x-2)

To solve this problem, follow these steps:

• Step 1: Expand the function
  We have $y = (3x + 1)(4x - 2)$. Expanding this, we get:

$y = 3x \cdot 4x + 3x \cdot (-2) + 1 \cdot 4x + 1 \cdot (-2)$

$y = 12x^2 - 6x + 4x - 2$

$y = 12x^2 - 2x - 2$

• Step 2: Find the derivative
  The derivative $y'$ of the function $y = 12x^2 - 2x - 2$ is:

$y' = \frac{d}{dx}(12x^2 - 2x - 2)$

$y' = 24x - 2$

• Step 3: Determine critical points
  Set the derivative equal to zero to find critical points:

$24x - 2 = 0$

$24x = 2$

$x = \frac{1}{12}$

• Step 4: Determine intervals of increase and decrease
  Evaluate the sign of $y' = 24x - 2$ on intervals relative to the critical point $x = \frac{1}{12}$:

Test values:

$y'(0) = 24(0) - 2 = -2$ (Negative, so decreasing)

- For $x > \frac{1}{12}$, choose $x = 1$:

$y'(1) = 24(1) - 2 = 22$ (Positive, so increasing)

Therefore, the function is decreasing for $x < \frac{1}{12}$ and increasing for $x > \frac{1}{12}$.

The correct answer is:

$\searrow:x<\frac{1}{12}\\\nearrow:x>\frac{1}{12}$