1. I see what you mean. They give $\displaystyle f(x)=x^2\sin(1/x^3)$ for x in (0,1], f(0)=0.

It's easy to check f'(x) is unbounded in this case. And then they conclude f'(x) is not integrable.....
It wouldnt be too hard to make such function with f(1)=1 and f(0)=0.

I'm confused right now. I think you may be right after all.

2. Originally Posted by Unbeatable0
That's not true. A derivative need not in general be integrable over a closed interval.
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3. Anyway. To correct my solution in that case, I'll just 'steal' your idea. Again we assume $\displaystyle f(x)\neq x$

since we only need to show the existence of $\displaystyle a,b$ with $\displaystyle f'(a)<1, f'(b)>1$ we need to show that $\displaystyle f'(x)\leq 1$ and $\displaystyle f'(x)\geq 1$ for all $\displaystyle x$ is impossible. Then we're done.

Now, $\displaystyle g(x)=f(x)-x$ is increasing/decreasing. And since $\displaystyle g'(x)\neq 0$for all $\displaystyle x$ we can't have $\displaystyle g(1)=0$.
Now again the intermediate value theorem guarantuees $\displaystyle f'(x),f'(y) \in (f'(a),f'(b))$such that $\displaystyle f'(x)f'(y)=1.$