2D graph of an inequality (of a 3 variable function domain)

[Guest]

Hey math mates! I have a (pretty dumb) question about inequalities, I'm preparing an partial exam for calculus and I'm struggling with this:

I have to find the domain of F(x,y)= arcsin ([tex]\frac{y}{x}[/tex]). These are basic exercises of one of the first topics, like an introduction to multivariable calculus.

Since the arcsin function is bounded between [-1;1], I put the restriction [tex]-1\le (\frac{y}{x}) \le 1[/tex] to the argument. Then, solving that, it finally is [tex]-x \le y \le +x[/tex]. They ask you to trace a graph and I find something like this:



But it has to be like this:



I just cannot realize how, analytically, deduce that the other side of the graph (corresponding to the reversed inequation: [tex]-x \ge y \ge +x[/tex]) also belongs to the domain. The book says that is either one or another of the possibilities, depending to the value of [tex]x[/tex] (negative or positive). Isn't it just implied in [tex]-x \le y \le +x[/tex] that the input can be [tex]x[/tex] positive or negative?

Because if you pull apart the double inequation [tex]-x \le y \le +x[/tex] into [tex]-x \le y[/tex] and [tex]y \le +x[/tex] it's also the same, isn't it? Just can't figure it out... i'm done. Can I be so dumb?


Cheers, Mark.

[HallsofIvy]

Hey math mates! I have a (pretty dumb) question about inequalities, I'm preparing an partial exam for calculus and I'm struggling with this:

I have to find the domain of F(x,y)= arcsin ([tex]\frac{y}{x}[/tex]). These are basic exercises of one of the first topics, like an introduction to multivariable calculus.

Since the arcsin function is bounded between [-1;1], I put the restriction [tex]-1\le (\frac{y}{x}) \le 1[/tex] to the argument. Then, solving that, it finally is [tex]-x \le y \le +x[/tex].

This is your basic mistake! If x is positive then multiplying both sides of [tex]-1\le \frac{y}{x}\le 1[/tex] by x gives [tex]-x\le y \le x[/tex]. However, if x is negative the multiplying by a negative number reverses the direction of the inequality. For example, -4< 0< 1 and multiplying each part by 2 gives -8< 0< 2 but multiplying by -2 gives 8> 0> -2.

If x is negative then multiplying each part of [tex]-1\le \frac{y}{x}\le 1[/tex] by x gives [tex]-x \ge y\ge x[/tex] or, equivalently, [tex]x\le y\le -x[/tex].