Question

Verify Rolle’s theorem for the following function:

`f(x) = e^(-x) sinx " on"  [0, pi]`

Answer 1

`f(x) = e^(-x). sin x`

1) f (x) is continuous on `[0,pi]` because `e^(-x)` and sin x are continuous function on its domain

2) `e^(-x)` and sin x is differentiable on `(0, pi)`

3) `f(0) = e^(-0). sin 0 = 0`

`f(pi) = e^(-pi). sin pi = 0`

4) Let c be number such that f'(x) = 0

`:. f'(x) = e^(-x). cos x _+ sin x. e^(-x) (-1)`

`:. f'(c) = e^(-c) (cos c -  sin c)`

`:. f'(c) = 0`

`e^(-c) (cos c - sin c) = 0`

`:.e^(-c) = 0`

`:. cos C - sin C = 0

tan C =1

`:. C = pi/4, (3pi)/4 , (5pi)/4,.....`

`:. pi/4 in [0, pi]`

∴ Rolle's theorem verify