Question

求函数y=1/（2-x）在x=-1处的切线斜率及方程

Answer 1

y'=-(2-x)'/(2-x)^2=1/(x-2)^2

切线斜率为：f'(-1)=1/(-1-2)^2=1/9

f(-1)=1/(2-(-1))=1/3

(y-1/3)/(x-(-1))=1/3

切线方程为：y=1/3 x+2/3