Question

已知函数f(x)=(ax-b)/(x^2+b)的图像在点M(-1,f(-1))处的切线方程为x+2y-5=0.则函数y=

Answer 1

将M代入切线方程求切点

-1+2f(-1)-5=0=>f(-1)=3

则M(-1,3)代入f(x)

f(1)=(-a-b)/(1+b)=3

f'(x)=[a(x^2+b)-2x(ax-b)]/(x^2+b)^2=(-ax^2+2bx+ab)/(x^2+b)^2

f'(-1)=(-a-2b+ab)/(1+b)^2=-1/2

a=-7

b=1