Question

f(x) = `x(x + 3)e^((–x)/2)` in [–3, 0]

Answer 1

We have, f(x) = `x(x + 3)e^((–x)/2)` 

Since polynomial function x(x + 3) and exponential function `"e"^((-x)/2)` are continuous and differentiable in R, given function f(x) is also continuous and differentiable in R

Also f(0) = f(–3) = 0

So, conditions of Rolle's theorem are satisfied.

Hence, there exists a real number c ∈ (–3, 0) such that f'(c) = 0

Now f(x) = `(x^2 + 3x)"e"^((-x)/2)`

∴ f'(x) = `(2x + 3)"e"^((-x)/2) - 1/2 "e"^((-x)/2) (x^2 + 3x)`

= `- 1/2 "e"^((-x)/2) (x^2 + 3x - 4x - 6)`

= `-1/2 "e"^((-x)/2)(x^2 - x - 6)`

So, f'(x) = 0

⇒ `- 1/2 "e"^((-x)/2) ("c" + 2)("c" - 3)` = 0

⇒ c = –2 ∈ (–3, 0)

Therefore, Rolle's theorem has been verified.