What is the least value of the function $y= (x-2) (x-4)^2 (x-6) + 6$? What is the least value of function:
$$y= (x-2)  (x-4)^2  (x-6) + 6$$
For real values of $x$ ?
Does $\frac{dy}{dx} = 0$, give the value of $x$ which will give least value of $y$?
Thanks in advance.

$$y= (x-2)  (x-4)^2  (x-6) + 6$$
$let , t = x-4$
$$y= t^2(t-2)(t+2)+ 6$$
$$y= t^4-4t^2+6 $$
$$y=(t^2-2)^2+2$$
$$y(min)=2$$ 
attained when $t^2=2$
$(x-4)^2=2$ which gives, $ x=4-\sqrt2$ and $ x=4+\sqrt2$
