verify Rolle's theorem for the function `f(x)=x(x+3)e^(-x/2)` in `[-3,0]`

Options

A. 0
B. `-1`
C. `-2`
D. `-3`

Answer

C

Explanation

Given `f(x)=x(x+3)e^(-(1//2)x)=(x^(2)+3)e^((-1/2)x)`
`:.f'(x)=(x^(2)+3x)e^((1//2)x),(-1/2)+(2x+3)e^(-(1//2)x)`
`=e^(-1//2)x{-1/2(x^(2)+3x)+2x+3}`
`=-1/2e^(-1//2)x){x^(2)-x-6}`
Since `f(x)` satisfies the Rolle's theorem.
`:.f'(c)=0`
`implies-1./2e^((-1//2)^(c)){c^(2)-c-6}=0`
`impliesc^(2)-c-6=0 [ :' 6^(-c//2)!=0]`
`impliesc=3, -2`
But `c=3 !in [-3,0]`
`:. c=-2`