设函数f（x）=x²sinα／3+x²cosα／2+tanα其中 α范围是[0,5π／12]求f′（1）的取值范围

显示全部楼层 · 2012-2-17 00:39:48

解：f(x)=x2sinα／3+x2cosα／2+tanαf'(x)=2xsinα／3+2xcosα／2=2x(sinα／3+cosα／2)=2x*√(1/9+1/4)*sin(α+arctan3/2)=√13/3*xsin(α+arctan3/2)f'(1)=√13/3*sin(α+arctan3/2)α范围是[0,5π/12]，故α+arctan(3/2)范围是[arctan3/2,5π/12+arctan3/2]因π/2>arctan(3/2)>arctan1=π/4故5π/12+π/4=2π/3<5π/12+arctan3/2<5π/12+π/2=11π/12则sin(5π/12+arctan3/2)...

千问 · 2012-2-17 00:39:48

f'(x)=x*cos(y) + (2*x*sin(y))/3f'(1)=cos(y) + 2sin(y)/3y的范围：[0,5π／12]得出f′（1）的取值范围。。。...