求函数f(x)=ax+ 1 x+b (a,b∈z),曲线y=f(x)在点(2,f(2))处的切线方程为y=3,求f

2025-06-25 16:30:39
推荐回答(1个)
回答1:

由题意得, f′(x)=a-
1
(x+b ) 2

∵在点(2,f(2))处的切线方程为y=3,
f′(2)=a-
1
(2+b) 2
=0
   ①
f(2)=2a+
1
2+b
=3
       ②
由①②解得,a=1,b=-1,
f(x)=x+
1
x-1