Real Roots of Polynomial Minimization Problem

[anemone]

For an integer $n\ge 2$, find all real numbers $x$ for which the polynomial $f(x)=(x-1)^4+(x-2)^4+\cdots+(x-n)^4$ takes its minimum value.

 

[castor28]

For an integer $n\ge 2$, find all real numbers $x$ for which the polynomial $f(x)=(x-1)^4+(x-2)^4+\cdots+(x-n)^4$ takes its minimum value.

[sp]
Since $x=\dfrac{n+1}{2}$ is an axis of symmetry, the point $x=\dfrac{n+1}{2}$ is either a minimum of a maximum, depending on the shape of the quartic.

However, the derivative $f'(x) = 4\left((x-1)^3+\cdots+(x-n)^3\right)$ is an increasing function (since it is a sum of increasing functions). Therefore, $f'(x)$ as exactly one zero, and $f(x)$ has only one critical point.

We conclude that $x=\dfrac{n+1}{2}$ is the unique global minimum.
[/sp]