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求函数y=根号2sin(2x-π/4)的单调递减区间,最小正周期和最值
人气:148 ℃ 时间:2019-11-24 20:25:36
解答
∵y=√2sin(x-π/4)
∴最小正周期为:[0,2π];
最大值√2,最小值-√2;
∴在x∈[2kπ+3π/4,2kπ+5π/4]上,y=√2sin(x-π/4)是单调递减;
在x∈[2kπ+5π/24,2k(π+2π+π/4]上y=√2sin(x-π/4)是单调递增.是y=√2sin(2x-π/4)
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