For $$x<0$$ , f(x) = $$2e^x$$ and it is a continuous function with approaching the value 2 as $$x\longrightarrow\ 0$$ 

For $$x\ge\ 0$$ , f(x) = $$2e^{-x}$$ and it is a continuous function with approaching the value 2 as $$x\longrightarrow\ 0$$ 

Therefore the function is continuous.

However, $$x\longrightarrow\ 0^-$$ f'(x) = 2 but $$x\longrightarrow\ 0^+$$ f'(x) = -2

Thus it is not differentiable at exactly 1 point