y = x^2 (x - 3)^2 increases for all values of | vedclass

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(A) Given function is $y = [x(x - 3)]^2 = (x^2 - 3x)^2$.To find the intervals where the function is increasing,we calculate the derivative $\frac{dy}{

Vedclass · Accepted Answer

$0 < x < 3/2$

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(A) Given function is $y = [x(x - 3)]^2 = (x^2 - 3x)^2$.To find the intervals where the function is increasing,we calculate the derivative $\frac{dy}{dx}$ and set it to be greater than zero.$\frac{dy}{dx} = 2(x^2 - 3x)(2x - 3) = 2x(x - 3)(2x - 3)$.For the function to be increasing,$\frac{dy}{dx} > 0$,so $2x(x - 3)(2x - 3) > 0$.The critical points are $x = 0$,$x = 3/2$,and $x = 3$.Testing the intervals on the number line:$1$. For $x \in (-\infty, 0)$,$\frac{dy}{dx} $2$. For $x \in (0, 3/2)$,$\frac{dy}{dx} > 0$.$3$. For $x \in (3/2, 3)$,$\frac{dy}{dx} $4$. For $x \in (3, \infty)$,$\frac{dy}{dx} > 0$.Thus,the function increases for $x \in (0, 3/2) \cup (3, \infty)$.Comparing this with the given options,the correct interval is $0 < x < 3/2$.

$y = x^2 (x - 3)^2$ increases for all values of $x$ lying in the interval :

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