Calculus f'(x) question

http://tinypic.com/r/16kxrbr/7
Hello.

I just drew a simple picture of f'(x), the derivative of f(x). Excuse my crappy drawing with paint.

What I wanted to know is, what is happening at f(3), and f(4) in the picture.

Is f(3) a lisp, just like f'(3) and is f(4) a vertical asymptote like f'(4) or does it depend..

Thank you.

f(3) will be well defined. The graph will have a slope of about 1 at f(3), and the derivative f'(3) is clearly defined. But the CONCAVITY of f(3) is not defined. That is, f"(3), the derivative of the derivative, will not be defined. That's because it's a cusp.

To be honest, I'm not exactly sure how I'd plot f(x) around x=3. It'd be REALLY difficult to show an undefined concavity with a well-defined derivative.

Around x=4, there is indeed an asymptote, but don't assume the asymptotic behavior is the SAME! It may or it may not be. Clearly f(4) can't be defined. But when you think about it, the function is increasing REALLY REALLY fast at f(3.99). It's ALSO increasing really, really fast at f(4.01). One way to show this is to make the discontinuity at x=4 go to +inf at the left, and come from -inf back into real numbers on the right. Basically, imagine that f'(x) = 1/x^2. That's kinda like your graph. Then f(x) will look kinda like -1/x.

At f(3) there is a cusp. The derivative of f(3) does not exist.

f(4) is a vertical asymptote. It is not continuous, and therefore their is not derivative at f(4).