# Rolles Theorem and Lagranges Theorem

In this section we will deal with some straight forward but quite useful applications of derivatives. We start with the Rolle’s theorem, a simple but powerful theorem having a lot of practical importance.

**(A) ROLLE’S THEOREM**

Let \(f\left( x \right)\) be a function defined on [*a*, *b*] such that

**(i)** it is continuous on [*a*, *b*]

**(ii)** it is differentiable on (*a*, *b*)

**(iii)** *f *(*a*) = *f* (*b*)

Then there exists a real number \(c \in \left( {a,b} \right)\) such that *f* '(*c*) = 0.

The geometrical interpretation of this theorem is quite straightforward. Consider an arbitrary curve *y* = *f *(*x*) and two points *x* = *a* and *x* = *b* such that *f *(*a*) = *f *(*b*).

Since *A* and *B* are joined by a continuous and differentiable curve, at least one point x = c will always exist in (a, b) where the tangent drawn is horizontal, or equivalently, *f '(c) = 0*. Convince yourself that no matter what curve joins *A* and *B*, as long as it is continuous and differentiable one such c will always exist.

From Rolle’s theorem, it follows that between any two roots of a polynomial *f *(*x*) will lie a root of the polynomial *f *'(*x*).

The (straightforward) proof of Rolle’s theorem is left as an exercise to the reader.

**(B) LAGRANGE’S MEAN VALUE THEOREM**

Let *f *(*x*) be a function defined on [*a*, *b*] such that

**(i) ** it is continuous on [*a*, *b*]

**(ii)** it is differentiable on (*a*, *b*).

Then there exists a real number \(c \in \left( {a,b} \right)\) such that

\[f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\]

To interpret this theorem geometrically, we take an arbitrary function *y* = *f *(*x*) and two arbitrary points *x* = *a* and *x* = *b* on it

We see that no matter what the curve between *R* and *P* is like, as long as it is continuous and differentiable, there will exist a \(c \in \left( {a,\,b} \right)\) such that the tangent drawn at *x* = *c* will have a slope equal to tan \(\theta \) i.e, the average slope from *x* = *a* to *x* = *b*.

For a rigorous proof of LMVT, consider the function

\[g\left( x \right) = f\left( x \right) - \left( {\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}} \right)x\]

Verify that *g*(*x*) satisfies all the three criteria of Rolle’s theorem on [a, b] so that

\(\qquad g'\left( c \right) = 0\) for at least one \(c \in \left( {a,\,b} \right)\)

or \(\begin{align}\qquad f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\end{align}\) for at least one \(c \in \left( {a,b} \right)\)

Notice that LMVT is an extension of the Rolle’s theorem. In fact, for *f *(*a*) = *f *(*b*), LMVT reduces to the Rolle’s theorem.

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