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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?

Mark
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  • Related : http://math.stackexchange.com/questions/532737/proving-any-product-of-four-consecutive-integers-is-one-less-than-a-perfect-squa and http://www.cut-the-knot.org/wiki-math/index.php?n=Algebra.ProductOfFourConsecutiveIntegers – lab bhattacharjee Dec 13 '13 at 15:11

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$\displaystyle(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)$ $\displaystyle=(x^2-5x+4)(x^2-5x+6)=y(y+2)=(y+1)^2-1\ge-1$

(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?)

lab bhattacharjee
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