Find Intervals of Increase and Decrease for y = (7x+3)(5x-2)

To solve this problem, we'll find the intervals of increase and decrease for the function $y = (7x + 3)(5x - 2)$.

First, let's expand the function:

$y = (7x + 3)(5x - 2) = 35x^2 + 15x - 14x - 6 = 35x^2 + x - 6$.

Next, compute the derivative $y'$:

$y' = \frac{d}{dx}(35x^2 + x - 6) = 70x + 1$.

To find critical points, set $y' = 0$:

$70x + 1 = 0$

$70x = -1$

$x = -\frac{1}{70}$.

Now, we need to determine the sign of $y'$ in the intervals around the critical point $x = -\frac{1}{70}$:

• For $x < -\frac{1}{70}$, choose a test point such as $x = -1$:
  $y' = 70(-1) + 1 = -70 + 1 = -69$, so $y' < 0$, indicating that the function is decreasing in this interval.
• For $x > -\frac{1}{70}$, choose a test point such as $x = 0$:
  $y' = 70(0) + 1 = 1$, so $y' > 0$, indicating that the function is increasing in this interval.

Putting it all together, we have:

The function is decreasing on the interval $x < -\frac{1}{70}$ and increasing on the interval $x > -\frac{1}{70}$.

Therefore, the intervals of increase and decrease are:

$\searrow:x < -\frac{1}{70} \\ \nearrow:x > -\frac{1}{70}$.