Solve (3x+1)(1-3x) > 0: Finding Values for Positive Function Output

To solve the problem, we analyze the function:

The function is given as $y = (3x + 1)(1 - 3x)$. This function is a quadratic expression in the factored form, which allows us to find the roots and analyze the intervals.

Step 1: Identify the roots.
Set each factor equal to zero:
$3x + 1 = 0$ leads to $x = -\frac{1}{3}$.
$1 - 3x = 0$ leads to $x = \frac{1}{3}$.

Step 2: Determine the sign in each interval divided by the roots.
The roots divide the real number line into the following intervals: $(-\infty, -\frac{1}{3})$, $(-\frac{1}{3}, \frac{1}{3})$, and $(\frac{1}{3}, \infty)$.

Step 3: Test the sign of $y$ in each interval:

• For $x \in (-\infty, -\frac{1}{3})$, choose $x = -1$:
  $(3(-1) + 1)(1 - 3(-1)) = (-2)(4) = -8$. So, $y < 0$.
• For $x \in (-\frac{1}{3}, \frac{1}{3})$, choose $x = 0$:
  $(3(0) + 1)(1 - 3(0)) = (1)(1) = 1$. So, $y > 0$.
• For $x \in (\frac{1}{3}, \infty)$, choose $x = 1$:
  $(3(1) + 1)(1 - 3(1)) = (4)(-2) = -8$. So, $y < 0$.

Thus, the function $y = (3x + 1)(1 - 3x)$ is positive for $x$ in the interval $-\frac{1}{3} < x < \frac{1}{3}$.

Therefore, the values of $x$ for which $f(x) > 0$ are $-\frac{1}{3} < x < \frac{1}{3}$.

The correct answer is: $-\frac{1}{3} < x < \frac{1}{3}$.