Restriction on Domain

[Jason76]

For the equation \(\displaystyle y = \pm(1 - x^{2})^{1/2}\) where the domain is the x values, also known as the independent variable (as opposed to the y values, the dependent variable)

If x is greater than \(\displaystyle 1\) than the answer becomes negative and negatives cannot exist in a square root (The 1/2 power is the same as a square root). However, the book also says that the domain cannot be less than \(\displaystyle -1\). However, negative numbers plugged into x don't produce negative numbers. Therefore, the can exist under a square root sign. So don't understand the book's logic.

 

[pka]

For the equation \(\displaystyle y = \pm(1 - x^{2})^{1/2}\) where the domain is the x values, also known as the independent variable (as opposed to the y values, the dependent variable)
If x is greater than \(\displaystyle 1\) than the answer becomes negative and negatives cannot exist in a square root (The 1/2 power is the same as a square root). However, the book also says that the domain cannot be less than \(\displaystyle -1\). However, negative numbers plugged into x don't produce negative numbers. Therefore, the can exist under a square root sign. So don't understand the book's logic.

For any number \(\displaystyle a\) the \(\displaystyle \sqrt{a}\) exist only if \(\displaystyle a\ge0\).

Therefore, \(\displaystyle 1-x^2\ge 0\) or \(\displaystyle -1\le x\le 1~.\)

 

[HallsofIvy]

For the equation \(\displaystyle y = \pm(1 - x^{2})^{1/2}\) where the domain is the x values, also known as the independent variable (as opposed to the y values, the dependent variable)

If x is greater than \(\displaystyle 1\) than the answer becomes negative and negatives cannot exist in a square root (The 1/2 power is the same as a square root). However, the book also says that the domain cannot be less than \(\displaystyle -1\). However, negative numbers plugged into x don't produce negative numbers. Therefore, the can exist under a square root sign. So don't understand the book's logic.

I don't understand your logic!

The x in that formula is squared so if x= -3, say, then \(\displaystyle x^2= 9\) and \(\displaystyle 1- x^2= -8\).

 

[Crimson Sunbird]

However, negative numbers plugged into x don't produce negative numbers. Therefore, the can exist under a square root sign. So don't understand the book's logic.

The square root is over \(\displaystyle 1-x^2\), not over \(\displaystyle x^2\), so it is the former we need to ensure is not negative.

If \(\displaystyle x<-1\), say \(\displaystyle x=-2\), you have \(\displaystyle 1-x^2=1-(-2)^2=1-4=-3\) which is negative – this is what matters. (The fact that \(\displaystyle x^2=(-2)^2=4\) is positive is not what matters.)

 

[Jason76]

Actually, I figured out the problem slightly after posting the problem. As someone noted, negative numbers that are squared are positive. So whether we plug in positive or negative into x, we still get positive. Finally, our next step is to evaluate the equation with the positive number (with the exception of 1) plugged in. In that case, we would always get a negative under a square root which isn't allowed.

 

[Deleted member 4993]

Actually, I figured out the problem slightly after posting the problem. As someone noted, negative numbers that are squared are positive. So whether we plug in positive or negative into x, we still get positive. Finally, our next step is to evaluate the equation with the positive number (with the exception of 1) plugged in. In that case, we would always get a negative under a square root which isn't allowed.

No... The absolute value of the number has to be less than or equal to 1. Or:

|x| ≤ 1 ...................or............................ -1 ≤ x ≤ 1

so x = - 0.9999 will work fine