# Differentiablity at $0$ of a function $f: \mathbb R \to \mathbb R$ which is twice differentiable in $\mathbb R \setminus \{0\}$

Let $f: \mathbb R \to \mathbb R$ be a function , twice differentiable in $\mathbb R \setminus \{0\}$ such that $f'(x)<0<f''(x) , \forall x <0$ and $f'(x)>0>f''(x) , \forall x >0$ ; then is it true that $f$ cannot be differentiable at $0$ ?

EDIT : My attempt : Suppose $f'$ exists at $0$ , then $f$ is differentiable everywhere , then as $f'(-1)<0<f'(1)$ , so by Darboux theorem , there is $c \in \mathbb R$ such that $f'(c)=0$ , but $f'(x) \ne 0 , \forall x \ne 0$ so $c=0$ , thus $f'(0)=0$ . I cannot make any further progress , Please help . Thanks in advance

Note that $f'$ is positive and decreasing on $(0,\infty)$. In particular, $f'(c)>f'(1)>0$ for any $c\in (0,1)$. Now suppose $h\in(0,1)$. By the mean value theorem, there is some $c\in (0,h)$ such that $f(h)-f(0)=hf'(c)>hf'(1)$. Thus $\frac{f(h)-f(0)}{h}>f'(1)$ for all positive $h<1$. Taking the limit as $h\to 0$, we find that $f'(0)\geq f'(1)>0$, if it exists. This contradicts your observation that $f'(0)=0$.
(Alternatively, instead of citing Darboux, you could just use a similar argument for negative $h$ to show that if you compute $f'(0)$ by a limit as $h$ approaches $0$ from below, $f'(0)\leq 0$.)