I need to determine if f '(0) exists.?

I need to explain this using the definition of the derivative, but I dont know what to do.

Consider the function:

f(x)= { x^2 +1 when x > or = to 0 & x+1 when x < 0} no, f'(0) doesnt exist even though it is continuous. this is because the function is non-differentiable whenever there is a corner, cusp, discontinuity and stuff and we can see by the graph that we imagine in our heads that this function has a corner at x=0.

f'(0) doesn't exist because the derivative is not the same when approaching along the positive axis and approaching from the negative axis. From the positive side, f'(0) = 2*0 or 0 while from the negative side, f'(0) = 1. Since the derivative diverges (isn't continuous), it does not exist by definition.

f'(x) = 1 where f(x) = x+1
and f'(x) = 2 * x where f(x) = x^2+1
(0)^2 + 1 = (0) + 1
1! = 2 * (0)

So, the function f'(0) has a value everywhere except x=0, because, even though the function is continuous, the two derivatives do not match approaching the intersection, they are not smooth, the come to a "point" at which the slope of the function cannot be determined. So you could say it has a derivative for left and right of 0 and is undefined at 0, or you could say there is no function. Just depends on how you approach it.