求函数y=2sin(x+π/3)在【0,π/2】上的最大最小值.令x+π/3=u,则此函数在区间【0,π/2】上为增函数,y=2sinu在区间【0,π/2】上为增函数,则y=2sin(x+π/3)在区间【0,π/2】为增函数,则最大值为x=π/2时,最
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![求函数y=2sin(x+π/3)在【0,π/2】上的最大最小值.令x+π/3=u,则此函数在区间【0,π/2】上为增函数,y=2sinu在区间【0,π/2】上为增函数,则y=2sin(x+π/3)在区间【0,π/2】为增函数,则最大值为x=π/2时,最](/uploads/image/z/1235967-15-7.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3D2sin%28x%2B%CF%80%2F3%EF%BC%89%E5%9C%A8%E3%80%900%2C%CF%80%2F2%E3%80%91%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E6%9C%80%E5%B0%8F%E5%80%BC.%E4%BB%A4x%2B%CF%80%2F3%3Du%2C%E5%88%99%E6%AD%A4%E5%87%BD%E6%95%B0%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%900%2C%CF%80%2F2%E3%80%91%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2Cy%3D2sinu%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%900%2C%CF%80%2F2%E3%80%91%E4%B8%8A%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E5%88%99y%3D2sin%28x%2B%CF%80%2F3%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B4%E3%80%900%2C%CF%80%2F2%E3%80%91%E4%B8%BA%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E5%88%99%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BAx%3D%CF%80%2F2%E6%97%B6%2C%E6%9C%80)
求函数y=2sin(x+π/3)在【0,π/2】上的最大最小值.令x+π/3=u,则此函数在区间【0,π/2】上为增函数,y=2sinu在区间【0,π/2】上为增函数,则y=2sin(x+π/3)在区间【0,π/2】为增函数,则最大值为x=π/2时,最
求函数y=2sin(x+π/3)在【0,π/2】上的最大最小值.
令x+π/3=u,则此函数在区间【0,π/2】上为增函数,y=2sinu在区间【0,π/2】上为增函数,则y=2sin(x+π/3)在区间【0,π/2】为增函数,则最大值为x=π/2时,最小值为x=0时 请问,此种做法到底哪里错了?
求函数y=2sin(x+π/3)在【0,π/2】上的最大最小值.令x+π/3=u,则此函数在区间【0,π/2】上为增函数,y=2sinu在区间【0,π/2】上为增函数,则y=2sin(x+π/3)在区间【0,π/2】为增函数,则最大值为x=π/2时,最
y=2sinu在区间【π/3,5π/6】上为增函数
令x+π/3=u,则此函数在区间【0,π/2】上为增函数,这里错了 U的区间应该是【0+π/3,π/2+π/3】,所以不是单调增函数。