Here, we observe that (a) f(x) is a polynomial, so it is continuous in the interval [0, 2]. (b) f′(x)=3x2−6x+2 exists for all x∈(0,2) So, f(x) is differentiable for all x∈(0,2) and (c) f(0)=0,f(2)=23−3(2)2+2(2)=0 ⇒f(0)=f(2) Thus, all the three conditions of Rolle's theorem are satisfied. So, there must exist c∈[0,2] such that f′(c)=0 ⇒ f′(c)=3c2−6c+2=0 ⇒ c=1±