3

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:

$$ \prod_{k=1}^{2n}(x-k) $$

roman
  • 5,391

2 Answers2

10

$$\begin{array}{rcl} (x-1)(x-2)(x-3)(x-4) &=& (x^2-5x+4)(x^2-5x+6) \\ &=& (x^2-5x+5)^2-1 \\ &=& ((x-2.5)^2-1.25)^2-1 \\ &\ge& -1 \end{array}$$

Equality is achieved when $(x-2.5)^2 - 1.25 = 0$.

Kenny Lau
  • 25,049
9

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.

MPW
  • 43,638