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曲线y=f(x)=ax-bx在点处的切线方程为7x-4y-12=0.则a.b的值分别为(
曲线y=f(x)=ax-bx在点处的切线方程为7x-4y-12=0.则a.b的值分别为(
发布时间:2021-02-20 12:50:27
曲线y=f(x)=ax-bx在点(2,f(2))处的切线方程为7x-4y-12=0,则a,b的值分别为( )
A、
a=1b=3B、
a=-1b=3C、
a=1b=-3D、
a=-1b=-3
网友回答
答案:
分析:
先求出切点坐标,然后根据曲线f(x)过切点以及在x=2处的导数等于切线的斜率建立方程组,解之即可.
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