in . Now, ′ f ( ) = x x Therefore, ′ f c ( ) = gives c c ) = which implies c = ± Now c = + ( , ) . Observe that − ∉ ( , ) and hence c = + satisfies the Rolle’s theorem. Rolle’s theorem can also be used to compute the number of roots of an algebraic equation in an interval without actually solving the equation.
Example . Without actually solving show that the equation x has only one real root in the interval ( , ) . Let f x ( ) = . Then f x ( ) is continuous in [ , ] and differentiable in ( , ) .
Now, ′ f ( ) = . If ′ then, - - Applications of Differential Calculus x = . Therefore, x = , − but ∉ Thus, ′ f ( ) > ∀∈ Hence by the Rolle’s theorem there do not exist a b ∈ such that, f a f b . Therefore the equation f x ( ) = cannot have two roots in the interval ( , ) .
But, f ( ) = −< and f ( ) = > tells us the curve y ( ) crosses the x -axis between and only once by the Intermediate value theorem. Therefore the equation x has only one real root in the interval ( , ) . As an application of the Rolle’s theorem we have the following, Example . Prove that between any two distinct real zeros of the polynomial a x a x there is a zero of the polynomial na x using the Rolle’s theorem.
Let P x a x a x ( ) = . Let α β be two real zeros of P x (