Let f(x) = 4sin(x). What is the absolute value of f'(x) <= ? By Mean Value Theorem, absolute value of (f(a)-f(b)) <= [blank] absolute value of (a-b) for all a and b.

Answer 1

f(x) = 4sinx

f'(x) = 4cosx So, l f'(x) l = l 4cosx l = 4 l cosx l ≤ 4(1) = 4, for all x.

By the MVT, there is at least one number, c, in the interval (a, b) such that f'(c) = [f(b) - f(a)] / (b - a).

So, f(b) - f(a) = f'(c) (b - a)

l f(b) - f(a) l ≤ 4 l b - a l