f(x)=sin(2x-π/6)-1 增减区间

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f(x)=sin(2x-π/6)-1增减区间f(x)=sin(2x-π/6)-1增减区间f(x)=sin(2x-π/6)-1增减区间依题意即-π/2+2kπ  增区间[-π/6+kπ,π/3+kπ]减

f(x)=sin(2x-π/6)-1 增减区间
f(x)=sin(2x-π/6)-1 增减区间

f(x)=sin(2x-π/6)-1 增减区间
依题意

-π/2+2kπ

  增区间[-π/6+kπ,π/3+kπ]
减区间 [π/3+kπ,5π/6+kπ] k∈Z

sin(2x-π/6)增区间
2kπ<=2x-π<=2kπ+π
2kπ+π<=2x<=2(k+1)π
kπ+π/2<=x<=(k+1)π
sin(2x-π/6)减区间
kπ<=x<=kπ+π/2

由此类函数周期公式T=2派/|w|,w即x前的系数,得T=派,再根据-派/6得图像往右移派/6个单位,再由-1得向下一个单位,最后根据图像求得单调区间