Analysis ?

Question

Suppose f : [0,2] →R is differentiable, f (0) = 0, f (1) = 2, and
f (2) = 2. Prove that
1. there is c1 such that f ′ (c1) = 0
2. there is c2 such that f ′ (c2) = 2
3. there is c3 such that f ′ (c3) = 3/2

Answer 1

The mean value theorem states that if f is continuous on [a, b] and differentiable on (a, b), then ∃c in (a, b) such that f'(c) = (f(b)-f(a))/(b-a).

#1: Let a=1, b=2, by MVT, ∃c1: f'(c1) = (2-2)/(2-1) = 0
#2: Let a=0, b=1, by MVT, ∃c2: f'(c2) = (2-0)/(1-0) = 2
#3: this one is trickier. Let g(x): [0, 1] → R be the function f(x+1)-f(x). Then g(0)=2, g(1)=0, 0<3/2<2, and g is a continuous function, so by the intermediate value theorem, there exists some a such that g(a) = 3/2. Let a satisfy g(a) = 3/2, b=a+1, then by MVT, ∃c3: f'(c3) = (3/2)/(1) = 3/2.

Answer 2

1 Mean Value Theorem
2 Mean Value Theorem
3 Prove that derivatives have the intermediate value property (note that derivatives need *not* be continuous!) and then use parts 1 and 2.