题目
求函数函数y=Asin(wx+φ))(A≠0,w>0)的单调区间
提问时间:2020-12-03
答案
求函数函数y=Asin(wx+φ))(A≠0,w>0)的单调区间
解析:∵函数y=Asin(wx+φ))(A≠0,w>0)
单调增区间:
2kπ-π/2<=wx+φ<=2kπ+π/2==>2kπ/w-(π+2φ)/(2w)<=x <=2kπ/w+(π-2φ)/(2w)
单调减区间:
2kπ+π/2<=wx+φ<=2kπ+3π/2==>2kπ/w+(π-2φ)/(2w)<=x <=2kπ/w+(3π-2φ)/(2w)
解析:∵函数y=Asin(wx+φ))(A≠0,w>0)
单调增区间:
2kπ-π/2<=wx+φ<=2kπ+π/2==>2kπ/w-(π+2φ)/(2w)<=x <=2kπ/w+(π-2φ)/(2w)
单调减区间:
2kπ+π/2<=wx+φ<=2kπ+3π/2==>2kπ/w+(π-2φ)/(2w)<=x <=2kπ/w+(3π-2φ)/(2w)
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