求下列函数的最大值,最小值及单调范围:1.y=3sin(2x+3分之π)+3
2.y=2sin(-3x+6分之π)-2
3.y=-2cos(2x-6分之π)+1
人气:392 ℃ 时间:2019-09-26 14:44:11
解答
唉... 1)正弦型函数因-1≤sin(2x+π/3)≤1,则0≤3sin(2x+π/3)+3≤6所以ymax=6,ymin=0单调增区间:-π/2+2kπ≤2x+π/3≤π/2+2kπ,即-5π/12+kπ≤x≤π/12+kπ单调减区间:π/2+2kπ≤2x+π/3≤3π/2+2kπ,即π/1...
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