Find Domains of (x-1)²-2: Analyzing Positive and Negative Regions

To find the positive and negative domains of the function $y = (x-1)^2 - 2$, we need to determine the points where the function intersects the x-axis, as these will mark changes in sign.

Step 1: Set the function equal to zero to find the roots.
$(x-1)^2 - 2 = 0$

Step 2: Move -2 to the other side and solve:
$(x-1)^2 = 2$

Step 3: Solve for $x$ by taking the square root of both sides:
$x - 1 = \pm \sqrt{2}$

Step 4: Solve for $x$ by isolating it:
$x = 1 \pm \sqrt{2}$

The roots are $x = 1 + \sqrt{2}$ and $x = 1 - \sqrt{2}$. These roots divide the x-axis into three parts.

Step 5: Evaluate the function behavior in each interval defined by these roots.

• For $x < 1 - \sqrt{2}$, pick a point such as nearly approaching zero value and test the sign.
• For $1 - \sqrt{2} < x < 1 + \sqrt{2}$, pick a midpoint value and test.
• For $x > 1 + \sqrt{2}$, pick a value greater than root for testing function positivity.

Step 6: Determine where the function is positive and negative:

• Within the interval $[1-\sqrt{2}, 1+\sqrt{2}]$, the function lies below the x-axis and is negative.
• Outside this interval, specifically $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$, the function lies above the x-axis and is positive.

The positive domain is $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$ and the negative domain is $1 - \sqrt{2} < x < 1 + \sqrt{2}$.

Therefore, the solution is:

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

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