已知a>0,函数f(x)=lnx-ax.(1)设曲线y=f(x)在点(1,f(1))处的切线L,若L与圆(x+1) 2 +y 2 =1相切,求a的值(2)求f(x)的单调区间;(3)求函数f(x)在(0,1]上的最大值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 16:02:12
![已知a>0,函数f(x)=lnx-ax.(1)设曲线y=f(x)在点(1,f(1))处的切线L,若L与圆(x+1) 2 +y 2 =1相切,求a的值(2)求f(x)的单调区间;(3)求函数f(x)在(0,1]上的最大值](/uploads/image/z/12890814-6-4.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%2C%E5%87%BD%E6%95%B0f%28x%29%3Dlnx-ax.%281%29%E8%AE%BE%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%281%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BFL%2C%E8%8B%A5L%E4%B8%8E%E5%9C%86%EF%BC%88x%2B1%29+2+%2By+2+%3D1%E7%9B%B8%E5%88%87%2C%E6%B1%82a%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E6%B1%82f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%EF%BC%9B%EF%BC%883%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%EF%BC%880%2C1%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
xݒN@_ŝ
&l$P&B$1QD}iq+xC
4sw;3q:<Ei)F,
nkM $
Ù[&z
(bpU$i9} BWp>-RT B
N8}6_fn⋹`wc:*A3+gXf6nY\UiHvH6wJwàW/s3
RH
UENTjux[qP[3hbkQS60Q6a2hL
8
b@n