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根号2sin(3x-π/4)的单增区间
人气:330 ℃ 时间:2020-07-12 08:31:13
解答
y=Asin(ax+b) (a>0,A>0)
递增区间[(2kπ)/a-π/(2a)-b/a,(2kπ)/a+π/(2a)-b/a]
递减区间[(2kπ)/a+π/(2a)-b/a,(2kπ)/a+3π/(2a)-b/a]
∴√2sin(3x-π/4)
递增区间[(2kπ)/3-π/12,(2kπ)/3,+π/4]
递减区间[(2kπ)/3,+π/4,(2kπ)/3+7π/12]
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