posted by Zisis on .
Suppose that function f:[0,2]->R is continuous at [0,2], differentiable at (0,2) and such that f(0)=0, f(1)=1 and f(2)=1.
a) Show that there is a x1 that belongs to (0,1) such that f'(x1)=1.
b) Show that there is a x2 that belongs to (1,2) such that f'(x2)=0.
c) Show that there is a x3 that belongs to (0,2) such that f'(x3)=2/3.
a) (f(1)-f(0))/(1-0) = 1
MVT says x1 exists
b) f(1)=f(2), so MVT or Rolle's Theorem says x2 exists
c) since differentiable, the IVT applies to the derivative, which must assume all values between 0 and 1 (which we know it attains from the two parts above).