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VNVDVI
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Longest streak: 40 days
VNVDVI
If a concept doesn't seem to make sense or an answer doesn't cover enough material, feel free to contact me for further help -- I'll happily help if I'm not occupied with school.
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karma
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688 Answers
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@vnvdvi VNVDVI wrote an answer to How do you find #int sinx -1/csc(x)dx #? .
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@vnvdvi VNVDVI wrote an answer to Is "betrayer" a word? .
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@vnvdvi VNVDVI wrote an answer to How do you find the domain of #f(x) = (x-4)/(x+3)#? .
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@vnvdvi VNVDVI wrote an answer to How do you find #lim 1+1/x# as #x->0^+#? .
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@vnvdvi VNVDVI wrote an answer to What is #(pi)/12 # radians in degrees? .
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@vnvdvi VNVDVI commented Yes, "m" is just the biggest value of |f^(n+1)(x)| on the given interval. on For #f(x)=x^-3#, #a=1#, and #2.5\lex\le3.5#, find the following?
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@vnvdvi VNVDVI commented The remainder problems are pretty tough, I assume my answer did not get to explain everything. The theorem itself is pretty loaded and requires finding many things [bound in the abs. val. of (n+1)th derivative, bound on |x-a|^(n+1) ] so it's a lengthy process. on For #f(x)=x^-3#, #a=1#, and #2.5\lex\le3.5#, find the following?
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@vnvdvi VNVDVI commented I made a trivial arithmetic error when taking the fourth derivative, my apologies! I calculated 6 * 60 as 600 rather than 360, so that caused problems with m. on For #f(x)=x^-3#, #a=1#, and #2.5\lex\le3.5#, find the following?
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@vnvdvi VNVDVI updated the answer to For #f(x)=x^-3#, #a=1#, and #2.5\lex\le3.5#, find the following? .
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@vnvdvi VNVDVI commented In my answer, I put "m" as "C," the bounding value on the absolute value of the fourth derivative over [2.5, 3.5]. I got C=0.98304. on For #f(x)=x^-3#, #a=1#, and #2.5\lex\le3.5#, find the following?
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