已知函数f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]求函数的单调区间第一步是怎么划出来的?不懂额~
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![已知函数f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]求函数的单调区间第一步是怎么划出来的?不懂额~](/uploads/image/z/3807583-7-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%282%E2%88%9A+2%29cosx%2F%5Bcos%28x%2F2%29-sin%28x%2F2%29%5D%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E7%AC%AC%E4%B8%80%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E5%88%92%E5%87%BA%E6%9D%A5%E7%9A%84%EF%BC%9F%E4%B8%8D%E6%87%82%E9%A2%9D%7E)
已知函数f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]求函数的单调区间第一步是怎么划出来的?不懂额~
已知函数f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]
求函数的单调区间
第一步是怎么划出来的?不懂额~
已知函数f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]求函数的单调区间第一步是怎么划出来的?不懂额~
cosx=cos²(x/2)-sin²(x/2)
所以f(x)=2√2[cos²(x/2)-sin²(x/2)]/[cos(x/2)-sin(x/2)]
=2√2[cos(x/2)+sin(x/2)]
=2√2[√2*sin(x/2+π/4)]
=4sin(x/2+π/4)
sinx增区间是(2kπ-π/2,2kπ+π/2)
减区间是(2kπ+π/2,2kπ+3π/2)
2kπ-π/2
首先用积化和差公式化解f(x),得到f(x)=√ 2cotx,函数f(x)单调区间就是三角函数cotx的单调区间。
∵f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]
=(2√ 2)[cos²(x/2)-sin²(x/2)]/[cos(x/2)-sin(x/2)]
=(2√ 2)[cos(x/2)+sin(x/2)]
∴f′(x)=√ 2[cos(x/2)-sin(x/2)]
令f′(x)=0,则 cos(x...
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∵f(x)=(2√ 2)cosx/[cos(x/2)-sin(x/2)]
=(2√ 2)[cos²(x/2)-sin²(x/2)]/[cos(x/2)-sin(x/2)]
=(2√ 2)[cos(x/2)+sin(x/2)]
∴f′(x)=√ 2[cos(x/2)-sin(x/2)]
令f′(x)=0,则 cos(x/2)-sin(x/2)=0,即tan(x/2)=1
∴ x=2kπ+π/2 , (k=0,±1,±2,±3,.......).
故f(x)在区间(2kπ,2kπ+π/2)单调增加,在区间(2kπ+π/2,2kπ+π)单调减少。
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