Find Decreasing Intervals for y = (3x+1)(1-3x): Quadratic Function Analysis

To solve for the intervals where the function $y = (3x+1)(1-3x)$ is decreasing, we employ the derivative approach.

Step 1: Simplify the Quadratic Function
Starting with the function:
$y = (3x+1)(1-3x)$
Expand the expression:
$y = 3x \cdot 1 - 3x \cdot 3x + 1 \cdot 1 - 1 \cdot 3x = 3x - 9x^2 + 1 - 3x$
Combine like terms:
$y = -9x^2 + 3x - 3x + 1 = -9x^2 + 1$.

Step 2: Find the Derivative
Differentiate $y = -9x^2 + 1$ with respect to $x$:
$\frac{dy}{dx} = -18x$.

Step 3: Determine Critical Points
Set the derivative equal to zero to find critical points:
$-18x = 0$
Solving gives:
$x = 0$.

Step 4: Sign Analysis of the Derivative
Check the sign of $\frac{dy}{dx} = -18x$ in the intervals determined by the critical points:

• For $x < 0$, choose $x = -1$: $-18(-1) = 18 > 0$, so increasing.
• For $x = 0$, the derivative is zero, indicating a critical point.
• For $x > 0$, choose $x = 1$: $-18(1) = -18 < 0$, so decreasing.

Thus, the function is decreasing in the interval $x > 0$.

Therefore, the correct answer is $x > 0$.