We have,
f(x) = x3(x – 1)2
Differentiate w.r.t x, we get,
f ‘(x) = 3x2(x – 1)2 + 2x3(x – 1)
= (x – 1)(3x2(x – 1) + 2x3)
= (x – 1)(3x3 – 3x2 + 2x3)
= (x – 1)(5x3 – 3x2)
= x2 (x – 1)(5x – 3)
For all maxima and minima,
f ’(x) = 0 = x2(x – 1)(5x – 3) = 0
= x = 0, 1,\(\frac{3}{5}\)
At x = \(\frac{3}{5}\)
f ’(x) changes from –ve to + ve
Since,
x = \(\frac{3}{5}\) is a point of Minima
At x = 1
f ‘(x) changes from –ve to + ve
Since,
x = 1 is point of maxima.