Solve (x-1/2)(-x+7/2) < 0: Finding Negative Function Values

Look at the following function:

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

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

$f(x) < 0$

To solve this problem, we'll begin by finding the roots of the quadratic equation $y = \left(x - \frac{1}{2}\right)\left(-x + 3\frac{1}{2}\right)$.

First, set each factor equal to zero:

• $x - \frac{1}{2} = 0$ gives $x = \frac{1}{2}$
• $-x + 3\frac{1}{2} = 0$ gives $x = 3\frac{1}{2}$

This means the roots of the quadratic are $x = \frac{1}{2}$ and $x = 3\frac{1}{2}$.

Next, analyze the intervals determined by these roots:

• Interval 1: $x < \frac{1}{2}$
• Interval 2: $\frac{1}{2} < x < 3\frac{1}{2}$
• Interval 3: $x > 3\frac{1}{2}$

Perform a sign test within these intervals:

• For $x < \frac{1}{2}$: Both $x - \frac{1}{2}$ and $-x + 3\frac{1}{2}$ are negative, thus their product is positive (negative times negative is positive).
• For $\frac{1}{2} < x < 3\frac{1}{2}$: The factor $x - \frac{1}{2}$ is positive and $-x + 3\frac{1}{2}$ is positive as well, thus product is positive (positive times positive is positive).
• For $x > 3\frac{1}{2}$: $x - \frac{1}{2}$ is positive, but $-x + 3\frac{1}{2}$ is negative, so the product is negative (positive times negative is negative).

Therefore, the quadratic function is negative for: $x > 3\frac{1}{2}$ and $x < \frac{1}{2}$.

The solution to the problem is: $x > 3\frac{1}{2}$ or $x < \frac{1}{2}$.