**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.