If $f$ is a function on $[a,b]$ and $f'(a)f'(b)>0$, then $f$ must vanish at least at one point in $(a,b)$. Is it true? If $f$ is a function on $[a,b]$ with $f(a)=f(b)=0$ and $f'(a)f'(b)>0$, then $f$ must vanish at least at one point in $(a,b)$.
Is it true?
 A: $f(x)= x- \left\lfloor x \right \rfloor-{1\over 2}$ is a counter example. It has a root as 0.5 and 1.5 but no root in between.
This happens because the given function is not continuous. If it's given to be continuous, you can show that there exists a $c_1$ and $c_2$ in $[a,b]$ by using the fact that the given function is differentiable at the end points such that $f(c_1)f(c_2)<0$ . Now by intermediate value theorem, this will have a root in between $( c_1 , c_2 )$ and therefore also in $[a,b]$
A: The conditions do not specify $a\neq b$. If $a=b$ there is no point in the (empty) open interval at which the conclusion of $f$ vanishing at "at least one point" can be true.
It might be that the context implies $a\neq b$, but in the absence of this, $a=b$ gives a counterexample even in the case where $f$ is differentiable in an open interval containing both $a$ and $b$ or continuous in $[a,b]$ (examples of potential additional conditions).
Other counterexamples are in some ways more significant, but I have put this just to show that some care has to be taken about basic assumptions - here, the same point can have two different names.
