Answer :

Condition (1):


Since, f(x)=x3- 7x2+16x-12 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)= x3- 7x2+16x-12 is continuous on [2,3].


Condition (2):


Here, f’(x)=3x2-14x+16 which exist in [2,3].


So, f(x)= x3- 7x2+16x-12 is differentiable on (2,3).


Condition (3):


Here, f(2)= 23- 7(2)2+16(2)-12=0


And f(3)= 33- 7(3)2+16(3)-12=0


i.e. f(2)=f(3)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(2,3) such that f’(c)=0


i.e. 3c2-14c+16=0


i.e. (c-2)(3c-7)=0


i.e. c=2 or c=7÷3


Value of


Thus, Rolle’s theorem is satisfied.


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