Rolle’s Theorem

In this section we shall review the Rolle’s theorem. The theorem is named after the seventeenth century French mathematician Michel Rolle (1652-1719).
(Rolle’s Theorem) : Let $f$ be a continuous function defined on $[a, b]$ and differentiable on $]a, b[$. If $f(a) = f(b)$, then there exists a number $x_o$ in $]a, b[$ such that $f^,(x) =0$
Geometrically, we can interpret the theorem easily. You know that since $f$ is continuous graph of $f$ is a smooth curve
Definition of Rolle's Theorem
As we that the derivative $f'(x)$ at some point $x_o$ gives the slope of the tangent at $(x_o,f(x_o))$ to the curve $y = f(x)$. Therefore the theorem states that if the end values $f(a)$ and $f(b)$ are equal, then there exists a point $x_o$ in $]a, b[$ such that the slope of the tangent at the point $P(x_o,f(x_o))$ is zero, that is, the tangent is parallel to $x-axis$ at that point. In fact we can have more than one point at which $f’(x) = 0$ as shown in Fig. This shows that the number $x_o$ in Theorem may not be unique.
The following example gives an application of Rolle’s theorem.
Example : Use Rolle’s theorem to show that there is a solution of the equation $\cot x=x$ in $ ] 0, \frac{\pi}{2} [ $ Solution : Here we have to solve the equation $ \cot x - x = 0$. We rewrite $ \cot x - x $ as $ \frac{\cot x - x \sin x}{\sin x}$. Solving the equation $ \frac{\cot x - x \sin x}{\sin x}=0$ in $] 0,\frac{\pi}{2} [$ is same as solving the equation $ \cot x - x \sin x = 0$. Now we shall see whether we can find a function f which satisfies the conditions of Rolle’s theorem and for which $f ' (x) = \cot x - x \sin x$. Our experience in differentiation suggests that we try $f(x) =x \cos x$. This function f is continuous in $[ 0, \frac{\pi}{2} ]$ differentiable in $] 0, \frac{\pi}{2} [$ and the derivative $f' (x) = \cot x - x \sin x$. Also $f(0)=0=f \left( \frac{\pi}{2} \right)$. Thus $f$ satisfies all the requirements of Rolle’s theorem. Hence, there exists a point $x_o$, in $] 0, \frac{\pi}{2} [$ such that $f'(x_o) = \cot x_o - x_o \sin x_o =0$. This shows that a solution to the equation $ \cot x - x = 0$ exists in $] 0, \frac{\pi}{2} [$