Question

If y = x^3(x - 2)^2, then the values of x for which y, increases, are:

Answer 1

Answer:

The answer is $(-\infty,\frac{6}{5}]\cup [2,\infty)$

Step-by-step explanation:

Given  $y=x^3(x-2)^2$

Differentiate w.r.t. $x$, using product rule

$\dfrac{dy}{dx}=x^3(2(x-2)+(x-2)^2(3x^2)$

     $=x^2(x-2)[2x+3(x-2)]\\=x^2(x-2)(5x-6)$

Now for y to be increasing

                    $\dfrac{dy}{dx}\ge 0$

   $\Rightarrow x^2(x-2)(5x-6)\ge 0\\\Rightarrow x\in (-\infty,\frac{6}{5}]\cup [2,\infty)$