Q. 485.0( 1 Vote )

# y = x (x – 3)2 decreases for the values of x given by :A. 1 < x < 3B. x < 0C. x > 0D. Given y = x (x – 3)2

y=x(x2-6x+9)

y=x3-6x2+9x

Applying the first derivative we get Applying the sum rule of differentiation, so we get Applying the power rule we get  By splitting the middle term, we get   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.

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