Verify Rolle’s theorem for each of the following functions:


Condition (1):


Since, f(x)=x2-5x+6 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)= x2-5x+6 is continuous on [2,3].


Condition (2):


Here, f’(x)=2x-5 which exist in [2,3].


So, f(x)= x2-5x+6 is differentiable on (2,3).


Condition (3):


Here, f(2)=22-5×2+6=0


And f(3)= 32-5×3+6=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. 2c-5=0


i.e.


Value of


Thus, Rolle’s theorem is satisfied.


1