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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