Verify Rolle’s theorem for each of the following functions:f(x) = x(

1.	Verify Rolle’s theorem for each of the following functions:f(x) = x(x+2)ex in [-2, 0]
Answer» 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.

Verify Rolle’s theorem for each of the following functions:f(x) = x(x+2)ex in [-2, 0]

Answer»

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.