Application of differentiation

(a) f(x) = (x-1)^3 / (x+1)^2 = x-5 + (12x+4)/(x+1)^2 (By long division)
= x-5 - 8/(x+1)^2 + 12/(x+1) (By partial fraction)
Hence f'(x) = 1 + 16/(x+1)^3 - 12/(x+1)^2 = (x-1)^2 (x+5) / (x+1)^3
f''(x) = -48/(x+1)^4 + 24/(x+1)^3 = 24(x-1)/(x+4)^4

(b)(i) x < -5 or -1 < x < 1 or x > 1 (for x = -1 is undefined)
(ii) -5 < x < -1
(iii) x > 1
(iv) x < -4 or -4 < x < 1 (for x = -4 is undefined)

(c) For f'(x) = 0, yields x = 1 or x = -5
f'(x) > 0 when -1 < x < 1 or x > 1, hence f(x) is increasing when x > 1.
{> 0 if x < -5
f'(x) {= 0 if x = -5 => f(x) is at maximum when x = -5.
{< 0 if x > -5
Max. point = (-5, f(-5)) = (-5, -27/2)

(d) Vertical asymptote : x = -1
Oblique asymptote : y = x+5

2012-03-12 22:22:43 補充：
Oblique asymptote should be y = x - 5