differentiation

Consider:

f'(x) = - 2 sin x + d[(1 + cos x)-1]/dx

= - 2 sin x - [1/(1 + cos x)2] d(1 + cos x)/dx

= - 2 sin x + sin x/(1 + cos x)2

= sin x [1/(1 + cos x)2 - 2]

Since π < x < 3π/2, sin x =/= 0 and therefore to find out the extreme points, we have:

1/(1 + cos x)2 = 2

(1 + cos x)2 = 1/2

1 + cos x = ±1/√2

cos x = - 1 + 1/√2 or -1 - 1/√2 (rejected)

Therefore for cos x = - 1 + 1/√2 and π < x < 3π/2, we have:

x = π + cos-1 (1 - 1/√2)

To check whether it is a min. or max. we should first find out that when:

x < π + cos-1 (1 - 1/√2), cos x < - 1 + 1/√2 since cos x is increasing when π < x < 3π/2.

So:

1 + cos x < 1/√2

(1 + cos x)2 < 1/2

1/(1 + cos x)2 > 2

1/(1 + cos x)2 - 2 > 0

sin x [1/(1 + cos x)2 - 2] < 0 since sin x < 0 for π < x < 3π/2.

f'(x) < 0

Similarly, we have f'(x) > 0 when x > π + cos-1 (1 - 1/√2)

Hence x = π + cos-1 (1 - 1/√2) is a minimum point of f(x) in π < x < 3π/2.

2010-11-30 22:38:32 補充：
When x = π + cos-1 (1 - 1/√2), cos x = 1/√2 - 1, therefore:

f(x) = 2 cos x + [1/(1 + cos x)]

= 2(1/√2 - 1) + √2

= 2√2 - 2