已知函数f(x)=ax^3-3/2x^2+1(x∈r),其中a>0(1)若a=1,求曲线y=f(x)在点(2,f(2))处的切线方程;(2)若在区间[-1/2,1/2]上,f(x)>0恒成立,求a的取值范围.
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![已知函数f(x)=ax^3-3/2x^2+1(x∈r),其中a>0(1)若a=1,求曲线y=f(x)在点(2,f(2))处的切线方程;(2)若在区间[-1/2,1/2]上,f(x)>0恒成立,求a的取值范围.](/uploads/image/z/6956796-12-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dax%5E3-3%2F2x%5E2%2B1%28x%E2%88%88r%29%2C%E5%85%B6%E4%B8%ADa%3E0%EF%BC%881%EF%BC%89%E8%8B%A5a%3D1%2C%E6%B1%82%E6%9B%B2%E7%BA%BFy%3Df%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%882%2Cf%EF%BC%882%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-1%2F2%2C1%2F2%5D%E4%B8%8A%2Cf%EF%BC%88x%EF%BC%89%3E0%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
已知函数f(x)=ax^3-3/2x^2+1(x∈r),其中a>0(1)若a=1,求曲线y=f(x)在点(2,f(2))处的切线方程;(2)若在区间[-1/2,1/2]上,f(x)>0恒成立,求a的取值范围.
已知函数f(x)=ax^3-3/2x^2+1(x∈r),其中a>0
(1)若a=1,求曲线y=f(x)在点(2,f(2))处的切线方程;
(2)若在区间[-1/2,1/2]上,f(x)>0恒成立,求a的取值范围.
已知函数f(x)=ax^3-3/2x^2+1(x∈r),其中a>0(1)若a=1,求曲线y=f(x)在点(2,f(2))处的切线方程;(2)若在区间[-1/2,1/2]上,f(x)>0恒成立,求a的取值范围.
(1) a=1 , f(x)=x^3-3/2x^2+1(x∈r), f(x)'=3x^2-3x
x=2 ,f(x)'=12-6=6 f(2)=3
切线方程为y=6x-9
(2)f(x)'=3ax^2-3x (a>0)
f(x)'=0 得 x=0 或 x=1/a
x在[-1/2,0] [1/a,∞)单增 [0,1/a]单减
a≥2时 ,1/a≤1/2 f(x)最小值为f(1/a)或f(-1/2)
f(1/a)=1/a^2-3/2*1/a^2+1=1-1/2*1/a^2>0 解得a>2^½ ∴a≥2
f(-1/2)= -a/8-3/8+1=-a/8+5/8>0 解得a<5
∴2≤a<5
0<a<2时 ,1/a>1/2 f(x)最小值为f(1/2)或f(-1/2)
f(1/2)=a/8-3/8+1=a/8+5/8>0 恒成立
f(-1/2)= -a/8-3/8+1=-a/8+5/8>0 解得a<5 ∴0<a<2
综上可得 a的取值范围 0<a<5