Rolle’s & Lagrange’s Mean Value Theorem

Rolle's Theorem

It is one of the most fundamental theorem of Differential calculus and has far reaching consequences. It states that if y = f (x) be a given function and satisfies,

(i) f (x) is continuous in [a , b]

(ii) f (x) is differentiable in (a , b )

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

Then$f^{\prime }(x)=0atleastonceforsomex\in (a,b)$

Note: (1) There can be more than one such c.

(2) Think! The conditions of Rolle's theorem are necessary or sufficient or both? The answer is conditions are only sufficient and necessary will be clear from the following examples:

(a) Let \begin{align}  f(x)=\dfrac{1}{x}+\dfrac{1}{l-x}\,\,if\,0<x<1 \\  f(0)=f(1)=0 \\ \end{align}

Here condition (i) is violated

However f '(x) = 0 if x = $\frac{1}{2}\in (0,1)$

By defining f(0) = 0 and f(1) = 3 we can see that the result is true when (i) and (iii) are violated.

(b) Let \text{f(x) = }\left\{ \begin{align}  \text{x,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{0 }\le \text{x}\le 1\text{ } \\  \text{1,}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{1}\le \text{x}<2\,\,\,\,\,\, \\ \end{align} \right..

Clearly (ii) does not hold in (0, 2) and even so $f^{\prime }\left(\frac{3}{2}\right)=0$

infact f '(x) = 0 for 1 < x < 2

(3) If f(x) is any polynomial then between any pair of roots of f (x) = 0 lies a root of f '(x) = 0

Illustration 1: If ax² + bx + c = 0, a, b, c $\in$ R. Find the condition that this equation would have at least one root in (0, 1).

Solution: Let f'(x) = ax² + bx + c

Integrating both sides,

$\Rightarrow$ $f(x)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+d$

$\Rightarrow$ f(0) = d and $f(1)=\frac{a}{3}+\frac{b}{2}+c+d$

Since, Rolle's theorem is applicable

$\Rightarrow$ f(0) = f(1) $\Rightarrow$ d = $\frac{a}{3}+\frac{b}{2}+c+d$

$\Rightarrow 2a+3b+6c=0$

Hence required condition is 2a + 3b + 6c = 0

Lagrange's Mean Value Theorem

This theorem is in fact the general version of Rolle's theorem. It says that if y = f(x) be a given function which is;

(i) Continuous in [a , b]

(ii) Differentiable in (a , b)

Then $f^{\prime }(x)=\frac{f(b)-f(a)}{b-a},$ $foratleastonceforsomex\in (a,b)$.

Let A $\equiv$ (a , f (a)) and B $\equiv$ (b , f (b)). Slope of Chord $AB=\frac{f(b)-f(a)}{b-a}$

Illustration 2: If a, b, are two numbers with a < b, show that a real number c can be found between a and b such the 3c² = b² + ab + a².

Solution: Consider the function f(x) = x³

It is continuous and differentiable in (a, b).

Hence by LMVT, there exist a point c such that a < c < b and $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

$\because f^{\prime }(x)=3x^2,weget3c^2=\frac{b^3-a^3}{b-a}=b^2+ba+a^2$.