Solve (x+1)(6-x): Finding Values Where Function is Positive

To solve this problem, we follow these steps:

• Step 1: Find the Roots
  Set $(x+1)(6-x) = 0$. Solving these linear equations, we get the roots:
  - $x+1 = 0$ gives $x = -1$.
  - $6-x = 0$ gives $x = 6$.
• Step 2: Determine the Intervals
  Based on the roots, the real line is divided into intervals: $(-\infty, -1)$, $(-1, 6)$, and $(6, \infty)$.
• Step 3: Test Each Interval for Sign
  - For $x \in (-\infty, -1)$: Choose $x = -2$. The expression $(x+1)(6-x) = (-2+1)(6+2) = (-1)(8) = -8 < 0$.
  - For $x \in (-1, 6)$: Choose $x = 0$. The expression $(x+1)(6-x) = (0+1)(6-0) = 1 \cdot 6 = 6 > 0$.
  - For $x \in (6, \infty)$: Choose $x = 7$. The expression $(x+1)(6-x) = (7+1)(6-7) = 8(-1) = -8 < 0$.

The solution, based on the interval where the product is positive, is when $-1 < x < 6$.

Therefore, the values of $x$ for which $f(x) > 0$ are $-1 < x < 6$.