I am familiar with why a derivative is undefined where the original function has a sharp turn, but when the derivative has a sharp turn, I don't understand what the original function did that caused such a sharp turn in the derivative. In other words, what does the original function look like for this derivative graphed?
Best Answer
You might think about the difference between the two functions
$$f(x) = \begin{cases} x^2, & \text{if $x \geq 0$} \\ -x^2, & \text{if $x <0$} \end{cases}$$
and
$$g(x)=x^3.$$
The graphs of both functions are similar. But the $f'(x)$ has a sharp turn at $x=0,$ while $g'(x)$ does not. The reason is that $g(x)$ is going to zero much faster than $f(x)$. So that $g(x)$ is almost flat by the time $x$ gets very close to zero and therefore the curvature of $g(x)$ is $0$ there. The second derivative makes the transition from negative to positive in a continuous fashion. So if you imagine the tangent line moving from left to right, when $x<0$, the line is rotating clockwise. When $x>0$, the line is rotation counter clockwise. As you move from left to right the clockwise rotation slows, stops and then reverses.
The situation is much different with $f(x)$. As you move from left to right, the tangent line is rotating clockwise at a certain speed (the second derivative is constant and equal to $-2.$) At $x=0$, the rotation instantly changes direction (the second derivative is now $2.$) Anyone riding the tangent line will get whiplash at $x=0.$
A similar thought experiment with the osculating circle might be enlightening.