f(x) = (x – 1)(x – 2)2

f’(x) = (x – 2)² + 2(x – 1)(x – 2)

= (x – 2)(x – 2 + 2x – 2)

= (x – 2)(3x – 4)

f’’(x) = (3x – 4) + 3(x – 2)

For maxima and minima,

f'(x) = 0

(x – 2)(3x – 4) = 0

x = 2, 4/3

Now

f’’(2) > 0

x = 2 is point of local minima

f’’(4/3) = – 2 < 0

x = 4/3 is point of local maxima

hence

local max value = f(4/3) = 4/27

local min value = f(2) = 0