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±31∈[0,2]