Understanding Higher-order Differentiability Conceptually I understand conceptually that a function $f\colon A\to\mathbf R$ is differentiable at a point $a\in A$ if it can be well approximated by a line there; more precisely, if we can find a constant $f'(a)$ such that $$f(a+h) = f(a) + f'(a)h+o(h).$$

My goal: I want to understand (intuitively) what it means when a function is twice differentiable, or thrice, etc, and likewise what it means when it isn't.

In a very loose sense, I have the intuition that a function is somehow smoother around a point if it has higher order differentiability there, and I suppose this somehow corresponds to the fact that it can be approximated well not only by a line but even better by a polynomial. (e.g. twice differentiability is  $f(a+h) = f(a) + f'(a)h + \tfrac12f''(a)h^2 + o(h^2)$, etc.). Does this mean polynomials canonically define the notion of "smoothness"?
Perhaps I can best get across what I'm asking for with an example. When I look at the graphs of
$$y=x|x| \qquad \text{and} \qquad y=x^3$$
for instance, I see that the latter grows quicker than the former, but around zero, they both basically look "smooth" to me. Yes I know that one is made of the absolute value function which is pointy, but if you just showed me these two pictures:
 
I wouldn't really feel that one is somehow "smoother" around zero than the other.
Is there a better way I can think about this?
 A: Here are some thoughts:
To study, as always, I use my favourite tool of Taylor's theorem(*):
$$ f(a+h) = f(a) + h f'(a) + \frac{h^2}{2} f''(a) +O(h^3)$$
Now, let's analyze $f''(a)$, if $f(a+h)$ is not differentiable, it means the left hand derivative and right hand derivative are not equal i.e(Or may be the limit themself don't exist, but let us forget the case for now):
$$ \lim_{h \to 0} \frac{f'(a+h) - f'(a)}{h} \neq \frac{ f'(a) - f'(a-h ) }{h}$$
Let's call the right hand 2nd derivative as $f_{R}$ and left hand 2nd derivative as $f_{L}$, this leads to 'two' series of the function locally. That is, for points to the right of $f$, we have
$$ f(a+h) = f(a) + hf'(a) + \frac{h^2}{2} f_{R}''(a) + O(h^3)$$
And another series for the left of $f$ as:
$$ f(a-h) = f(a) - h f'(a) + \frac{h^2}{2} f_{L}''(a) + O(h^3)$$
Now, here's the deal, the second order derivative and higher terms only become really relevant (for most nice functions) after $h>1$, this because if $h<1$ then $h^2 <h$, so it turns out that if it is only the second derivative and above which of a function which doesn't match, the Taylor series approximates well.. but outside that bound, we need to be careful and using the piecewise definition.

*: The link is to an article I've written on it, I highly suggest reading it if you want a real insight into the theorem.
**: Right hand derivative: $ \frac{f'(a+h)-f'(a)}{h} $ and the left hand derivative $ \frac{f'(a) - f'(a-h)}{h}$
***: If we restrict ourselves to the differentiability of second derivatives, then we can think of it as the convexity of the graph suddenly changing.
A: I think that there is a sense in which polynomials can indeed be viewed as "defining the notion of smoothness" because a function whose $n$th derivative is identically $0$ is a polynomial of degree $n-1$.
A: I think it would be difficult to deduce that $x|x|$ is not twice differentiable at $x=0$ simply by looking at its graph, but this is how I would think about it. Every function $y=f(x)$ can trivially be parameterised with respect to the $x$-coordinate, meaning that we can define its displacement function as $s(t)=(t,f(t))$, and its velocity function as $v(t)=(1,f'(t))$. The second derivative then has a slightly more concrete interpretation as the rate of change of the vertical component of the velocity function.
In the case of $y=x^3$, $s(t)=(t,t^3)$ and $v(t)=(1,3t^2)$. In the below animation, the velocity vector is represented by the red arrow, and the vertical component of that velocity vector is also shown separately in green:

Notice how the green arrow slows down smoothly before it gets to zero, and then starts speeding up again. This is in constrast to how $y=x|x|$ behaves around $0$:

Since the derivative of $x|x|$ is $|x|$, the green vector arrow is moving downwards at a constant rate, and then it instantly changes direction and goes up at the same rate when $x>0$. Looking at how the red and green vector arrows suddenly pop up gives us a clue about the differentiability of $x|x|$ at $0$. Of course, this animation is not entirely convincing, and things get much harder when you start talking about third and fourth derivatives, but this is how like to think about intuitively. @Fakemistake's car analogy also helped me in understanding this.
