Challenge problem to the community - - - domain and range

[checkitagain]

What are the domain and range of the following:
(Note: It is a relation that is not a function.)



$y^2(x^2 - 1) = x^4$

 

[dextercioby]

Since there's no one to one correspondence between a value for x and a value for y, you have 2 different mappings (functions in their own right). y_1 (x) and y_2 (x) with the same maximal domain (R-{+-1}). y_1 is minus 1 times y_2, for any allowable value of x.

 

[checkitagain]

Since there's no one to one correspondence
between a value for x and a value for y, you have 2 different mappings
(functions in their own right). y_1 (x) and y_2 (x) with the

$> >$ same maximal domain (R-{+-1}). $< <$

y_1 is minus 1 times y_2, for any allowable value of x.

.



I take it you are only addressing the domain . . . for now.


With your notation, does (R - {+-1}) mean $(-\infty, -1) \ \ \cup \ \ (-1, 1) \ \ \cup \ \ (1, \infty) \ ?$


Or, does it mean $(-\infty, -1) \ \ \cup \ \ (1, \infty) \ ?$



I am asking you for your clarification before I comment further on the
possible/alleged correctness of your take on it. (I want to comment,
but I am waiting on some more information.)

 

[dextercioby]

I was too lazy to write down the LaTex code, but here goes

$$\mathbb{R} - \{\pm 1\}$$.

 

[checkitagain]

I was too lazy to write down the LaTex code, but here goes

$$\mathbb{R} - \{\pm 1\}$$.

What about x = -1/2 or x = 1/2, for instance?


Would those work or not?

 

[dextercioby]

No, y would be imaginary.

 

[checkitagain]

No, y would be imaginary $> > \ \ regarding \ \ x \ = -1/2 \ \ and \ \ x \ = 1/2 \ \ < < .$

(R - {+-1}) means $(-\infty, -1) \ \ \cup \ \ (-1, 1) \ \ \cup \ \ (1, \infty) \ ?$


Or, does it mean $(-\infty, -1) \ \ \cup \ \ (1, \infty) \ ?$

So, then, we must eliminate the first choice of interval notation
in the above quote box. Is the second choice of interval notation
in the quote box correct?


Can x = 0 give y as a real value or not?