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Existence of a root between 2 given points

by sara_87
Tags: existence, points, root
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sara_87
#1
Mar4-12, 10:09 AM
P: 773
1. The problem statement, all variables and given/known data

Show that there exists one root int (0,2) of the following function:

f(x)=(1-x^2)^2-√((1-x^2)*(1-1/2*x^2))

2. Relevant equations



3. The attempt at a solution

I first found:
f(0)=0
and
f(2)=7.268

But, i don't know what to do now. I'm not sure if it has something to do with the mean value theorem or not, or if i have to differentiate or not.

Any ideas will be very much appreciated.
Thank you.
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HmBe
#2
Mar4-12, 10:19 AM
P: 45
Well you've found a root at x=0, f(0)=0, but this isn't in the interval you're interested in, so it's best to try another value.

With finding a root in an interval your best bet is to try and find an a and a b in the interval such that f(a)<0 and f(b)>0, then by the intermediate value theorem there will be a root.

You've already got your b, such that f(b)>0, so now you just need to find your a.
sara_87
#3
Mar4-12, 10:23 AM
P: 773
thank you.
so, i can take b=0.1
f(0.1)=-0.0124

but, this show that there exists at least 1 root in the interval, but i need to show that there exists exactly one root.
how would i go from there?
doesn't have something to do with the derivative?

MathematicalPhysicist
#4
Mar4-12, 10:25 AM
P: 3,243
Existence of a root between 2 given points

Well if I read your function correctly (if I am not mistaken you're a PhD student, then why don't you use TeX?), it's:
[tex]f(x)=(1-x^2)^2-\sqrt{(1-x^2)(1-\frac{1}{2} x^2)}[/tex]

If this is your function then we can see that at x=1 the function vanishes, and you also know that at the boundaries at x=0 it equals 0 and at x=2 you get a positive function.

Now you should prove that for x in (0,1] there isn't another zero as in f(x) is always positive in this interval (non-negative), it's just a matter of playing with inequlities.

If you prove that f had the same sign all over the interval (0,1] then now you should show that if it were negative in (0,1) then in (1,2) it's a positive.

I'll show you the second part:
for x in (1,2) we have (1-x^2)^2>(1-4)^2=9
[tex]\sqrt{(1-x^2)(1-1/2 x^2)} < \sqrt{1/2}[/tex]
so you get [tex]f(x)>9-\sqrt{1/2}>0[/tex].

Hopefully, I didn't get it wrong (with me it's always a possibility).
MathematicalPhysicist
#5
Mar4-12, 10:30 AM
P: 3,243
Wait a minute, you want to show there exists exactly one, or that exists such root.

Cause I can now that there exist more than one root.
sara_87
#6
Mar4-12, 10:40 AM
P: 773
thank you,
but I don't understand, where did you get the 4 from in:
(1-4)^2 ?

also, what do we do if we assume that we don't know that x=1 is a root?
sara_87
#7
Mar4-12, 10:40 AM
P: 773
we want to show that there exists exactly 1 root
HmBe
#8
Mar4-12, 10:43 AM
P: 45
Quote Quote by sara_87 View Post
also, what do we do if we assume that we don't know that x=1 is a root?
I'd follow the method I said above for finding a root, to show that there is only one root I'd just do a rough sketch of the function in the interval you're interested in, because you're right - with a trickier function (something you're more likely to use IVT on) you won't know where the root is, so the inequality thing is much harder to do.

However, without looking at your mark scheme, I wouldn't know if this is enough. A sketch usually suffices for us though.
sara_87
#9
Mar4-12, 10:47 AM
P: 773
I have plotted the graph on MATLAB, and i can see that there is exactly one root (namely x=1)
but we have to use another method (not a graphical method). We were told that finding the derivative is a clue, but i don't see why.
sara_87
#10
Mar4-12, 10:48 AM
P: 773
I'm not sure if Rolle's theorem comes into this content?
HmBe
#11
Mar4-12, 11:10 AM
P: 45
Can you show the function is even, and thus symmetrical? Then it will have another root in (-2,0).

Then you could show using the derivative that the root at x=0 is repeated, and because the root is quartic can only have one more root...
sara_87
#12
Mar4-12, 11:35 AM
P: 773
Yes, the function is even since
f(-x)=f(x)
(since x appears only as x^2).

But, I don't understand the next part:
The derivative is

f'(x)= [itex]-4 x \left(1-x^2\right)-\frac{-x \left(1-x^2\right)-2 x \left(1-\frac{x^2}{2}\right)}{2 \sqrt{\left(1-x^2\right) \left(1-\frac{x^2}{2}\right)}}[/itex]

should i solve f'(x)=0?
HmBe
#13
Mar4-12, 11:39 AM
P: 45
Well I suppose you could cheat a little, showing that f'(0)=0 and that f(0)=0 (which you've already done) shows there is a repeated root at x=0, then you don't have to solve anything.
sara_87
#14
Mar4-12, 11:46 AM
P: 773
Yes, I think that's what we are supposed to do.

but, how does showing that f'(0)=0 and that f(0)=0 show that there is a repeated root at x=0?
HmBe
#15
Mar4-12, 11:53 AM
P: 45
I was pretty sure the definition of a repeated root was that if there is a root at x=a, that is f(a)=0, and f'(a)=0, then the root is repeated. Thinking about it, it seems to make sense, but I couldn't find anything on the internet to verify this.
Ray Vickson
#16
Mar4-12, 11:53 AM
Sci Advisor
HW Helper
Thanks
P: 5,081
Quote Quote by HmBe View Post
Well you've found a root at x=0, f(0)=0, but this isn't in the interval you're interested in, so it's best to try another value.

With finding a root in an interval your best bet is to try and find an a and a b in the interval such that f(a)<0 and f(b)>0, then by the intermediate value theorem there will be a root.

You've already got your b, such that f(b)>0, so now you just need to find your a.
There may be a problem with this approach in this case: as written, the function f(x) does not exist (in the reals) for 1 < x < sqrt(2). So, application of the intermediate-value method needs to be done with great care.

PS: I assume the OP means
f(x) = (1-x^2)^2 - sqrt[(1-x^2)(1 - (x^2/2))], not (1-x^2)^2 - sqrt[(1-x^2)(1 - 1/(2x^2))].

RGV
HmBe
#17
Mar4-12, 11:58 AM
P: 45
Oh yes, just realised I made a pretty big mistake when I plotted the function, and that it's not continuous, so applying IVT would be very tricky.
sara_87
#18
Mar4-12, 12:02 PM
P: 773
Oh right, i see.
If we assume we have a different function that IS continuous in (0,2) and that the IVT does apply, then we can use that f(0)=0 and f'(0)=0
But, what do you mean by:
''because the root is quartic can only have one more root''?


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