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.