Condition (1):

Since, f(x)=(x-1)(x-2)(x-3) is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= (x-1)(x-2)(x-3) is continuous on [1,3].

Condition (2):

Here, f’(x)= (x-2)(x-3)+ (x-1)(x-3)+ (x-1)(x-2) which exist in [1,3].

So, f(x)= (x-1)(x-2)(x-3) is differentiable on (1,3).

Condition (3):

Here, f(1)= (1-1)(1-2)(1-3) =0

And f(3)= (3-1)(3-2)(3-3) =0

i.e. f(1)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(1,3) such that f’(c)=0

i.e. (c-2)(c-3)+ (c-1)(c-3)+ (c-1)(c-2)=0

i.e. (c-3)(2c-3)+(c-1)(c-2)=0

i.e. (2c²-9c+9)+(c²-3c+2)=0

i.e. 3c²-12c+11=0

i.e.

i.e. c=2.58 or c=1.42

Value of c=1.42ϵ(1,3) and c=2.58ϵ(1,3)

Thus, Rolle’s theorem is satisfied.