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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