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求函数y=2sin(3x+3/π)的周期,单调区间,最大值与最小值,并分别写出取到最大值与最小值时自变量x的集合
人气:109 ℃ 时间:2019-12-06 07:59:22
解答
T=2π/W 所以T=2π/3单调增区间:2kπ - π/2 ≤ 3x + π/3 ≤ 2kπ + π/2 即:2kπ/3 - 5π/18 ≤ x ≤ 2kπ/3 + π/18单调减区间:2kπ + π/2 ≤ 3x + π/3 ≤ 2kπ + 3π/2即:2kπ/3 + π/18 ≤ x ≤ 2kπ3 + ...
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