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Easy to understand, hard to solveby moham_87
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#1
Jan804, 03:56 PM

P: 13

hi
that is my first time, and i have questions for my exam. * Show that if a function "f" is continuous and has no zeros on an interval, then either f(x)>0 or f(x)<0 for every "x" in the interval (Intermediate Value Theorem) I solved it in that way: since no zeros in the interval then the interval will be (inf,0)U(0,inf) so f(x) will not be 0.... but that can be possible, and didn't know how to do it??? please i need the answer in maximum 2 days any efforts will be appreciated thank you 


#2
Jan804, 04:33 PM

Sci Advisor
P: 875

I always had trouble with these. I always would just think, "that's true by definition. How can I prove it?"
First, assume that it is not true. Then, for some value of x on the interval, f(x) and f(x+dx) will have opposite signs, so that: f(x)f(x+dx)<0 (positive * negative is always <0) using the basic definition of a derivitive, f'(x)=[f(x+dx)f(x)]/dx you can solve for f(x+dx)=dx*f'(x)+f(x) substituting, you get: f(x)*[dx*f'(x)+f(x)]<0 Take the limit as dx>0 and: f(x)*f(x)<0 The left side of the equation is a square, it can not be <0 So assuming the thing you are trying to prove is false yields an impossibility. I think this is good. Njorl PS I know I used sloppy notation, dx instead of delta x. I forgot how to make deltas. 


#3
Jan804, 07:09 PM

Sci Advisor
HW Helper
P: 2,537

Nice approach, but just because the function is continuous, doesn't mean that it's differentiable.
Lets say we have f(x) > 0 and f(y) < 0 with x,y on the interval. let a = min (x,y) and b=max(x,y) Clearly (a,b) is contained in the interval. Therefore f is continuous on [a,b]. It's also clear that 0 is between f(a) and f(a) Then by applying the intermediate value theorem we get hat there must be some point x on the interval [a,b] such that f(x)=0, but this contradicts the hypothesis that [tex]f(x) \neq 0[/tex] on the interval. 


#4
Jan904, 05:44 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,558

Easy to understand, hard to solve
How you would do this depends upon what prior theorems you have to work with. Have you had the "intermediate value theorem"? That says that if f(x) is continuous on [a,b] then, on [a,b], f takes on all values between f(a) and f(b).



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