问题
选择题
已知函数f(x)满足f(2x-1)=
|
答案
∵f(2x-1)=
f(x)+x2-x+2,1 2
再边对x求导,∴2f'(2x-1)=
f'(x)+2x-1.令x=1,1 2
∴2f'(1)=
f'(1)+1.1 2
∴f'(1)=2 3
∴y=f(x)在(1,f(1))处的切线斜率为k=
.2 3
又在f(2x-1)=
f(x)+x2-x+2中令x=1,得f(1)=41 2
∴函数y=f(x)在(1,f(1))处的切线方程为y-4=
(x-1),2 3
即2x-3y+10=0.
故选B.