Solve (2x-1/2)(x-2.25): Finding Negative Domain Regions

Find the positive and negative domains of the following function:

$y=\left(2x-\frac{1}{2}\right)\left(x-2\frac{1}{4}\right)$

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

$f(x) < 0$

To solve this problem, we'll follow these steps:

• Step 1: Identify the roots of the quadratic function.
• Step 2: Analyze the intervals between these roots to determine where the function is negative.
• Step 3: Write the final solution as the interval where $f(x) < 0$.

Now, let's work through each step:
Step 1: Determine the roots by solving each factor for zero:
- $2x - \frac{1}{2} = 0 \Rightarrow 2x = \frac{1}{2} \Rightarrow x = \frac{1}{4}$.
- $x - 2\frac{1}{4} = 0 \Rightarrow x = 2\frac{1}{4}$.
Thus, the roots are $x = \frac{1}{4}$ and $x = 2\frac{1}{4}$.

Step 2: Analyze the intervals determined by the roots $x = \frac{1}{4}$ and $x = 2\frac{1}{4}$:

• Interval 1: $x < \frac{1}{4}$
• Interval 2: $\frac{1}{4} < x < 2\frac{1}{4}$
• Interval 3: $x > 2\frac{1}{4}$

Step 3: Test each interval:

• For interval 1 $x < \frac{1}{4}$: Choose $x = 0$. $f(0) = \left(2(0) - \frac{1}{2}\right)(0 - 2\frac{1}{4}) = \left(-\frac{1}{2}\right) \left(-2\frac{1}{4}\right) = \frac{9}{8} > 0$.
• For interval 2 $\frac{1}{4} < x < 2\frac{1}{4}$: Choose $x = 1$. $f(1) = \left(2(1) - \frac{1}{2}\right)(1 - 2\frac{1}{4}) = \left(\frac{3}{2}\right)(-\frac{5}{4}) = -\frac{15}{8} < 0$.
• For interval 3 $x > 2\frac{1}{4}$: Choose $x = 3$. $f(3) = \left(2(3) - \frac{1}{2}\right)(3 - 2\frac{1}{4}) = \left(\frac{11}{2}\right)\left(\frac{3}{4}\right) = \frac{33}{8} > 0$.

Therefore, the solution to $f(x) < 0$ is found in the interval $\frac{1}{4} < x < 2\frac{1}{4}$.