Find the minium of function Find the minimum of $f(x)=(x-1)(x-2)(x-3)(x-4)$ without using the calculus, I know it's easy to find it using the derivative, but I need to fiugre out how to solve it without it. 
I know that the minimum are between $(1,2)\vee(3,4)$ becouse of chart and $f(x)=x^4-10x^3+35x^2-50x+24$, have anyone idea how to solve it?

$\displaystyle(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)$
$\displaystyle=(x^2-5x+4)(x^2-5x+6)=y(y+2)=(y+1)^2-1\ge-1$
(setting $x^2-5x+4=y$)
Now the equality occurs if $\displaystyle x^2-5x+4=y=-1\iff x^2-5x+5=0$
Observe that the discriminant of the last equation is $>0$  which is extremely crucial here(why?)
