
#1
Mar412, 10:09 AM

P: 774

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)=(1x^2)^2√((1x^2)*(11/2*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. 



#2
Mar412, 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. 



#3
Mar412, 10:23 AM

P: 774

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? 



#4
Mar412, 10:25 AM

P: 3,173

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)=(1x^2)^2\sqrt{(1x^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 (nonnegative), 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 (1x^2)^2>(14)^2=9 [tex]\sqrt{(1x^2)(11/2 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). 



#5
Mar412, 10:30 AM

P: 3,173

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. 



#6
Mar412, 10:40 AM

P: 774

thank you,
but I don't understand, where did you get the 4 from in: (14)^2 ? also, what do we do if we assume that we don't know that x=1 is a root? 



#7
Mar412, 10:40 AM

P: 774

we want to show that there exists exactly 1 root




#8
Mar412, 10:43 AM

P: 45

However, without looking at your mark scheme, I wouldn't know if this is enough. A sketch usually suffices for us though. 



#9
Mar412, 10:47 AM

P: 774

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. 



#10
Mar412, 10:48 AM

P: 774

I'm not sure if Rolle's theorem comes into this content?




#11
Mar412, 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... 



#12
Mar412, 11:35 AM

P: 774

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(1x^2\right)\frac{x \left(1x^2\right)2 x \left(1\frac{x^2}{2}\right)}{2 \sqrt{\left(1x^2\right) \left(1\frac{x^2}{2}\right)}}[/itex] should i solve f'(x)=0? 



#13
Mar412, 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.




#14
Mar412, 11:46 AM

P: 774

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? 



#15
Mar412, 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.




#16
Mar412, 11:53 AM

HW Helper
Thanks
P: 4,670

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



#17
Mar412, 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.




#18
Mar412, 12:02 PM

P: 774

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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