Question

in the lecture on mean value theorem (MVT) , it was said that if f'(x)=0 then the function f is constant in the interval a 12 years ago

Answer 1

The statement in the lecture is that if f'=0, f is constant. It wasn't specified as being constant over some specific interval, but implicitly it is constant over the entire domain for which f exists and f'=0.

The way we prove this is to use this form of the MVT:\[f(b)=f(a)+f'(c)(b-a)\]This is simply a rearrangement of the original form in which the MVT was presented, but it allows us to see at once that if f'=0, then the part after the plus sign goes away and we're left with f(b)=f(a). And here's the piece that answers your question: this is true for any two points a and b that we choose. Therefore f has to be constant.

Answer 2

It is possible for f(a) = f(b) and c to be a critical point such that f'(c)=0 .The function can then be a non const

ex : a parabola : y=x^2 in the domain (a,b)=(-1,1)
f(-1)=1 f(1)=1
take c=0; f'(c)=2x=0

f(a)=f(b) & f'(c)=0 but function is not const in (-1,1)

Answer 3

In your example, f'=0 at a particular point. This is not enough to satisfy the corollary of MVT we are discussing. The condition is that f'=0 over the relevant domain. Prof. Jerison did not state the condition in exactly these terms, but it is clear from the context that the condition requires f'=0 at every point in the relevant domain, not just at some individual point.

You may have consciously or unconsciously rejected this interpretation on the basis that when f'=0 at every point in the domain, the conclusion that f is constant seems too obvious to require proof. This is the main difficulty students have with the MVT: it tells us something that doesn't seem to require proof. Yet it does require proof, and it's absolutely crucial ("the key to everything," as Prof. Jerison says), because MVT provides the theoretical foundation on which the Fundamental Theorem of Calculus (which comes up a little later) is built.