Differentiability at point of inflection Point of inflection literally means that the slope does not change at that point so shouldn't all points of inflection be compulsarily differentiable?
I don't quite understand point of inflection. It's a point where the concavity of graph changes, but that seems too vague. The slope is constant at that point, but how can a slope be constant at a 'POINT'?? Don't we need nearby points for the slope? What exactly is so special about inflection point when all the points that are differentiable already have same left hand and right hand derivative? When x is tending to 0, don't all differentiable points have equal slope on left and right side? So why is inflection point any different?
Hope you got my question!
 A: As also illustrated in this answer, it is incorrect to assume that the condition $f''(x) = 0$ is the sole criteria to determine whether a point is a point of infection. I will quote the same example just to make this point.
Consider the function:
$$f(x)=\cases{ -x^2 & $x\le 0$ \\ x^2 & $x>0$ }$$
If you sketch the graph of the function, you can note quite clearly that the concavity of the above function changes at $x = 0$. However, the second derivative is not continuous at that point.
Therefore, your argument that a point of inflection implies that the slope "does not change" is also quite incorrect. While it is true that if the second derivative is continuous at a point $x = a$ and the second derivative exists at the point, then $f''(a) = 0$, we cannot simply conclude a point is a point of inflexion by noting the double derivative vanishes. You can also consider the graphs of the functions $y = x^{2n}$ for $2 \le n$ where the double derivative vanishes at $x=0$ but the concavity does not change.
