Answer :

Condition (1):


Since, f(x)= x3+3x2-24x-80 is a polynomial and we know every polynomial function is continuous for all xϵR.


f(x)= x3+3x2-24x-80 is continuous on [-4,5].


Condition (2):


Here, f’(x)= 3x2+6x-24 which exist in [-4,5].


So, f(x)= x3+3x2-24x-80 is differentiable on (-4,5).


Condition (3):


Here, f(-4)= (-4)3+3(-4)2-24(-4)-80=0


And f(5)= (5)3+3(5)2-24(5)-80=0


i.e. f(-4)=f(5)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(-4,5) such that f’(c)=0


i.e. 3c2+6c-24=0


i.e. c=-4 or c=2


Value of c=2 ϵ(-4,5)


Thus, Rolle’s theorem is satisfied.


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