Calculus is a critical part of maths. When we review about functions in calculus, we come across two basic nevertheless essential theorems: mean value theorem as well as Rolle's theorem, determined by concepts like: differentiability and continuity of an function in calculus.

Specially, Rolle's theorem plays an essential role in calculus. Rolle's theorem is a special case regarding mean value theorem.
It had been discovered by Erika Rolle, that 's the reason, this theorem is termed so.
Rolle's Theorem Statement:

consider a real-valued function f(x) that is continuous on closed interval [a, b] and differentiable on open interval (a, b). Then f(a) = f(b), there is a number "c" in the given interval (a, b), so that f '(c) = 0.

If $f(x)$ is a function defined on $[a,b]$ such that:

1) $f(x)$ is continuous on $[a,b]$

2) $f(x)$ is derivable on (a,b) and

3) $f(a) = f(b)$

Then there exists atleast one value of x = c such that a< c< b, for which f'(c) = 0
Since the function f(x) is continuous on [a,b], it is bounded in [a,b]. Further the function f(x) attains its bounds atleast once.

Let M and m be the lower upper bound and greatest lower bound of f(x) in [a,b]

Now two different cases arise:

Case 1: Let M = m

Then f(x) = m = M = f(a) = f(b)

i.e.,the function f(x) reduces to a constant over [a,b] and consequently f'(x) = 0 for all x $\in$ [a,b]. Hence the theorem.

Case 2: Let M $\neq$ m

Since f(a) = f(b), atleast one of the bounds M and m rent from f(a) and f(b)

Again as every function, that is continuous on a closed interval, attain its lowest upper

bound, there exists a real number say c $\in$ [a,b] such that f(c) = M

Further as f(a) $\neq$ M $\neq$ f(b), c is different from both a and b.

That is c $\in$ (a,b)

Since, f(c) is the greatest value of the function, f(c+h) - f(c) $\leq$ 0, both when h > 0 and when h < 0. Thus

$\frac{f(c+h) -f(c)}{h}$ $\leq$ 0 for h > 0    ........1

$\frac{f(c+h) -f(c)}{h}$ $\geq$ 0 for h < 0    ..........2

By data f(x) is differentiable in the open interval (a,b). In particular f(x) is derivable at c $\in$ (a,b).

Therefore now taking the limit of equation 1 and equation 2 as h -> 0, we get,

lim$_{h->0}$ $\frac{f(c+h)-f(c)}{h}$ = f'(c)$\leq$ 0  for h > 0

lim$_{h->0}$ $\frac{f(c+h)-f(c)}{h}$ = f'(c)$\geq$ 0  for h < 0

But the relations f'(c) $\leq$ 0 and f'(c) $\geq$ 0 hold good only if f'(c) = 0

Hence there is a point c $\in$ (a,b) at which f'(x)= 0.
Given below are some examples on rolle's theorem.

Example 1: Verify rolle's theorem for the function f(x) = x$^{2}$ - 6x + 8, in the interval [2,4]

Solution: The given function is

f(x) = x$^{2}$ - 6x + 8

and the interval is [2,4].

Since f(x)= x$^{2}$ - 6x + 8 is a polynomial, it is continuous for every x, in particular in [2,4]

Now f'(x) = 2x - 6, exists in (2,4)

Also f(2)= 0 = f(4)

Therefore f(x) satisfies all the conditions of rolle's theorem.

Therefore there exists atleast one point c $\in$(2,4)such that f'(c)= 0

Now f'(x)= 2x - 6 =0

x = 3

Therefore the required value of c is 3 $\in$ (2,4)

Hence the theorem is verified.

Example 2: Verify Rolle's theorem for

f(x)= (x-a)$^{3}$(x-b)$^{4}$

Solution: The given function

f(x) = (x-a)$^{3}$(x-b)$^{4}$

is a polynomial in x on expansion by binomial theorem. Therefore it is continuous for all $x \in R$.

Therefore f(x) is continuous on [a,b].

Consider, f'(x)= (x-a)$^{3}$.4(x-b)$^{3}$ + (x-b)$^{4}$.3(x-a)$^{2}$

= $(x-a)^{2}(x-b)^{3}[4(x-a)+3(x-b)]$

= $(x-a)^{2}(x-b)^{3}(7x-4a-3b)$

which exists.

i.e., has a unique and definite value for any x $\in$ (a,b).

Therefore f(x) is derivable in (a,b)

Now f(a) = 0 = f(b)

Therefore f(x) satisfies all the conditions of Rolle's theorem.

There exists atleast one value of x \in (a,b) such that f'(x)= 0

Now f'(x)= 0

$(x-a)^{2}.(x-b)^{3}$ (7x-4a-3b) = 0

7x - 4a - 3b = 0

(x $\neq$ a, x $\neq$ b as we require the values in (a,b))

x = $\frac{4a + 3b}{7}$

Therefore f'(x) = 0 when x = $\frac{4a + 3b}{7}$ clearly this value of x $\in$ (a,b) because it divides a and b internally in the ratio 4: 3 .

Example 3: Verify rolle's theorem for the function

f(x)  = $log$ $[\frac{x^{2}+ab}{x(a+b)}]$ in [a,b], 0 $\in$ [a,b]

Solution: The function can be written as f(x) = $log (x^{2}+ab) - log x - log(a+b)$

Since it is a composite of continuous functions in [a,b] is a continuous function in [a,b].
Now consider

$f'(x)$ =$\frac{2x}{x^{2}+ab}$ - $\frac{1}{x}$

This does not become infinite or indeterminate for (a,b)

Therefore f(x) is derivable in (a,b)

Again f(a) =  0 = f(b)

The condition of the theorem is satisfied. To find the value of x $\in$ (a,b), for which

f'(x)=0, consider,

f'(x) =0

$\frac{2x}{x^{2}+ab}$ - $\frac{1}{x}$ = 0

$2x^{2} - x^{2} - ab = 0$

x$^{2}$ = ab

x = $\pm \sqrt{ab}$

of these two values of x, $\sqrt{ab}$$\in$ (a,b) being the geometric mean of a and b.