MHB Real Roots of Polynomial Minimization Problem

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The polynomial \( f(x)=(x-1)^4+(x-2)^4+\cdots+(x-n)^4 \) achieves its minimum value at \( x=\dfrac{n+1}{2} \) for integer \( n \ge 2 \). This point serves as an axis of symmetry for the quartic function. The derivative \( f'(x) = 4\left((x-1)^3+\cdots+(x-n)^3\right) \) is an increasing function, indicating that there is only one critical point. Consequently, \( x=\dfrac{n+1}{2} \) is confirmed as the unique global minimum of the polynomial. Thus, the minimum value of the polynomial occurs exclusively at this point.
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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.
 
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anemone said:
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.
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