f ′(x) exist for every value of x in the interval[-3,0]
Hence, f(x) is differentiable and hence, continous in the interval [−3,0]
Also, we have f(−3)=f(0)=0⇒All the three conditions of Rolle's theorem are satisfied.
Since, the value x=−2 lies in the open interval [−3,0] the Rolle's theorem is verified.
