Why is the domain and range of x^2+y^2=1 this answer?

Question

anonymous · Answer

$$D = {-1\le1 \le1}$$

anonymous · Answer

maybe 
$$-1\leq x\leq 1$$

turingtest · Answer

you mean$$-1\le x\le 1$$what if x=2 ?
what would y be?

anonymous · Answer

$$R = -1 \le y \le 1$$

anonymous · Answer

ohh I meant that

anonymous · Answer

yes you are right, it is the unit circle

anonymous · Answer

I get so confused with domains and ranges D:

anonymous · Answer

like why is it -1 less than or equal to x less than or equal to 1?

turingtest · Answer

what if $$|y|>1$$what would x be?

anonymous · Answer

i don't know.

turingtest · Answer

try it with x=2
what would y be?

anonymous · Answer

what do you mean?

experimentx · Answer

can't call it a function though ... it's a closed curve.

anonymous · Answer

I just don't understand why it's greater than or less than.. o.o

turingtest · Answer

@milliex51 can you solve the above for y ?
an experimX is correct as usual, but that doesn't really matter for what we're talking about

turingtest · Answer

solve$$x^2+y^2=1$$for $y$ please :)

anonymous · Answer

$$y = +/- \sqrt{-x}+1$$

anonymous · Answer

-x^2

turingtest · Answer

not quite...

anonymous · Answer

really?

anonymous · Answer

or do I close bracket the 1 too?

turingtest · Answer

$$y=\pm\sqrt{1-x^2}$$

anonymous · Answer

ohh..

turingtest · Answer

$$x^2+y^2=1$$$$y^2=1-x^2$$$$y=\pm\sqrt{1-x^2}$$now what if x=2 ?
what do you get for y ?

experimentx · Answer

now if we redefine it x^2+y^2=1 by
y = sqrt(1-x^2) &
y = -sqrt(1-x^2) on a field of real numbers than ..

turingtest · Answer

^echo, echo.... :P

experimentx · Answer

TuringTest was thinking exactly same as me

anonymous · Answer

isn't it the same from my answer?
um, x=2 would be y=non real?

turingtest · Answer

and if y is not real, then x=2 is not in the domain, right?

anonymous · Answer

right.

turingtest · Answer

what about any $$|x|>1$$?
what will y be?

anonymous · Answer

how do I solve that?

turingtest · Answer

I asking you if you plug in any number greater than 1 in for x into the formula we have for y, what will y e?

turingtest · Answer

*what will y be?

anonymous · Answer

i don't know D:

turingtest · Answer

ok, look at it this way...

turingtest · Answer

what is under the radical in$$y=\pm\sqrt{1-x^2}$$cannot be negative, right?
otherwise y would be imaginary.
do we agree?

anonymous · Answer

yes, yup.

turingtest · Answer

so that mean that we have a requirement that$$1-x^2\ge0$$right?

anonymous · Answer

yup. or else it will be undefined

anonymous · Answer

or imaginary

turingtest · Answer

so solve that requirement for x and you will get your domain

anonymous · Answer

but how? ahhh

turingtest · Answer

$$1-x^2\ge0$$$$1\ge x^2$$perhaps here is where you get a little confused...$$1\ge x\ge -1$$

anonymous · Answer

ohh.. I transpose the negative x^2

turingtest · Answer

remember that x^2 is always positive, so all that matters is that$$|x|\le1$$and y will be real

anonymous · Answer

how about when I have to look at a graph?

anonymous · Answer

Like:|dw:1333035682317:dw|

anonymous · Answer

what signs am I going to use? o.o

turingtest · Answer

|dw:1333035736039:dw|we are saying that x can take on any value in the black region, right?