Condition (1):

Since, f(x)=(x-2)⁴(x-3)³ is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= (x-2)⁴(x-3)³ is continuous on [2,3].

Condition (2):

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

So, f(x)= (x-2)⁴(x-3)³ is differentiable on (2,3).

Condition (3):

Here, f(2)= (2-2)⁴(2-3)³=0

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

(c-2)³(c-3)²(7c-18)=0

i.e. c=2 or c=3 or c=18÷7

Value of

Thus, Rolle’s theorem is satisfied.