Question

设函数f（x）=ax+1/（x+b）（a.b∈z），曲线y=f（x）在点（2，f（2））处的切线方程式为y=3，

求f（x）解析式（要详细过程Orz)

Answer 1

f'(x)=a-1/(x+b)^2
So f'(2)=a-1/(b+2)^2=0
Then we have:
a*(b+2)^2=1
Also, we know f(2)=3
that is:2*a+1/(b+2)=3
then 2/(b+2)^2+1/(b+2)-3=0
Solve the eq.,we have:b=-1,or b=-8/3
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