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

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

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 ($$\frac{y}{x}$$). 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 $$-1\le (\frac{y}{x}) \le 1$$ to the argument. Then, solving that, it finally is $$-x \le y \le +x$$. 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: $$-x \ge y \ge +x$$) also belongs to the domain. The book says that is either one or another of the possibilities, depending to the value of $$x$$ (negative or positive). Isn't it just implied in $$-x \le y \le +x$$ that the input can be $$x$$ positive or negative?

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

Cheers, Mark.
Guest

### Re: 2D graph of an inequality (of a 3 variable function doma

Guest wrote: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 ($$\frac{y}{x}$$). 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 $$-1\le (\frac{y}{x}) \le 1$$ to the argument. Then, solving that, it finally is $$-x \le y \le +x$$.

This is your basic mistake! If x is positive then multiplying both sides of $$-1\le \frac{y}{x}\le 1$$ by x gives $$-x\le y \le x$$. 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 $$-1\le \frac{y}{x}\le 1$$ by x gives $$-x \ge y\ge x$$ or, equivalently, $$x\le y\le -x$$.

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