• Support PF! Buy your school textbooks, materials and every day products Here!

Domain: deceivingly simple, turns evil

  • #1
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?
 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
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=1*1

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, a condidtion we do not have in the inequality case.
 
  • #3
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.
 
  • #4
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
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.
 
  • #5
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 :wink: .
 
  • #6
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
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.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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.
 

Related Threads on Domain: deceivingly simple, turns evil

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
6
Views
1K
Replies
19
Views
8K
  • Last Post
Replies
1
Views
2K
Replies
5
Views
5K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
71K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
9
Views
4K
Top