Find Increasing Intervals: Analyzing y = (3x+1)(1-3x)

Let's solve this problem step-by-step:

The function is given by:

$y = (3x + 1)(1 - 3x)$

We can first find the derivative $y'$ to determine where the function is increasing:

• First, expand the product: $y = 3x + 1 - 9x^2 - 3x$
• Simplify: $y = -9x^2 + 0x + 1$

Now the function looks like this quadratic form $y = -9x^2 + 1$.

Next, compute the derivative:

$\frac{dy}{dx} = -18x$

To find the critical points, set $\frac{dy}{dx} = 0$:

$-18x = 0$

Solving, we find $x = 0$.

Now analyze the sign of $\frac{dy}{dx} = -18x$ around this critical point:

• For $x < 0$, $-18x > 0$. Therefore, the function is increasing.
• For $x > 0$, $-18x < 0$. Therefore, the function is decreasing.

Therefore, the solution is that the function is increasing on the interval:

$x < 0$.