求曲线f(x)=x-3x+2x过原点的切线方程
人气:230 ℃ 时间:2019-10-17 03:34:17
解答
f'(x)=3x^2-6x+2,设切点的斜率为K 1.当切点是原点时,K=f'(0)=2 切线方程为:y=2x 2.当切点不是原点时,设切点是(X0,Y0) Y0=X0^3-3X0^2+2X0,K=f'(X0)=3X0^2-6X0+2 K=Y0/X0=X0^2-3X0+2 X0=3/2,K=-1/4 切线方程为:y=-...
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