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已知曲线y=13x3+43.则过点P(2.4)的切线方程为x-y+2=0.或4x-y-4=
已知曲线y=13x3+43.则过点P(2.4)的切线方程为x-y+2=0.或4x-y-4=
发布时间:2021-02-20 12:50:30
已知曲线y=13x3+43,则过点P(2,4)的切线方程为x-y+2=0,或4x-y-4=0
x-y+2=0,或4x-y-4=0
.
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答案:
分析:
设出曲线过点P切线方程的切点坐标,把切点的横坐标代入到导函数中即可表示出切线的斜率,根据切点坐标和表示出的斜率,写出切线的方程,把P的坐标代入切线方程即可得到关于切点横坐标的方程,求出方程的解即可得到切点横坐标的值,分别代入所设的切线方程即可.
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