# prove there is a c_1

• Apr 24th 2010, 08:16 PM
alice8675309
prove there is a c_1
Suppose f: [0,2]-> R is differentiable. f(0)=0, f(1)=2 and f(2)=2. Prove that

1. There is $c_1$ such that f'( $c_1$) =0
2. There is $c_2$ such that f'(c_2)=2 and
3. There is $c_3$ such that f'( $c_3$=1/2
• Apr 24th 2010, 08:48 PM
For (1) and (2), use the mean value theorem on the intervals [1,2] and [0,1] respectively. Since a derivative always has the intermediate value property (Darboux's theorem), it follows that there is a $c_3$ between $c_1$ and $c_2$ where the derivative is 1/2.
For (1) and (2), use the mean value theorem on the intervals [1,2] and [0,1] respectively. Since a derivative always has the intermediate value property (Darboux's theorem), it follows that there is a $c_3$ between $c_1$ and $c_2$ where the derivative is 1/2.