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How can I show f(x)=x2cosx=0 has exactly 2 solutions? 
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#1
Dec703, 08:01 PM

P: 364

How can I show f(x)=x^{2}cosx=0 has exactly 2 solutions? I can show it has at least two by using the fact that f is even and continuous and f(0)<0 and f(1)>0 so there's at least one solution in 0<x<1 and thus one solution in 1<x<0. I'm trying to derive a contradiction by assuming there's a third solution but I can't seem to find one.



#2
Dec703, 08:23 PM

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PF Gold
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One of the cuter ways to prove a function has no zeroes in a certain interval (e.g. [1, ∞) )is to prove that it's left endpoint is positive and that it's increasing throughout the interval
Another thing that might help is Rolle's theorem; if f(a) and f(b) are both zero, then f'(c) is zero for some c between a and b. 


#3
Dec703, 09:06 PM

P: 364

I can take it from here. Thanks 


#4
Dec703, 09:58 PM

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How can I show f(x)=x2cosx=0 has exactly 2 solutions?
To simplify things a little bit...
If you assume there are more than 2 solutions, there must be more than 2 positive solutions, and zero can't lie between those two solutions. 


#5
Dec703, 11:33 PM

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#6
Dec803, 12:13 AM

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You already noticed that it's an even function. If the 3rd solution is negative, that means there's a 4th solution that's positive. (or, you can just play with the two negative solutions, if you prefer)
Basically, the thing I wanted to get at is saying "Well, if there are three solutions, then by Rolle's theorem 2x = sin x has 3 solutions" is a non sequitor; in principle Rolle's theorem could return the same number three times. Given the actual problem at hand, it will give you the same number twice... but there's still that logical detail to prove that rolle's theorem doesn't give the same number a third time. 


#7
Dec803, 12:38 PM

P: 364

Ok well in that case there are at least two positive solutions. I think you made a mistake when you said there will be more than 2 positive solutions.
I figured that Rolle's theorem gives distinct x by applying it to disjoint intervals. If c, c_{p} are distinct positive solutions with c<c_{p}, then Rolle's theorem guanantees an x for each interval [c_{p},c], [c,c], [c,c_{p}] and each x will not be an endpoint of the interval. I found it necessary however to consider both positive and negative solutions. 


#8
Dec803, 05:05 PM

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Yes, that's what I meant; if there's a third solution, there must be a second positive solution!
Consider this: you know the only solution to 2x+sinx = 0 is x=0... so no solution can exist in the range [c, c_{p}]. 


#9
Dec803, 08:17 PM

P: 364

I should have seen that myself. 


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