Let g(x)=∣x∣. Concerning the derivative at x=0, which statement is correct?
a) g′(0)=0 because the cusp is symmetric at x=0
b) g′(0)=1 since ∣x∣ behaves like x for small x
c) g′(0) does not exist because the left–hand derivative is −1 while the right–hand derivative is 1
d) g′(0) does not exist because the function is discontinuous at 0