Question

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

Answer 1

解：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

追问

不好意思，题目当中第一个是x³

追答

哦。那好办多了。
解：f(x)=x^3*sinα／3+x²cosα／2+tanα
f'(x)=x^2*sinα+x*cosα
f'(1)=sinα+cosα=√2*(√2/2*sinα+√2/2*cosα)=√2sin(α+π/4)
α范围是[0,5π/12]，故α+π/4范围是[π/4,2π/3]
sin(π/4)=√2/2<sin(2π/3)=√3/2
则√2/2≤sin(α+π/4)≤1
故1≤f'(1)=√2sin(α+π/4)≤√2

Answer 2

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。。。