Let us assume that all functions are continuous. I was teaching my calculus students the other day. We were talking about what points of non-differentiability look like. Two ways a function can fail to be differentiable at a point is if it looks like $y=|x|$ or like a Brownian motion (think of $x\sin x$ for instance), where the derivative oscillates too much. However, I do not have an intuition about $C^1$ functions and how they differ from $C^i$ functions for higher $i$. An example that I know is the function $$f(x)=x^2,x\geq 0\mbox{ and }f(x)=-x^2,x\leq 0.$$ The graph of this actually looks smooth to me. So the question rephrased may be:
how can one visually tell the difference between $C^1$ functions and $C^2$ functions in a straight forward way.
Although this is for undergrads, I wouldn't mind a more advanced answer.
Best Answer
It will be hard to distinguish $C^1$ and $C^2$ functions visually in general. One thought is that if $f'$ is continuous whereas $f''$ has a point of discontinuity then the signed radius of curvature will have so too, the signed radius of curvature being $$ R=\frac{\left(1+f'(x)^2\right)^{3/2}}{f''(x)} $$ where normally $|R|$ is referred to as the radius of curvature. In your example this signed radius of curvature jumps from $-\frac{1}{2}$ to $+\frac{1}{2}$ showing that the center of the osculating circle jumps from $(0,-0.5)$ to $(0,+0.5)$ at $x=0$:
For $g(x)=x^3$ something different happens. Though the radius of curvature changes sign at $x=0$ it changes via limits of $-\infty$ and $+\infty$ since $g''(x)=0$. In general, a $C^2$ curve has to 'straighten out any curvature' before curving in a different direction.
Also sudden changes in curvature will not be allowed such as for instance the curve $h(x)=x^4$ for $x<0$ and $h(x)=x^2$ for $x\geq 0$ which has $h''(x)=12x^2$ for $x<0$ and $h''(x)=2$ for $x>0$ so that $h''(x)$ has the ambiguity of choosing between the value $0$ and $2$ for $x=0$.
Nevertheless this is a very subtle property when looking at the curves.