let $f(k)$ be a function determined as the average value of $h(x) = sin(x-k)$ on the interval $[0,\pi]$. Show that $f$ is a continuous function of $k$ and determine the maximum and minimum values of $f$.
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So $$f(k)=\frac{1}{\pi}\int_0^{\pi} {\sin(x-k) dx}$$ Therefore $$f(k)=\frac{1}{\pi}\left(\cos(k)-\cos(\pi-k)\right)=\frac{2\cos k}{\pi}$$ Hint from here: 1) Differentiability implies continuiuty and 2) Set derivative to 0 for min and max (or just use common sense about what the cosine function does).
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1+1 for the common sense. It would be really weird to use calculus to get the minimum and maximum ._. – chubakueno Apr 10 '14 at 23:56
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1@chubakueno yeah I should have added "use common sense for continuity too and disregard everything else I said." – Apr 11 '14 at 00:02
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Wow thanks I didn't even consider that!! – breaane Apr 11 '14 at 02:21
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lol just took in the "average value" part of the question.. – breaane Apr 11 '14 at 02:28
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1nice do you get it now? :) average value means you take the integral over the region and then divide by the length of the region. – Apr 11 '14 at 02:29
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Yeah!! When I read it again I thought about the whole area underneath a curve and I kinda figured it out! You saved my life man :') – breaane Apr 11 '14 at 02:36
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Alright cool u go kill those tests girl. – Apr 11 '14 at 02:40
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Haha I'll try :) God bless!! – breaane Apr 11 '14 at 02:42