Answer :

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.


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