Verify Rolles theorem for the function: `f(x)=x(x+3)e^(-x//2)` on `[-3,\ 0]` .

Answer

N/a

Explanation

We have, `f(x) = x(x+3)e^(-x//2)`
(i) `f(x)` is a continuous function. [since it is a combination of polynomical functions `x(x+3)` and an exponential function `e^(-x//2)` which are continuous functions]
So, `f(x) = x(x+3)e^(-x//2)` is continuous in `[-3,0]`.
(ii) `:. f(x) = (x^(2)+3x).(d)/(dx)e^(-x//2)+e^(-x//2).(d)/(dx)(x^(2)+3x)`
`= (x^(2)+3x).e^(-x//2).(-(1)/(2))+e^(-x//2).(2x+3)`
`=e^(-x//2)[2x+3-1/2.(x^(2)+3x)]`
` =e^(-x//2)[(4x+6-x^(2)-3x)/(2)]`
`=e^(-x//2).1/2[-x^(2)+x+6]`
`= (-1)/(2)e^(-x//2)[x^(2)-x-6]`
`= (-1)/(2)e^(-x//2)[x^(2)-3x+2x-6]`
`= (-1)/(2)e^(-x//2)[(x+2)(x-3)]` which exists in `(-3,0)`.
Hence, `f(x)` is differentiable in `(-3,0)`.
`:. f(-3) = -3(-3+3)e^(-3//2) = 0`
and `f(0) = 0(0+3)e^(-0//2) = 0`
`rArr f(-3) = f(0)`
Since, conditions of Rolle's theorem are satisfied.
Hence, there exists a real number c such that `f(c) = 0`
`rArr -1/2e^(-c//2)(c+2)(c-3) = 0`
`rArr c = - 2,3` where `-2 in (-3,0)`
Therefore, Rolle's theorem has been verified.