# Rolle's Theorem

• Feb 27th 2010, 09:25 AM
MatthewD
Rolle's Theorem
I'm really stuck on this problem...

If I let the function:

f : II be continuous on I and differentiable on the open set I
for I := [0,1]

I know that theres at least 1 point t [0, 1] such that f(t) = t.

Now I need to use Rolle’s Theorem to show that if f'(x) is not equal to 1 in (0, 1), then there is exactly one such point t

I've tried 3 different proofs for this, but none of them are giving me uniqueness of t. Please help!
• Feb 27th 2010, 11:12 AM
tonio
Quote:

Originally Posted by MatthewD
I'm really stuck on this problem...

If I let the function:

f : II be continuous on I and differentiable on the open set I
for I := [0,1]

I know that theres at least 1 point t [0, 1] such that f(t) = t.

Now I need to use Rolle’s Theorem to show that if f'(x) is not equal to 1 in (0, 1), then there is exactly one such point t

This cannot be correct since $\displaystyle f(x)=x$ is a counterexample...Perhaps it should be that $\displaystyle f'(x)\neq 0$ in $\displaystyle (0,1)$ ? Because then it is easy to prove the point t must be unique...
Check this.

Tonio

I've tried 3 different proofs for this, but none of them are giving me uniqueness of t. Please help!

.
• Feb 27th 2010, 11:21 AM
MatthewD
It may be a typo on the worksheet...

How would I go about proving it if it was 0 instead? Either way, my proof methods weren't getting me anywhere.
• Feb 27th 2010, 11:27 AM
tonio
Quote:

Originally Posted by MatthewD
It may be a typo on the worksheet...

How would I go about proving it if it was 0 instead? Either way, my proof methods weren't getting me anywhere.

Assume $\displaystyle t_1\,,\,t_2\in[0,1]\,,\,\,t_1\neq t_2$ are s.t. $\displaystyle f(t_i)=t_i\,,\,\,i=1,2$. and apply directly Rolle's theorem to f in the interval $\displaystyle [t_1,t_2]$ ...

Tonio
• Feb 27th 2010, 11:34 AM
MatthewD
I'm a little confused as to why f(x) = x would be a counterexample (if there isn't a mistake)... because f'(x) = 1, so we don't have a problem since we are looking at $\displaystyle f'(x) \neq 1$...? or am I missing something here
• Feb 27th 2010, 05:59 PM
tonio
Quote:

Originally Posted by MatthewD
I'm a little confused as to why f(x) = x would be a counterexample (if there isn't a mistake)... because f'(x) = 1, so we don't have a problem since we are looking at $\displaystyle f'(x) \neq 1$...? or am I missing something here

My bad: I misread your first post and thought it said "if f'(x) IS EQUAL to 1 then there is a unique..." .

Tonio
• Feb 28th 2010, 01:25 AM
HallsofIvy
Right. If f'(x) is NOT 1 in [0, 1] f(x)= x is NOT a counter example and the statement as given is true.

Apply Rolle's theorem to the function g(x)= f(x)- x. If there are two points, x1 and x2, such that f(x1)= x1 and f(x2)= x2, then g(x1)= g(x2)= 0. Therefore by Rolle's theorem...