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 [-3,6].
Condition (2):
Here, f’(x)=2x-5 which exist in [-3,6].
So, f(x)= x2-5x+6 is differentiable on (-3,6).
Condition (3):
Here, f(-3)=(-3)2-5×(-3)+6=30
And f(6)= 62-5×6+6=12
i.e. f(-3)≠f(6)
Conditions (3) of Rolle’s theorem is not satisfied.
So, Rolle’s theorem is not applicable.