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
展开全部
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
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
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询