Verify Rolle’s theorem for each of the following functions on the indicated intervals :

f(x) = x2 + 5x + 6 on [ – 3, – 2]

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 + 5x + 6 on [ – 3, – 2]


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( – 3) = ( – 3)2 + 5( – 3) + 6


f( – 3) = 9 – 15 + 6


f( – 3) = 0


f( – 2) = ( – 2)2 + 5( – 2) + 6


f( – 2) = 4 – 10 + 6


f( – 2) = 0


f( – 3) = f( – 2), Rolle’s theorem applicable for function ‘f’ on [ – 3, – 2].


Let’s find the derivative of f(x):





f’(x) = 2x + 5


We have f’(c) = 0 cϵ( – 3, – 2), from the definition given above.


f’(c) = 0


2c + 5 = 0


2c = – 5



C = – 2.5ϵ( – 3, – 2)


Rolle’s theorem is verified.


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