Question

设函数f(x)=g(x)+x²,曲线y=g(x)在点(1,g(1))处的切线方程为y=2x+1,则曲线y=f(x)在点(1,f(1))处的切线的斜率为(　　)

(A)2    (B)-   (C)4    (D)-

Answer 1

C解析:因为曲线y=g(x)在点(1,g(1))处的切线方程为y=2x+1,

所以g′(1)=2.

又f′(x)=g′(x)+2x,

故曲线y=f(x)在点(1,f(1))处的切线的斜率为f′(1)=g′(1)+2=4.