a Max-Min Problem

Here's the problem:

A function f is everywhere differentiable.
We are asked to find the minimum value of y = f(x), on -1 ≤ x ≤ 1.
We are given that f(0) = 1.
How would you proceed?

>I always start by differentiating, to see if the derivative is zero.
Okay, let's start that way:

  1. We have y = f(x), and, to solve dy/dx = 0, we first let y = u ex.
  2. Then dy/dx = du/dx ex + u ex = 0 gives du/dx + u = 0 ... since ex ≠ 0.
  3. But du/dx + u = 0 has solutions:   u = C e-x.
  4. Then y = f(x) = u ex = C, a constant.
  5. Since we're given f(0) = 1, then C = 1 .... and that must be the minimum value we seek.
>That's it?
That's it. See anything wrong with our solution?
>Uh ... looks okay to me.
But we found a minimum and we didn't even know the function f.

>But you determined that f(x) = C. Have you forgotten already?
Are you saying that the only everywhere-differentiable functions are constant on [-1,1]?
We got that by step #4.

>Hmmm ... that's interesting, isn't it?
So where's the error in our argument?

>Our argument? It ain't mine!
In steps 2 and 3 we're assuming that dy/dx = 0 everywhere, not just at the point where a minimum occurs.

>So it ain't surprising that f(x) = constant, eh?
You're getting schmarter.