Conjecture
Let $f$ be a continuous function from [a, b] to [a, b], and is differentiable
on (a, b).
If f is surjective then there exists x $\in$ (a, b) such that $|f'(x)| = 1$
Any counter example for this conjecture ?
**Addition after Kavi Rama Murthy'answer **, we can improve the problem by: If $f(a)\leq f(b)$ and f is surjective then there exists x $\in$ (a, b) such that $f'(x)= 1$
Suppose this is not true. By IVP of derivative either $|f'(x)| >1$ for all $x$ or $|f'(x)| <1$ for all $x$. In the first case $|f(b)-f(a)| =|b-a||f'(x)| >|b-a|$ for some $x$ but this contradicts the fact that $f$ maps $[a,b]$ into itself. In the second case choose $x$ and $y$ such that $f$ attains its maximum, say $M$, at $x$ and minimum, say $m$ at $y$. Then $M-m =|y-x| |f'(y)| < |y-x|\leq b-a$ for a suitable $y$ which is again contradiction since $f$ is surjective.
Consider the function $g: [a,b] \to \mathbb{R}$ defined by $g(x) = f(x) - x$ for all $x \in [a,b].$ Then, $g$ must be continuous and differentiable over $(a,b).$
We need to show that, there exists a point $x \in (a,b)$ such that $g'(x) = 0.$
Now, this would follow from the intermediate value property of derivatives.
