Minimizing the maximum of |x2−x−k|Let k be a real number. Also, let M(k) be the maximum value of f(x)=|x2−x−k| in 0≤x≤1. What are all real numbers k which minimize M(k)?

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kinoNoidae1wj · Accepted Answer

Let g(x)=x2−x−k=(x−12)2−k−14. The graph of g(x) is a parabola with the vertex at (12,−k−14). Since f(x)=|g(x)|, the maximum of f(x) at [0,1] is attained when x∈{0,12,1}. We have f(0)=f(1)=|k| and f(12)=|k+14|. So in order to minimize M(k) we have to minimize M′(k)=max{|k|,|k+14|}. It is easy to check that M′(−18)=18 and M′(k)≥18 for each k.

ysnlm8eut · Accepted Answer

Recall that for any given&amp;nbsp;n≥1, among the polynomials of degree n with leading coefficient 1,&amp;nbsp;12n−1Tn(x)&amp;nbsp;is the one of which the maximal absolute value on [-1,1] is minimal which is equal to&amp;nbsp;12n−1.We have&amp;nbsp;maxx∈[0,1]|x2−x−k|=maxy∈[−1,1]|(y+12)2−y+12−k|&amp;nbsp;=14maxy∈[−1,1]|y2−1−4k|&amp;nbsp;≥14⋅122−1&amp;nbsp;=18.Also,&amp;nbsp;18&amp;nbsp;is attainable at e.g.&amp;nbsp;k=−18&amp;nbsp;(corresponding to&amp;nbsp;y2−1−4k=12T2(y)=y2−12). We claim that&amp;nbsp;k=−18&amp;nbsp;is the unique solution to&amp;nbsp;maxy∈[−1,1]|y2−1−4k|=12. Indeed, by letting&amp;nbsp;g(y)=|y2−1−4k|, from&amp;nbsp;g(0)≤12&amp;nbsp;and&amp;nbsp;g(1)≤12, we have&amp;nbsp;|1+4k|≤12&amp;nbsp;and&amp;nbsp;|4k|≤12&amp;nbsp;which results in&amp;nbsp;k=−18.

Minimizing the maximum of \(\displaystyle{\left|{x}^{{2}}-{x}-{k}\right|}\) Let k be

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