The function f(x) = `x(x + 3)e^(-(x)/2)` satisfies all the conditions of Rolle's theorem on [– 3, 0]. Find the value of c such that f'(c) = 0.

#### Solution

The function f(x) satisfies all the conditions of Rolle's theorem on [– 3, 0] such that f'(c) = 0.

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

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

∴ f'(x) = `d/dx[(x^2 + 3x)e^(-x/2)]`

= `(x^2 + 3x).d/dx(e^(-x/2)) + e^(-x/2).d/dx(x^2 + 3x)`

= `(x^2 + 3x).e^(-x/2).d/dx(-x/2) + e^(-x/2) xx (2x + 3 xx 1)`

= `(x^2 + 3x).e^(-x/2) xx -(1)/(2) + e^(-x/2)(2x + 3)`

= `e^(-x/2)[(2x + 3) - (x^2 + 3x)/2]`

= `e^(-x/2)[(4x + 6 - x^2 - 3x)/2]`

= `(e^(-x/2))/2(6 + x - x^2)`

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

∴ f'(c) = `(e^(-c/2))/2(3 - c)(2 + c)`

∴ f'(c) = 0 gives `(e^(-c/2))/2(3 - c)(2 + c)` = 0

∴ (3 – c)(2 + c) = 0 ...`[∵ (e^(-c/2))/2 ≠ 0]`

∴ (3 – c) = 0 or (2 + c) = 0

∴ c = 3 or c = – 2

But `3 notin ( - 3, 0)`

∴ c ≠ 3

Hence, c = – 2.