easy to understand, hard to solve

[moham_87]

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

[Njorl]

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.

[NateTG]

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 f(x) \neq 0 on the interval.

[HallsofIvy]

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).