Question

已知函数f(x)满足axf(x)=b+f(x)(a,b≠0),f(1)=2且f '(0)=1/2,求f(x)的解析式

Answer 1

axf(x)=b+f(x)

f(x)=b/(ax-1)

f(1)=b/(a-1)=2

b=2a-2

f'(x)=-b/(ax-1)²

所以f'(0)=-b=1/2

b=-1/2

a=(b+2)/2=3/4

所以f(x)=-1/[2(3x/4-1)]=-2/(3x-4)