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.