# Math Help - Differentiable Func.

1. ## Differentiable Func.

Given that $f$ is function that's differentiable on the interval $[a,b]$ where $f'(x) \neq 1\,\,\, \forall x \in [a,b]$.

1.) Prove $f$ has at most 1 fixed point on the interval $[a,b]$. (ie., show there's at most 1 point $x \in [a,b]$ where $f(x) = x$

2.) By giving specific examples, show it's possible that $f$ either has no fixed points or that $f$ has 1 fixed point on $[a,b]$.

2. Originally Posted by Ideasman
2.) By giving specific examples, show it's possible that $f$ either has no fixed points or that $f$ has 1 fixed point on $[a,b]$.
what are the restrictions here? it seems to me you can come up with a lot of trivial examples for this. example, f(x) = 2 on the interval [0,1]

3. Suppose that there are two point in [a,b] such that $f\left( {x_1 } \right) = x_1 \,\& \,f\left( {x_2 } \right) = x_2$.

We can assume that $x_1 \le x_2$ so by the mean value theorem
$\left( {\exists c \in \left( {x_1 ,x_2 } \right)} \right)\left[ {f'(c) = \frac{{f\left( {x_2 } \right) - f\left( {x_1 } \right)}}{{x_2 - x_1 }} = 1} \right]$.

4. I thought f'(x) =! 1, but you have that f'(c) = 1..

I could see a contradiction if x_1 was strictly less than x_2, but it can also be equal, so perhaps I'm overlooking it. Then, if x_1 were equal to x_2, there'd be a contradiction. But you have the stipulation that it can be equal to it.

And to Jhevon: that was the exact question. I'm not sure if I can use a trivial example such as that, as this was a big question and I doubt it'd be that easy.

5. Originally Posted by Ideasman
I thought f'(x) =! 1, but you have that f'(c) = 1..

I could see a contradiction if x_1 was strictly less than x_2, but it can also be equal, so perhaps I'm overlooking it. Then, if x_1 were equal to x_2, there'd be a contradiction. But you have the stipulation that it can be equal to it.
yes, that's the contradiction. if we have more than 1 fixed point, we end up having something they told us we couldn't have in the first place, thus we must have at most one fixed point, as the theorem says