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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

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- Oct 18, 2017

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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.