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May 1, 2020 at 4:24 comment added DSM Sorry about the oversight, and thanks for catching that. It is $t=\arg\max_{0\leq x\leq 1}(x + \sin(\pi x)]$ instead of $t=\arg\min_{0\leq x\leq 1}(x + \sin(\pi x)]$. Hope it helps.
May 1, 2020 at 4:21 history edited DSM CC BY-SA 4.0
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Apr 30, 2020 at 16:44 comment added nomadofnowhere313 Thanks for the answer! I tried to test this theory on Desmos, and, unless I am mistaken, $f(x)=\sin(x)$ does not have a max/min at any $a$ over $[0,t]$. But I am certain that this $f$ does not satisfy the statement. I calculated $t=-\frac{\cos^{-1}\left(-\frac{1}{\pi}\right)}{\pi}$, which means $[0,t]$ should actually be written $[t,0]$. Is this a case of me not understanding your answer or of an error in the theory?
Apr 30, 2020 at 8:18 history answered DSM CC BY-SA 4.0