Solve (x+1)(6-x) > 0: Finding Positive Function Values

To find the intervals where $f(x) = (x + 1)(6 - x) > 0$, follow these steps:

• Step 1: Find the roots by setting each factor to zero.
  The roots occur at $x + 1 = 0$ and $6 - x = 0$. Solving these gives $x = -1$ and $x = 6$.
• Step 2: Determine the sign of the function in the intervals defined by these roots: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.
• Step 3: Evaluate the sign of the product in each interval:
  - For $x \in (-\infty, -1)$, choose $x = -2$. Then, both factors $(x + 1)$ and $(6 - x)$ are negative, making their product positive.
  - For $x \in (-1, 6)$, choose $x = 0$. Then $(x + 1)$ is positive and $(6 - x)$ is positive, making their product positive.
  - For $x \in (6, \infty)$, choose $x = 7$. Then $(x + 1)$ is positive, but $(6 - x)$ is negative, making their product negative.

Thus, the intervals where $f(x) > 0$ are $x < -1$ and $x > 6$.

The solution to the problem is $x > 6$ or $x < -1$.