Question

Find the point of local maximum or local minimum, if any, of the following function, using the first derivative test. Also, find the local maximum or local minimum value, as the case may be:

f(x) = x³(2x \[-\] 1)³.

Answer 1

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, `1/4`

At x = 1/4 f'(x) changes from −ve to + ve

Since, x = `1/4` is a point of minima

Hence, local min value f`(1/4) = -1/512`