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Apr 28, 2018 at 0:24 history closed JMP
Claude Leibovici
Ethan Bolker
Xander Henderson
CommunityBot
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Apr 27, 2018 at 4:29 review Close votes
Apr 27, 2018 at 13:51
Feb 21, 2018 at 8:30 review Close votes
Feb 21, 2018 at 9:54
Apr 14, 2013 at 22:59 comment added Stefan Smith you accepted an elegant answer that doesn't require calculus, but if you know a little calculus, it is easy: let $f(x) = 4x -x^4$. $f(x) \to -\infty$ as $|x| \to \infty$, so $f$ achieves a maximum value at some $x_0$. $f'(x_0) = 0$. $f'(x) = 4 - 4x^3$ so $x_0 = 1$. $f(1) = 3$ is the maximum value of $f$.
Apr 14, 2013 at 9:45 answer added gukoff timeline score: 0
Apr 14, 2013 at 9:05 answer added clark timeline score: 15
Apr 14, 2013 at 9:04 vote accept CommunityBot
Apr 14, 2013 at 8:53 answer added bryan.blackbee timeline score: 2
Apr 14, 2013 at 8:44 answer added Phil Wang timeline score: 7
Apr 14, 2013 at 8:43 answer added AlexM timeline score: 0
Apr 14, 2013 at 8:36 answer added learner timeline score: 12
Apr 14, 2013 at 8:34 history asked user53386 CC BY-SA 3.0