设f(x)在x=2处连续,则,则曲线y=f(x)在点(2,f(2))处的切线方程为_____________.

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问题 设f(x)在x=2处连续,则,则曲线y=f(x)在点(2,f(2))处的切线方程为_____________.

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答案y-[*](x-2)

解析=-1得f(2)=,且
-1==-2f’(2).
则f’(2)=,曲线y=f(x)在点(2,f(2))处的切线方程为y-(x-2).
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