Finding Positive Domains: Solve (x-1/2)(-x+7/2) > 0

Question

Find the positive and negative domains of the function below:

y=(x12)(x+312) y=\left(x-\frac{1}{2}\right)\left(-x+3\frac{1}{2}\right)

Determine for which values of x x the following is true:

f(x) > 0

Step-by-Step Solution

To solve this problem, we'll determine when the product (x12)(x+312) (x - \frac{1}{2})(-x + 3\frac{1}{2}) is positive. This involves finding the roots of the equation and testing the intervals between these roots:

Step 1: **Determine the roots of the factors.**
- The first factor x12=0 x - \frac{1}{2} = 0 gives the root x=12 x = \frac{1}{2} .
- The second factor x+312=0 -x + 3\frac{1}{2} = 0 gives the root x=312 x = 3\frac{1}{2} .

Step 2: **Identify intervals based on these roots.**
- The roots divide the x x -axis into three intervals: x<12 x < \frac{1}{2} , 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} , and x>312 x > 3\frac{1}{2} .

Step 3: **Analyze the sign of the function in each interval.**
- For x<12 x < \frac{1}{2} :
- x12<0 x - \frac{1}{2} < 0 and x+312>0 -x + 3\frac{1}{2} > 0 , so the product is negative.
- For 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} :
- Both x12>0 x - \frac{1}{2} > 0 and x+312>0 -x + 3\frac{1}{2} > 0 , so the product is positive.
- For x>312 x > 3\frac{1}{2} :
- x12>0 x - \frac{1}{2} > 0 and x+312<0 -x + 3\frac{1}{2} < 0 , so the product is negative.

Therefore, the intervals where y>0 y > 0 are 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} .

This matches the given correct answer choice: 12<x<312 \frac{1}{2} < x < 3\frac{1}{2} .

Answer

\frac{1}{2} < x < 3\frac{1}{2}