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设函数f(x)=sinx-cosx+x+1 0≤x<2π.求函数的单调区间与极值
人气:493 ℃ 时间:2019-08-17 20:24:38
解答
函数f(x)的导函数为:cosx+sinx+1
由cosx+sinx+1>0得
根2*sin(x+π/4)+1>0
sin(x+π/4)>-根2/2
所以0根2*sin(x+π/4)+1>0sin(x+π/4)>-根2/2怎么运算来的?sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(sinxcosπ/4+sinπ/4cosx)=√2sin(x+π/4)因为cosx+sinx+1>0 所以 √2sin(x+π/4)+1>0 所以 sin(x+π/4)>-√2/2
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