Question

Verify Rolle’s theorem for each of the following functions:

f(x) = x(x+2)e^x in [-2, 0]

Answer 1

Condition (1):

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

⇒ 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 c ϵ (-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.