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Andrej111111
@Andrej111111
July 2022
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Критические точки функции y=2x+sin2x
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aln20162016
Первая производная равна 2cos2x +2. Это выражение равно нулю при х =Пи/2. 2Cos(Пи) +2 = -2 + 2 =0. Значит критическая точка (Пи/2, 0).
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