$f(x)$ is a differentiable and strictly increasing function on the open interval $(a,b)$.
Prove or disprove that $f'(c)>0\; \forall c\;\in (a,b)$.
I was trying to apply Lagrange's derivative theorem, but the condition that the function must be continuous on the closed interval $[a,b]$ is not given. Im assuming the answer should be false, but I can't prove it's false or give a counterexample. Can we find a way to apply Lagrange's theorem? Then for sure, it's a true affirmative.
$f:x\to x^3$ on $(-1,1)$ is a counter-example for your statement, as $f'(0)=3\times(0)^2=0$ .
The true implication is that if $f$ is differentiable on an open interval $(a,b)$ and strictly increasing, then $f'\geq0$ and there exists no open interval $(x,y)\subset(a,b)$ such that $\forall{t}\in(x,y) : f'(t)=0$, which means that $f'$ is never constant at zero on an open interval of $(a,b)$, since if it was the case, we'd get that $f$ is constant on that interval which would contradict the strict monotony.
In fact, suppose by contradiction that there exists $(x,y)\subset(a,b)$ on which $f$ is always zero, then let $t,h\in(x,y)$ such that $t\neq h$. By the mean value theorem, there exists $c\in(x,y)$ such that : \begin{equation} f'(c)=\frac{f(t)-f(h)}{t-h} \end{equation} Since $f$ is always zero on $(x,y)$, so in particular $f'(c)=0$, hence $f(t)=f(h)$ which contradicts the strict monotony of $f$.
