已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若函...已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若
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![已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若函...已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若](/uploads/image/z/3990934-46-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%EF%BC%9Dpx%EF%BC%8Dp%2Fx%EF%BC%8D2inx%2C%EF%BC%88%EF%BC%89%E8%8B%A5p%EF%BC%9D2%2C%281%29%E6%B1%82%E6%9B%B2%E7%BA%BFf%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%881%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%2C%282%29%E8%8B%A5%E5%87%BD...%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%EF%BC%9Dpx%EF%BC%8Dp%2Fx%EF%BC%8D2inx%2C%EF%BC%88%EF%BC%89%E8%8B%A5p%EF%BC%9D2%2C%281%29%E6%B1%82%E6%9B%B2%E7%BA%BFf%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%EF%BC%881%EF%BC%89%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%2C%282%29%E8%8B%A5)
已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若函...已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若
已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若函...
已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若函数f(x)在气定义域内为曾函数,求正实数p的取值范围.
已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若函...已知函数f(x)=px-p/x-2inx,()若p=2,(1)求曲线f(x)在点(1,f(1))处的切线方程,(2)若
(1)f(x)=2x-2/x-2lnx
f(1)=0
f'(1)=2+2/1^2-2/1=2
切线方程:y-0=2(x-1)即y=2x-2
(2)f(x)定义域:x>0
f'(x)=p+p/x^2-2/x=(px^2-2x+p)/x^2
因为f(x)单增,所以分子恒≥0
p>0且4-4p^2≤0
p≥1
p=2, f(x)=2x-2/x-2lnx, f'(x)=2+2/x^2-2/x, 把x=1代入, f(1)=0, f'(1)=2, 切线方程y=2x-2
因为x的定义域为x>0, 只要f'(x)>0函数就是增函数, 即p+p/x^2-p/x>0, 解得p>0
(1).由p=2,f(x)=2x-2/x-2lnx,f'(x)=2+2/x2-2/x,f'(1)=2.f(1)=-2ln1 因为该点是切点 又可设切线L:y-f(1)=f'(1)(x-1),代入得L:y=2x-2-2ln1。 (2).f'(x)>=0,即p+p/x2-2/x>=0,又由x>0得px2+p-2x>=0,解得p>=2x/(x2+1), 设g(x)=2x/(x2+1),(x>0)求导解出最大值点x=1值为1,代入上式解得p>=1