Let f(x)=⌊x⌋, x∈[1,2]. Clearly, f(x)=1 for all x∈(1,2), and thus f′(x)=0 for all x∈(1,2). But if someone applies the Mean Value Theorem, then there exists c∈(1,2) such that
f′(c)=f(2)−f(1)/2−1=2−1/1=1≠0.
Analyze the reason why the contradiction appears.

Question

Let f(x)=⌊x⌋, x∈[1,2]. Clearly, f(x)=1 for all x∈(1,2), and thus f′(x)=0 for all x∈(1,2). But if someone applies the Mean Value Theorem, then there exists c∈(1,2) such that
f′(c)=f(2)−f(1)/2−1=2−1/1=1≠0.
Analyze the reason why the contradiction appears.

anonymous · Answer

So, using MVT is an illegal operation because f(x) is not continuous at the interval [1,2]. What else should I note?

anonymous · Answer

that is good enough
it must be differentiable on the open interval and continuous on the closed interval
since it isn't, no MVT

anonymous · Answer

Thanks @satellite73