Answer :

Condition (1):


Since, f(x)=x(x+2)ex is a combination of exponential and polynomial function which is continuous for all xϵR.


f(x)= x(x+2)ex is continuous on [-2,0].


Condition (2):


Here, f’(x)=(x2+4x+2)ex which exist in [-2,0].


So, f(x)=x(x+2)ex is differentiable on (-2,0).


Condition (3):


Here, f(-2)= (-2)(-2+2)e-2 =0


And f(0)= 0(0+2)e0=0


i.e. f(-2)=f(0)


Conditions of Rolle’s theorem are satisfied.


Hence, there exist at least one cϵ(-2,0) such that f’(c)=0


i.e. (c2+4c+2)ec =0


i.e. (c+√2)2=0


i.e. c=-√2


Value of c=-√2 ϵ(-2,0)


Thus, Rolle’s theorem is satisfied.


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