Domain: deceivingly simple, turns evil

[StephenPrivitera]

I wonder what went wrong here:
Find the domain of f(x)=sqrt(1-sqrt(1-x²))
So of course the term under the radical must be greater than or equal to zero. For neatness I ignore the equal part.

1-sqrt(1-x²)>0
1>sqrt(1-x²)
both left and right are positive, so
1>1-x²
x²>0

But it seems to me that the domain is actually [-1,1]

You can do this alternatively by finding the domain of the innermost radical, which is [-1,1], and intersecting it with the set of all x such that sqrt(1-x²) is in the domain of h(y)=sqrt(1-y). Since sqrt(1-x^2)>=0 for all x in its domain, the intersection is just [-1,1]. So what happened wrong with the first approach?

 

[quantumdude]

Originally posted by StephenPrivitera
1-sqrt(1-x²)>0
1>sqrt(1-x²)
both left and right are positive, so
1>1-x²

You cannot square both sides of an inequality, because that implies that both sides are equal!

See, it works for an equation as follows:

(1-x²)^1/2=1
((1-x²)^1/2)²=1²

Note carefully that what we are really doing is:

(1-x²)^1/2*(1-x²)^1/2[/color]=1*1[/color]

But remember our rule that "whatever you do to one side, you have to do to the other side".

Question: How is it that we can square both sides, when we are multiplying the left by (1-x²)^1/2 and the right by 1?

Answer: Because in the beginning we asserted that those two quantities are equal[/color], a condidtion we do not have in the inequality case.

 

[StephenPrivitera]

I suspected that might have been the case. But then I thought, if we write a>b as a=b+c for some c in R, where a,b,c are greater than 0, then
a²=b²+(2bc+c²)=b²+d. d is positive and in R. So, by my definition, a²>b².

 

[quantumdude]

Originally posted by StephenPrivitera
I suspected that might have been the case. But then I thought, if we write a>b as a=b+c for some c in R, where a,b,c are greater than 0, then
a²=b²+(2bc+c²)=b²+d. d is positive and in R. So, by my definition, a²>b².

Yeah, but you squared an equation, and then drew an inference about an inequality, which is perfectly valid. At no point did you perform the invalid step of squaring the inequality.

 

[StephenPrivitera]

Yes, but doesn't this show that if a>b and a>0 and b>0 then a²>b²?
So, since 1>sqrt(1-x²)
and 1>0 and sqrt(1-x²)>0 for all x in it's domain
then 1>(1-x²).
I suspect that the problem lies in the fact that I have to comment
sqrt(1-x²)>0 for all x in it's domain.
I believe that when I get to x²>0, this means that the inequality holds not for all x but for for all x in the domain of sqrt(1-x²). I must be restricting the values of x when I decide to square each side.
I understand that in general you cannot square an inequality. For example, -5>-10 does not imply 25>100. But as far as I can tell squaring an inequality is ok if both the left and right are positive. Of course, I've been wrong before .

 

[quantumdude]

Originally posted by StephenPrivitera
Yes, but doesn't this show that if a>b and a>0 and b>0 then a²>b²?

Ah, you're right, it does.

But something is going screwy, because I can just as easily prove that the domain is [-1,1] if I only use:

1>(1-x²)^1/2

Because if I multiply the right by (1-x²) and the right by 2, then I certainly can preserve the inequality:

2>=(1-x²)

and get x²=<1, which leads to [-1,1]. In fact, you can pretty much make the domain anything you want if that inequality is all you use. Just multiply the left by some number a>0, and the right by some number 0<b<a.

 

[HallsofIvy]

Irrespective of squaring or not squaring, you determined the condition 1- [sqrt](1-x^2) is positive. In order for that to make sense you still need to determine the conditions under which
[sqrt](1-x^2) EXISTS- i.e. when 1- x^2>= 0.