) . Therefore, P P α β = . Since P x ( ) is continuous in [ , ] α β and differentiable in ( , α β by an application of Rolle’s theorem there exists γ α β ∈ ( , ) such that ′ P ( ) γ . Since, ′ P x na x which completes the proof.
Example . Prove that there is a zero of the polynomial, in the interval ( , ) given that and are the zeros of the polynomial x Applying the above example . with P x α β and observing ′ P x Q x , (say). This implies that there is a zero of the polynomial Q x ( ) in the interval ( , ) .
For verification, Q ( ) = = − Q ( ) = > From this we may see that there is a zero of the polynomial Q x ( ) in the interval ( , ) . Remark There are functions for which Rolle’s theorem may not be applicable. ( ) For the function f x | |, [ , ] ∈− Rolle’s theorem is not applicable, even though = = because f x ( ) is not differentiable at x = . ( ) For the function, when when even though f = , Rolle's theorem is not applicable because the function f x ( ) is not continuous at x = .
( ) For the function f x sin , ∈ Rolle’s theorem is not applicable, even though ( ) is continuous in the closed interval , π and differentiable in the open interval