If f(x) = 3 for x < 0 and f(x) = 2x for x ≥ 0, is f(x) differentiable at x = 0? State and justify why/why not.
The Attempt at a Solution
Obviously, since f(x) is not continuous and the limit doesn't exist as x[itex]\rightarrow[/itex]0, the function shouldn't be differentiable at that point. But I can't justify this because I don't really understand why not. Simply put, I get that the limit doesn't exist at 0 and, hence, the function is discontinuous there, but I don't see why that invalidates the idea that the gradient of the line f(x) = 2x at x = 0 is 2. Further, as far as I can tell (and I have a strong feeling I'm going wrong here somewhere), f'(x) = 0 for x < 0 and f'(x) = 2 for x ≥ 0, which seems to suggest that the derivative of f(x) for all values of x ≥ 0 is 2. What am I missing here?
As always, any help would be much appreciated, thanks.