# Differentiability of a piecewise polynomial function which is continuous everywhere.

Let $$P(x)$$ and $$Q(x)$$ be any two polynomials. Consider a function $$f:\mathbb{R}\rightarrow\mathbb{R}$$, where $$\mathbb{R}$$ is the set of real numbers, given by $$f(x)=\begin{cases}P(x): x\le a\\Q(x):x> a\end{cases}$$ such that $$P(a)=Q(a)$$, then $$f$$ is continuous on $$\mathbb{R}$$. Also $$f$$ is differentiable on the set $$\mathbb{R}$$ except posssibly at $$a$$. Further suppose that $$\displaystyle\lim_{x\rightarrow{a}}f'(x)$$ does not exist. Then can we make a conclusion that $$f$$ is not differentiable at $$a$$? If yes then I want a rigorous proof, otherwise a counter example.

My effort and understanding: I know, that in general, for a function which is continuous on $$\mathbb{R}$$ and differentiable everywhere except possibly at a certain point, $$f$$ may or may not be differentiable at that point even though limit of derivative at that point does not exists.

Consider a function $$f(x)=\begin{cases} x^2\sin{\frac{1}{x}}:x\ne 0\\0:x=0\end{cases}$$ . This function is differentiable on $$\mathbb{R}$$, however $$\displaystyle\lim_{x\rightarrow{0}}f'(x)$$ does not exist.

Since $$P(x)$$ and $$Q(x)$$ are polynomials, then they are continuously differentiable. Hence $$f$$ must be continuously differentiable on $$\mathbb{R}\backslash\{a\}$$ for sure. As a result, we immediately have that $$\lim_{x\to a^-}f'(x)$$ and $$\lim_{x\to a^+}f'(x)$$ must exist. Since the limit in both directions exist, then the only reason why $$\lim_{x\to a}f'(x)$$ does not exist is if we have a jump discontinuity at $$a$$. But the derivatives cannot have jump discontinuities or removable discontinuities. Hence $$f'(a)$$, $$\lim_{x\to a^-}f'(x)$$ and $$\lim_{x\to a^+}f'(x)$$ cannot all exist. We conclude that $$f$$ cannot be differentiable at $$a$$.
If $$f$$ was differentiable at $$a$$, then $$f'(a)=P'(a)=Q'(a)$$. That's because, assuming that the derivative exists, $$f'(a)=\lim_{h\to0^-} \frac{f(a+h)-f(a)}{h}=\lim_{h\to0^-} \frac{P(a+h)-P(a)}{h}=P'(a)=\lim_{x\to a^-}P'(x)=\lim_{x\to a^-}f'(x)$$ and, since $$P(a)=Q(a)=f(a)$$, $$f'(a)=\lim_{h\to0^-} \frac{f(a+h)-f(a)}{h}=\lim_{h\to0^+} \frac{Q(a+h)-Q(a)}{h}=Q'(a)=\lim_{x\to a^+}Q'(x)=\lim_{x\to a^+}f'(x)$$ and this implies that $$\lim_{x\to a^-}f'(x)=\lim_{x\to a^+}f'(x)$$. Therefore if $$\lim_{x\rightarrow{a}}f'(x)$$ does not exist, then $$f$$ cannot be differentiable at $$a$$.
Observe that this is actually true whenever $$P$$ is continiously differentiable on $$(a-\epsilon, a]$$ and $$Q$$ is continiously differentiable on $$[a,a+\epsilon)$$, even without $$P$$ and $$Q$$ being polynomials