As title states I need to:
Find $\min[f(x)]$ where $f(x) = (x-1)(x-2)(x-3)(x-4)$ without using derivatives
Since I i'm restricted to not use the derivatives I've started to play with the function in different ways. After some experiments I've noticed the following:
Let $$ g(x) = (x-1)(x-3) $$
and
$$ h(x) = (x-2)(x-4) $$
Then i tried to find vertexes with $x_v = -{b \over 2a}$ and calculate values of $g(x)$ and $h(x)$ in $x_v$ points and they appear to be the minimum values for $f(x)$. I've also checked this for $p(x) = (x-1)(x-2)(x-5)(x-6)$ and a lot of other similar polynomials. All of them are symmetric with respect to some $x$.
Based on the above the $\min[f(x)] = -1$ and $\min[g(x)] = -4$ but I'm not sure why this worked. Could someone explain me what happened? I would also appreciate if anyone could tell whether there exists a general way of finding minimum for even power polynomials of the following kind:
$$ \prod_{k=1}^{2n}(x-k) $$
Hint: You can exploit symmetry. Note the zeroes are equally spaced, so their center is $$\tfrac{1+2+3+4}{4}=\tfrac{10}{4}=\tfrac{5}{2}$$ This means you can write $$f(x)=(x-\tfrac52-\tfrac32)(x-\tfrac52-\tfrac12)(x-\tfrac52+\tfrac12)(x-\tfrac52+\tfrac32)$$ $$=\left((x-\tfrac52)^2-(\tfrac32)^2\right)\left((x-\tfrac52)^2-(\tfrac12)^2\right)$$ $$=(z-\tfrac94)(z-\tfrac14)$$ where $z=(x-\tfrac52)^2$. Now you can find the $z_0 \geq 0$ which minimizes this polynomial using symmetry (vertex of the parabola). Then $f(z_0)$ is the desired quantity.