# Analysis- if f assumes max/min for x in (a,b) prove f'(x) = 0

1. Nov 14, 2006

### sarahr

hello,

i need to prove:
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if a differentiable function f:(a,b) ----> R (reals) assumes a max or a min at some x element of (a,b), prove that f'(x) = 0. why is this assertion false when [a,b] replaces (a,b)?
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-I'm stumped at where to start...because if i knew that f was continuous on [a,b] and f(a) = f(b), this would be Rolle's thm...which i think i could prove...

-or if f(a) did not equal f(b), then maybe i could get somewhere by mean value thm... (that is, if f was cont on [a,b])

-BUT i dont know that f is cont on [a,b]...

-the only thing i know given what I have is that since f is differentiable on (a,b) then its derivative f'(x) has the intermediate value property....can this get me anywhere?

-any help on how to get started on this proof would be greatly appreciated! thanks for reading through my lengthy post! :)

2. Nov 14, 2006

### StatusX

I don't understand the question. Are you saying that there is some x such that f(x)<=f(y) for all y in (a,b) (or >=). This does not imply f is constant. Just take f(x)=x^2 on (-1,1). Rather than try to guess what you mean, I'll let you clarify it.

3. Nov 14, 2006

### sarahr

no, the question does not say that f(x)<=f(y) (or vice versa.) what is typed between the two lines is the exact question from my text. thanks!

4. Nov 14, 2006

### StatusX

I'm sorry, I thought it meant f'(x)=0 for all x in (a,b), I didn't see x was the element where the max or min occurred. Then you can just use the definition of the derivative. You can show the left limit must be >=0 and the right must be <=0 (or the other way around, depending on if it's a max or a min), and so since these limits must be equal, they must be 0.

5. Nov 14, 2006

### NateTG

Differentiable is stronger than continuous. (i.e. for a function to be differentiable, it must also be continuous.)

You might consider a nice simple example, like:
$$f(x)=x$$
on
$$(0,1)$$
and
$$[0,1]$$

6. Nov 14, 2006

### sarahr

oh! this is very clear now. thanks so much!

sarah.