Browse Questions

# $y=x(x-3)^2$ decreases for the values of x given by:

$(A)\;1< x <3\quad(B)\;x < 0\quad(C)\;x>0\quad(D)\;0< x < \large\frac{3}{2}$

Toolbox:
• Let $f(x)$ be a function defined on $(a,b)$ if $f'(x) <0$ for all $x \in (a,b)$ except for a finite number of points where $f'(x)=0$, then $f(x)$ is decreasing on $(a,b)$
Step 1
$y=x(x-3)^2$
differentiating w.r.t $x$ we get,
( apply product rule)
$y'=x[2(x-3)]+(x-3)^2.1$
$y'=2x(x-3)+(x-3)^2$
$=2x^2-6x+x^2-6x+9$
$= 3x^2-12x+9$
Step 2
If it is a decreasing function, then $y'<0$
$\Rightarrow 3x^2-12x+9 <0$
$\Rightarrow 3(x^2-4x+3) < 0$
$\Rightarrow 3(x-3)(x-1)<0$
This clearly implies the function decreases when
$1 < x < 3$
Hence A is the correct answer.