Q. 655.0( 1 Vote )

# Verify the Rolle’

Given: f(x) = x(x – 1)2

f(x) = x (x2 + 1 – 2x)

f(x) = x3 + x – 2x2 in [0,1]

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

On expanding f(x) = x(x – 1)2, we get f(x) = x3 + x – 2x2

Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x R

f(x) = x3 + x – 2x2 is continuous at x [0,1]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = x3 + x – 2x2

Since, f(x) is a polynomial and every polynomial function is differentiable for all x R

f(x) is differentiable at [0,1]

Hence, condition 2 is satisfied.

Condition 3:

f(x) = x3 + x – 2x2

f(0) = 0

f(1) = (1)3 + (1) – 2(1)2 = 1 + 1 – 2 = 0

Hence, f(0) = f(1)

Hence, condition 3 is also satisfied.

Now, let us show that c (0,1) such that f’(c) = 0

f(x) = x3 + x – 2x2

On differentiating above with respect to x, we get

f’(x) = 3x2 + 1 – 4x

Put x = c in above equation, we get

f’(c) = 3c2 + 1 – 4c

, all the three conditions of Rolle’s theorem are satisfied

f’(c) = 0

3c2 + 1 – 4c = 0

On factorising, we get

3c2 – 3c – c + 1 = 0

3c(c – 1) – 1(c – 1) = 0

(3c – 1) (c – 1) = 0

(3c – 1) = 0 or (c – 1) = 0

So, value of

Thus, Rolle’s theorem is verified.

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