I've been trying this question and I tried using Langrange's Mean Value Theorem but I don't know the derivative will help in getting the answer. Help? The options are:
a)$ f(101) = f(202) + 8 $
b)$ f(101)= f(200) +2 $
c)$ f(101) = f(201) +1 $
d) None of these
Write this as follows:$$|f(x+h) - f(x)| \le 7 |h|^{201}$$Or, if you see where this is going,$$\left| \frac{f(x+h) -f (x)}{h}\right| \le 7 |h|^{200}$$In other words,$$-7|h|^{200} \le \frac{f(x+h) -f (x)}{h} \le 7 |h|^{200}$$Take the limit as $h \to 0$. By squeeze theorem, it's zero. It is, therefore, a constant function because $f'(x) = 0$.