continuous and monotonous implies almost differentiable Let f be a strictly monotonic positive valued continuous function defined on $[a,b]$ such that $f(a) < a$ and $f(b)>b$ where $b>a>0$ then prove that there exist some $c\in (a,b)$ such that $f’(c)>1$ My approach: I know that if $f$ is strictly monotonous and continuous then $f$ is differentiable almost everywhere. As graph of $f$ crosses $y=x$, at some $c\in (a,b)$, $f’(c)>1.$ I am not fully convinced with my approach. Is this correct or I need differentiable condition in the problem

If you mean to assert that there exists $c$ such that $f$ is differentiable at $c$ and $f'(c)>1$ then what you're trying to prove is false. (It's probaby true under some weaker notion of "$f'(c)>1$".)

Say $K\subset [0,1]$ is the Cantor set and $g:[0,1]\to[0,1]$ is the "Cantor-Lebesgue function". Then $g(0)=0$, $g(1)=1$, $g$ is continuous and non-decreasing, $g'(x)=0$ for every $x\in[0,1]\setminus K$, and if $x\in K$ it's not hard to see that $g$ is not differentiable at $x$ (because if $x\in K$ then for every $n\ge1$ we have $x\in[j/3^n,(j+1)/3^n]$ with $g((j+1)/3^n)-g(j/3^n)=1/2^n$).

So if you define $$f(t)=g(t)-\frac13+\frac23t$$then $f$ is continuous and strictly increasing, $f(0)<0$, $f(1)>1$, but $f'=2/3$ at every point where $f$ is differentiable.