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.