We have,
f(x) = x3(2x – 1)3
Differentiate w.r.t x, we get,
f‘(x) = 3x2(2x – 1)3 + 3x3(2x – 1)2.2
= 3x2(2x – 1)2(2x – 1 + 2x)
= 3x2(4x – 1)
For the point of local maxima and minima,
f’(x) = 0
= 3x2(4x – 1)= 0
= x = 0, \(\frac{1}{4}\)
At x = \(\frac{1}{4}\)
f’(x) changes from –ve to + ve
Since,
x = \(\frac{1}{4}\) is a point of Minima
Hence, local min value f (\(\frac{1}{4}\)) = \(-\frac{1}{512}\)