Proving local minimum

[trap]

Suppose that c1 < c2 and that f takes on local maxima at c1 and c2. Prove that if f is continuous on [c1, c2], then there is at least one c in (c1, c2) at which f takes on a local minimum.

This question seems common sense, but does anyone know how to actually prove this?

[Tide]

Hint: Sketch the graph of f ' (x). Since f ' ' (c1) < 0 and f ' ' (c2) < 0 the graph must decrease as you move away from x = c1 and toward x = c2. Likewise, the graph must increase as you move away from x = c2 and toward x = c1. Therefore, f ' (x) must pass from a positive value to a negative value somewhere in (c1, c2).

[Hurkyl]

Don't forget that f might not be differentiable... though looking at derivatives might jump start your understanding of the problem.


Trap: do you know anything, in general, about the minima of continuous functions on closed intervals?

[trap]

thanks everyone for the responds.

I do not know about the minima of continuous functions on closed intervals, can u provide an answer to this? thanks :smile:

[Hurkyl]

Any continuous function on a compact set (such as a closed interval [a, b]) has a global minimum and maximum.

So, for your problem, it would suffice to prove that a and b can't be the (only) global minimum.


I don't know if you've had this theorem yet, though.

[trap]

m....I haven't learnt about global min/max theorems, but thanks anyways for your help and explainations.