Find Domain of y=-(x-2½)²+½: Analyzing Input Values

To find the positive and negative domains of the function $y = -\left(x - 2.5\right)^2 + 0.5$, we analyze when $y$ is greater than and less than zero.

• Step 1: Solve for the positive domain ( y > 0 ).

We need to solve the inequality:
-\left(x - 2.5\right)^2 + 0.5 > 0 .

Rearrange this to:
-\left(x - 2.5\right)^2 > -0.5 .

Remove the negative sign by multiplying by $-1$ (which flips the inequality sign):
\left(x - 2.5\right)^2 < 0.5 .

Taking the square root of both sides gives:
|x - 2.5| < \sqrt{0.5} .
This implies:
-\sqrt{0.5} < x - 2.5 < \sqrt{0.5} .

Solve for $x$:
2.5 - \sqrt{0.5} < x < 2.5 + \sqrt{0.5} .

Step 2: Solve for the negative domain ( y < 0 ).

From the inequality:
-\left(x - 2.5\right)^2 + 0.5 < 0 .

Rearrange to:
-\left(x - 2.5\right)^2 < -0.5 .

Again, multiply by $-1$:
\left(x - 2.5\right)^2 > 0.5 .

Taking the square root gives:
|x - 2.5| > \sqrt{0.5} .
This implies:
x - 2.5 < -\sqrt{0.5} or x - 2.5 > \sqrt{0.5} .

Solving gives:
x < 2.5 - \sqrt{0.5} or x > 2.5 + \sqrt{0.5} .

Recall $\sqrt{0.5} = \frac{\sqrt{2}}{2}$, so:
The positive domain is: 2.5 - \frac{\sqrt{2}}{2} < x < 2.5 + \frac{\sqrt{2}}{2} .
The negative domain is: x < 2.5 - \frac{\sqrt{2}}{2} or x > 2.5 + \frac{\sqrt{2}}{2} .

Therefore, the correct answer based on the choices provided is:

x<0:2\frac{1}{2}-\frac{\sqrt{2}}{2} or x > 2\frac{1}{2} + \frac{\sqrt{2}}{2}

x > 0 : 2\frac{1}{2} - \frac{\sqrt{2}}{2} < x < 2\frac{1}{2} + \frac{\sqrt{2}}{2}