(文)曲线y=x3-x+3在点(1,3)处的切线方程为( ) A.2x+y+1=0 B.2x-y+1=0 C.2x-y-1=0 D.x-2y+1=0
问题描述:
(文)曲线y=x3-x+3在点(1,3)处的切线方程为( )
A. 2x+y+1=0
B. 2x-y+1=0
C. 2x-y-1=0
D. x-2y+1=0
答
∵y=f(x)=x3-x+3,
∴f′(x)=3x2-1.
设所求切线的斜率为k.
∵点(1,3)在y=f(x)的图象上,是切点,
∴k=f′(1)=3×12-1=2,
∴所求曲线的切线方程为:y-3=2(x-1),
即2x-y+1=0;
故选:B.