# 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

Given $y = x (x - 3)\textsuperscript{2}$

$\\ { \Rightarrow y=x(x\textsuperscript{2}-6x+9)}\\ { \Rightarrow y=x\textsuperscript{3}-6x\textsuperscript{2}+9x}\\$

Apply first derivative and get
$\frac{d y}{d x}=\frac{d\left(x^{3}-6 x^{2}+9 x\right)}{d x}$
Apply sum rule of differentiation and get
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(6 \mathrm{x}^{2}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(9 \mathrm{x})}{\mathrm{dx}}$
Apply power rule and get
$\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}-12 \mathrm{x}+9 \\\Rightarrow \frac{d y}{d x}=3\left(x^{2}-4 x+3\right)$
Now, split middle term and get
$\\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3\left(\mathrm{x}^{2}-3 \mathrm{x}-\mathrm{x}+3\right) \\\Rightarrow \frac{d y}{d x}=3(x(x-3)-1(x-3)) \\\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=3((\mathrm{x}-3)(\mathrm{x}-1))$

Now, $\frac{\mathrm{d} y}{\mathrm{dx}}=0$ gives us

x=1, 3

The points divide this real number line into three intervals

$(i) \text{In the interval } (-\infty, 1), f'(x) > 0 \\ \therefore \text{ } f(x) \text{ is increasing in } (-\infty, 1) (ii) \text{In the interval } (1, 3), f'(x) \leq 0 \\ \therefore \text{ } f(x) \text{ is decreasing in } (1, 3) (iii) \text{In the interval } (3, \infty), f'(x) > 0 \\ \therefore \text{ } f(x) \text{ is increasing in } (3, \infty)$

So, the interval on which the function decreases is (1,3) i.e., 1<x<3

So the correct answer is option A.