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

Explanation

Since a polynomial function as well as an exponential function in constinuous and the product of two continuous function is continuous, it follows that `f(x)` is continuous on the given interval `[-3, 0]`
Now, `f'(x) = (2x + 3) e^(-(x//2)) - (1)/(2) e^(-(x//2)) (x^(2) + 3x)`
`= e^(-(x//2)) ((x + 6 - x^(2))/(2))`
Which is clearly finite for all values of x in `[-3, 0]`.
Thus, all the conditions of Rolle's theorem are satisfied.
So, there must exist `c in [-3, 0]` such that `f'(c) = 0`
But, `f'(c) = 0 hArr e^(-(c//2)) ((c + 6 - c^(2))/(2)) = 0 hArr c + 6 - c^(2) = 0`
`hArr (3 -c) (c +2) = 0 hArr c = 3 or c = -2`
Thus, `c = -2 in [-3, 0]` such that `f'(c) = 0`
Hence, Rolle's theorem is verified.