I have a function ( which I don't know what you call it when it has 2 parts like this):

f(x) = { -1/2 (x^2 - 3) if x < -3

{ 3/2 ( x + 1 ) if x >= -3

I have to determine from first principles using the definition of the left-derivative and right derivative of f, whether the derivative of f is defined at x=-3

So for first equation

lim x-> -3 f(x) for both left and right limits using

f(x)-f(c)/(x-c) = -1/2 (x-3) = 3

and for second equation

lim x-> -3 f(x) for both left and right limits gives 3/2

Since the derivatives exist but are 2 different values, does this mean the function is not continuous at -3?

Thanks