Find Intervals of Increase and Decrease for y=(5x-1)(4x-1/4)

To find the intervals of increase and decrease for the function $y = (5x - 1)\left(4x - \frac{1}{4}\right)$, we perform the following steps:

• Step 1: Differentiate the function using the product rule.
• Step 2: Find the critical points by setting the derivative to zero.
• Step 3: Determine the intervals where the derivative is positive or negative to infer increasing and decreasing behavior.

Now, let's work through each step in detail:

Step 1: Differentiate the function.
Using the product rule, consider $u = 5x - 1$ and $v = 4x - \frac{1}{4}$. The derivative of the function is:

$y' = \frac{d}{dx}\left[(5x - 1)\left(4x - \frac{1}{4}\right)\right] = (5x - 1)\frac{d}{dx}\left(4x - \frac{1}{4}\right) + \left(4x - \frac{1}{4}\right)\frac{d}{dx}(5x - 1)$

$= (5x - 1) \cdot 4 + \left(4x - \frac{1}{4}\right) \cdot 5 = 20x - 4 + 20x - \frac{5}{4}$

$= 40x - \frac{21}{4}$

Step 2: Find the critical points.
Set the derivative to zero:

$40x - \frac{21}{4} = 0$

Solving for $x$, multiply both sides by 4 to clear the fraction:

$160x - 21 = 0$

$x = \frac{21}{160}$

Step 3: Analyze the sign of the derivative around the critical point to determine increasing or decreasing intervals.
Choose a test point in each interval defined by the critical point $x = \frac{21}{160}$.

• For $x < \frac{21}{160}$, choose $x = 0$ and check the sign of $y' = 40(0) - \frac{21}{4} = -\frac{21}{4}$ which is negative, indicating $y$ is decreasing.
• For $x > \frac{21}{160}$, choose $x = 1$ and check the sign of $y' = 40(1) - \frac{21}{4} = \frac{139}{4}$ which is positive, indicating $y$ is increasing.

Thus, the function decreases for $x < \frac{21}{160}$ and increases for $x > \frac{21}{160}$.

Therefore, the intervals are $\searrow:x<\frac{21}{160}$ and $\nearrow:x>\frac{21}{160}$.

The correct choice is:

$\searrow:x<\frac{21}{160}\\\nearrow:x>\frac{21}{160}$