求 设函数f（x）=ax-x分之b 曲线y=f(x)在点（2,f(2))处的切线方程为7x-4y-12=0

f'(x)=a+b/x^2
f'(2)=a+b/4=7/4
并且(2,2a-2/b)在直线7x-4y-12=0上即
14-4*(2a-2/b)-12=0
∴a=3/4,b=4或者a=2,b=-1
f(x)=3x/4-4/(3x)或者2x+1/x
f（x）=ax-b/x
一阶导数 f'(x)=a+bx^(-2)
曲线y=f(x)在点（2,f(2))处的切线方程为7x-4y-12=0
f(2)=y=7/4x-3 |x=2 =1/2
{ 2a-b/2=1/2
a+b/4=7/4
==>a=1 b=3
所以f（x）=x-3/x