Answer :

Condition (1):


Since, f(x)=x2-4x+3 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)=x2-4x+3 is continuous on [1,3].


Condition (2):


Here, f’(x)=2x-4 which exist in [1,3].


So, f(x)=x2-4x+3 is differentiable on (1,3).


Condition (3):


Here, f(1)=(1)2-4(1)+3=0


And f(3)= (3)2-4(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. 2c-4=0


i.e. c=2


Value of c=2 ϵ(1,3)


Thus, Rolle’s theorem is satisfied.


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