Question

曲线f(x)=2x²-1在点(1,f(1))处的切线方程为

Answer 1

点(1,f(1)),可由方程得为f(1)=2*1^2-1=1,为点(1,1)，f(x)的导函数为f'(x)=2*2*x-1=4*x-1,所以在点(1,1)处的斜率为f'(1)=4*1-1=3,切线方程为y-1=3*(x-1)
==&g俯担碘杆鄢访碉诗冬涧t;
y=3*x-2