Langrange's Mean Value Theorem question

[Ryuzaki]

Homework Statement

Find a function f on [-1,1] such that :-

(a) there exists c \in (-1,1) such that f'(c) = 0 and
(b) f(a) \neq f(b) for any a\neq b \in [-1,1]

Homework Equations

Lagrange's Mean Value Theorem (LMVT), which states that if f:[a,b]-->ℝ is a function which is continuous in [a,b] and differentiable in (a,b), there exists some c\in (a,b) such that f'(c) = (f(b) - f(a))/(b-a).

The Attempt at a Solution

If f was said to be differentiable in (-1,1), then it's quite easy to prove that such a function cannot exist, using LMVT. But here, nothing is said about the continuity or differentiability of f in (-1,1). The function could just be differentiable at the point x = c.

I'm quite sure that such a function cannot exist. However, this is purely intuitive, and I'm unable to prove it. Can someone please hint on a proof, or an example if there's any?

Thank you.

 

[haruspex]

If f was said to be differentiable in (-1,1), then it's quite easy to prove that such a function cannot exist, using LMVT.

I don't see that. In fact, I don't see the relevance of LMVT here. Were you told to use it somehow? Unless I've misunderstood something, there's quite an easy (and differentiable) solution.

 

[Ryuzaki]

Thank you, haruspex for your reply. No, I wasn't told to use LMVT, but that was the only way I could think of.

Well, what I had in mind was assuming that such a function f exists, which is continuous in [-1,1] and differentiable in (-1,1), and f(a) not equals f(b), for any a not equals b, in [-1,1] (Sorry for my abstinence from symbols, I'm typing from my phone). Then by LMVT, there would exist some c such that

f'(c) = (f(1) - f(-1))/(1-(-1))

Now, if f'(c) were to be 0, then essentially f(-1) = f(1).

But we assumed that, f(a) not equals f(b), for any a not equals b, yet we got f(-1) = f(1). Hence, our assumption is wrong, and therefore such a function cannot exist.

But the above proof relies on the assumption that f is continuous in [-1,1] and differentiable in (-1,1). But this need not be the case. From the question, what I understand is that f could very well be differentiable ONLY at x = c.

Could you share your thoughts on this, haruspex? Please don't give out the solution, as I'd like to have the pleasure of solving it on my own. Thanks.

 

[haruspex]

Thank you, haruspex for your reply. No, I wasn't told to use LMVT, but that was the only way I could think of.

Well, what I had in mind was assuming that such a function f exists, which is continuous in [-1,1] and differentiable in (-1,1), and f(a) not equals f(b), for any a not equals b, in [-1,1] (Sorry for my abstinence from symbols, I'm typing from my phone). Then by LMVT, there would exist some c such that

f'(c) = (f(1) - f(-1))/(1-(-1))

Now, if f'(c) were to be 0, then essentially f(-1) = f(1).

Sure, there exists some c s.t. f'(c) = (f(1) - f(-1))/(1-(-1)), but why should f'(c)=0 for that c? f could quite easily be 0 somewhere else.
Think of some point at which f' is zero. Now extend f left and right satisfying f(a)≠f(b).