Q. 2 D5.0( 1 Vote )

# Verify Rolle’s th

Answer :

First let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a,b].

b) The function ‘f’ needs differentiable on the open interval (a,b).

c) f(a) = f(b)

Then there exists at least one c in the open interval (a,b) such that f’(c) = 0.

Given function is:

f(x) = x(x – 1)2 on [0,1]

Since, given function f is a polynomial it is continuous and differentiable everywhere i.e, on R.

Let us find the values at extremums:

f(0) = 0(0 – 1)2

f(0) = 0

f(1) = 1(1 – 1)2

f(1) = 02

f(1) = 0

f(0) = f(1), Rolle’s theorem applicable for function ‘f’ on [0,1].

Let’s find the derivative of f(x):

Differentiating using UV rule,

f’(x) = ((x – 1)2×1) + (x×2×(x – 1))

f’(x) = (x – 1)2 + 2(x2 – x)

f’(x) = x2 – 2x + 1 + 2x2 – 2x

f’(x) = 3x2 – 4x + 1

We have f’(c) = 0 cϵ(0,1), from the definition given above.

f’(c) = 0

3c2 – 4c + 1 = 0

ϵ(0,1)

Rolle’s theorem is verified.

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