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.