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
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![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](/uploads/image/z/8568620-44-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dx%26sup2%3Bsin%CE%B1%EF%BC%8F3%2Bx%26sup2%3Bcos%CE%B1%EF%BC%8F2%2Btan%CE%B1%E5%85%B6%E4%B8%AD+%CE%B1%E8%8C%83%E5%9B%B4%E6%98%AF%5B0%2C5%CF%80%EF%BC%8F12%5D%E6%B1%82f%E2%80%B2%EF%BC%881%EF%BC%89%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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
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
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
f(x)=x²sinα/3+x²cosα/2+tanα
f'(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)=sin(5π/12)*cos(arctan3/2)+cos(5π/12)*sin(arctan3/2)
=(√6+√2)/4*2/√13+(√6-√2)/4*3/√13=(√6-√2/2)/√13
sin(arctan3/2)=3/√13>(√6-√2/2)/√13=sin(5π/12+arctan3/2)
故√13/3*(√6-√2/2)/√13≤f'(1)=√13/3*sin(α+arctan3/2)≤√13/3*3/√13
也即√6/3-√2/6≤f'(1)≤1
f'(x)=x*cos(y) + (2*x*sin(y))/3
f'(1)=cos(y) + 2sin(y)/3
y的范围:[0,5π/12]
得出f′(1)的取值范围。。。