y = x(x-3)2 decreases for the values of x given by:

(a) 1<x<3

(b) x<0

(c) x>0

(d) 0<x<3/2

Solution:

Given y = x(x-3)2

dy/dx = 2x(x-3) + (x-3)2

= (x-3)(2x + x-3)

= (x-3)(3x-3)

= 3(x-3)(x-1)

To find the critical points, put dy/dx = 0

=> (x-3)(x-1) = 0

=> x = 1, 3

dy/dx < 0 

x∈ (1, 3)

Hence option a is the answer.

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