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 maxima and minima,
f'(x) = 0 – (x – 1)2(x + 1)(5x + 1) = 0
x = 1, – 1, – 1/5
Now,
f’’(1) = 0
x = 1 is inflection point
f’’(– 1) = – 4× – 4 = 16 > 0
x = – 1 is point of minima
f’’(\(-\frac{1}{5}\)) = – 5(36/25) x 4/5
= – 144/25 < 0
x =\(-\frac{1}{5}\) is point of maxima
Hence,
local max value = f(\(-\frac{1}{5}\)) = \(\frac{3456}{3125}\)
local min value = f(–1) = 0
