# Find the po

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,

At x = f ’(x) changes from –ve to + ve

Since, x = is a point of Minima

Hence, local min value f =

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