y = x (x – 3)2 decreases for the values of x given by :
Given y = x (x – 3)2
⇒ y=x(x2-6x+9)
⇒ y=x3-6x2+9x
Applying the first derivative we get
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Applying the sum rule of differentiation, so we get
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Applying the power rule we get
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By splitting the middle term, we get
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Now
gives us
x=1, 3
These points divide the real number line into three intervals
(-∞, 1), (1,3) and (3,∞)
(i) in the interval (-∞, 1), f’(x)>0
∴ f(x) is increasing in (-∞,1)
(ii) in the interval (1,3), f’(x)≤0
∴ f(x) is decreasing in (1,3)
(iii) in the interval (3, ∞), f’(x)>0
∴ f(x) is increasing in (3, ∞)
Hence the interval on which the function y = x (x – 3)2 is decreasing is (1,3) i.e., 1<x<3
So the correct option is option A.