Given as f(x) = – (x – 1)3(x + 1)2
f’(x) = – 3(x – 1)2(x + 1)2 – 2(x – 1)3(x + 1)
= – (x – 1)2(x + 1) (3x + 3 + 2x – 2)
= – (x – 1)2(x + 1) (5x + 1)
f’’(x) = – 2(x – 1)(x + 1)(5x + 1) – (x – 1)2(5x + 1) – 5(x – 1)2(x – 1)
For the maxima and minima, f'(x) = 0
– (x – 1)2(x + 1) (5x + 1) = 0
x = 1, – 1, – 1/5
f’’ (1) = 0
x = 1 is the inflection point
f’’(– 1) = – 4× – 4 = 16 > 0
x = – 1 is the point of minima
f’’ (– 1/5) = – 5(36/25) × 4/5 = – 144/25 < 0
x = – 1/5 is the point of maxima
Thus, local max value = f (– 1/5) = 3456/3125
The local min value = f (– 1) = 0
