You don't say what your difficulty is and it seems fairly clear to me. Perhaps it would make more sense to phrase it a little differently. Suppose f'(c)= 0 for all c in I. Then for any two numbers, a and b, in I, the mean value theorem says "there exist c in I such that . From that it follows that f(a)- f(b)= 0(a- b)= 0 so that f(a)= f(b). That is, for any two numbers a and b in I, f(a)=n f(b).

Suppose f'(c)> 0 for all c in I. For b>a, by the mean value theorem there exist c in I such that so that f(b)- f(a)= f'(c)(b-a) f'(c)> 0 by hypothesis and b- a> 0 because b> a. Therefore f'(c)(b- a)> 0 so f(b)- f(a)> 0 and f(b)> f(a).