Answer :

Condition (1):

Since, f(x)= x2-5x+6 is a polynomial and we know every polynomial function is continuous for all xϵR.

f(x)= x2-5x+6 is continuous on [-3,6].

Condition (2):

Here, f’(x)=2x-5 which exist in [-3,6].

So, f(x)= x2-5x+6 is differentiable on (-3,6).

Condition (3):

Here, f(-3)=(-3)2-5×(-3)+6=30

And f(6)= 62-5×6+6=12

i.e. f(-3)≠f(6)

Conditions (3) of Rolle’s theorem is not satisfied.

So, Rolle’s theorem is not applicable.

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