I came acroos the following function today which gave different results for its differentiation. Here is it
$$f(x)=\lfloor x\rfloor\text{ at }x \in [1,2]$$.
In the above function, the value of f(x) is always 1. Lets consider f'(x) =0, for all x $\in$ (1,2). But when I use Mean Value Theorem(MVT), I get,
$$f'(c) = \frac{f(2)-f(1)}{2-1} = 1 \neq 0$$
My guess: Floor function creates discontinuity, while MVT checks for values in a continuous function. Is the contradictory answer justified by my guess?.
Thanks