作业帮设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)的取值范围
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![设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)的取值范围](/uploads/image/z/2624767-7-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%28sin%CE%B8%2F3%29x%5E3%2B%28%E2%88%9A3cos%CE%B8%2F2%29x%5E2%2Btan%CE%B8%2C%E5%85%B6%E4%B8%AD%CE%B8%E2%88%88%5B0%2C5%CF%80%2F12%5D%2C%E5%88%99%E5%AF%BC%E6%95%B0f%27%281%29%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)的取值范围
设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)
的取值范围
设函数f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,其中θ∈[0,5π/12],则导数f'(1)的取值范围
(f(x)=(sinθ/3)x^3+(√3cosθ/2)x^2+tanθ,
f'(x))=3(sinθ/3)x^2+2(√3cosθ/2)x
=(sinθ)x^2+(√3cosθ)x
f'(1)=sinθ+√3cosθ=2sin(θ+π/3)
θ∈[0,5π/12],
θ+π/3∈[π/3,3π/4],
f'(1)=sinθ+√3cosθ=2sin(θ+π/3)∈[√2/2,1],
则导数f'(1)的取值范围[√2/2,1],
http://wenku.baidu.com/view/1e97b48402d276a200292eaf.html 第十六页……
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