Q. 2 A5.0( 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) = x2 – 8x + 12 on [2,6]


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(2) = 22 – 8(2) + 12


f(2) = 4 – 16 + 12


f(2) = 0


f(6) = 62 – 8(6) + 12


f(6) = 36 – 48 + 12


f(6) = 0


f(2) = f(6), Rolle’s theorem applicable for function ‘f’ on [2,6].


Let’s find the derivative of f(x):





f’(x) = 2x – 8


We have f’(c) = 0 cϵ(2,6), from the definition given above.


f’(c) = 0


2c – 8 = 0


2c = 8



C = 4ϵ(2,6)


Rolle’s theorem is verified.


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