Proving Uniqueness of t with Rolle's Theorem

[MatthewD]

Homework Statement

Let the function:


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


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

Homework Equations

I know that there's at least 1 point t ∈ [0, 1] such that f(t) = t.

The Attempt at a Solution

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

 

[snipez90]

Did you mean that there is exactly one t for which f(t) = t in your problem statement? As it stands, it's reading as if f'(t) does not equal 1 in (0,1), then there is exactly one point for which it does not equal 1, which makes no sense.

Assuming you meant the former statement. You're right in that there is at least one, which is due to the simplest of fixed point theorems. Now if there were two such points, say x_1 and x_2 consider g(x) = f(x) - x and apply Rolle's theorem to get a contradiction.

 

[Dick]

Suppose there are two points t1 and t2 such that f(t1)=t1 and f(t2)=t2. Apply Rolle's theorem to the function g(t)=f(t)-t on the interval [t1,t2].

 

[MatthewD]

So this is how I've been starting, as you said, and I think I'm confusing myself... can't I only apply Rolle's Theorem if f(x1) = f(x2)?

I have:
suppose f'(x) not= 1 on [0,1] and suppose there exists 2 pts, x1 and x2 in [0,1] such that
f(x1) = x1
f(x2) = x2
and let some function g be defined as g(x):=f(x)-x for x in [x1,x2]

To apply Rolle's Theorem, don't I need g(x1) = g(x2)?

I appreciate your help, I'm just still confused as to how I can actually apply Rolle's :(

 

[VeeEight]

But what is g(x₁)? It is f(x₁) - x₁ = 0, since x₁ is a fixed point of f. Similarly for g(x₂). So Rolle's Theorem tells you something about g.

 

[MatthewD]

Which is clearly a contradiction! I realized my mistake: at the end I kept using Rolle's to say there exists a point c where f'(c)=0 instead of looking at the new function. It seems so easy now... I'm sorry! Thank you for your time, I really appreciate it!