作业帮已知函数y=acos(2x+π/6)+b的定义域是[-π/3,2π/3],值域是,[-3,1],试着确定f(x)=bsin(ax+π/3)的单调区间
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![已知函数y=acos(2x+π/6)+b的定义域是[-π/3,2π/3],值域是,[-3,1],试着确定f(x)=bsin(ax+π/3)的单调区间](/uploads/image/z/1994467-67-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0y%3Dacos%282x%2B%CF%80%2F6%29%2Bb%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AF%5B-%CF%80%2F3%2C2%CF%80%2F3%5D%2C%E5%80%BC%E5%9F%9F%E6%98%AF%2C%5B-3%2C1%5D%2C%E8%AF%95%E7%9D%80%E7%A1%AE%E5%AE%9Af%28x%29%3Dbsin%28ax%2B%CF%80%2F3%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4)
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已知函数y=acos(2x+π/6)+b的定义域是[-π/3,2π/3],值域是,[-3,1],试着确定f(x)=bsin(ax+π/3)的单调区间
已知函数y=acos(2x+π/6)+b的定义域是[-π/3,2π/3],值域是,[-3,1],试着确定f(x)=bsin(ax+π/3)的单调区间
已知函数y=acos(2x+π/6)+b的定义域是[-π/3,2π/3],值域是,[-3,1],试着确定f(x)=bsin(ax+π/3)的单调区间
根据x的取值来确定.