Verify Rolle’s th

Condition (1):

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

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

Condition (2):

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

So, f(x)=x(x+2)e^x 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)e^c =0

i.e. (c+√2)²=0

i.e. c=-√2

Value of c=-√2 ϵ(-2,0)

Thus, Rolle’s theorem is satisfied.