# Thread: Real Analysis

1. ## Real Analysis

Let f:[0,2] -> R be continuous on [0,2] and differentiable on (0,2) with f(x)=0 and f(1)=f(2)=1

Show that there is a c(1) in (0,1) with f ’(c(1)) = 1

Show that there is a c(2) in (1,2) with f ‘(c(2)) = 0

Show that there is a c(3) in (0,2) with f ‘ (c(3)) = 1/3

2. Originally Posted by dhhnerd
Let f:[0,2] -> R be continuous on [0,2] and differentiable on (0,2) with f(x)=0 and f(1)=f(2)=1

Show that there is a c(1) in (0,1) with f ’(c(1)) = 1

Show that there is a c(2) in (1,2) with f ‘(c(2)) = 0

Show that there is a c(3) in (0,2) with f ‘ (c(3)) = 1/3
These are just applications of the Mean Value Theorem, I think for #3 you want to say $f(c_3) = 1/2$.

In that case just by MVT we get:
1) $f(1)-f(0)=f'(c_1)(1-0)\implies f'(c_1)=1$
2) $f(2)-f(1)=f'(c_2)(2-1)\implies f'(c_2)=0$
3) $f(2)-f(1)=f'(c_3)(2-0)\implies f'(c_3)=1/2$