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已知三次函数y=f (x)有三个零点x1,x2,x3,且在点(xi,f(xi))处的切线斜率为ki(i=1,2,3),则1/k1+1/k2+1/k3=_____
人气:191 ℃ 时间:2019-10-10 00:55:50
解答
记f(x)=a(x-x1)(x-x2)(x-x3)f'(x)=a(x-x1)(x-x2)+a(x-x2)(x-x3)+a(x-x1)(x-x3),记p=a(x1-x2)(x2-x3)(x1-x3)k1=f'(x1)=a(x1-x2)(x1-x3)=p/(x2-x3)k2=f'(x2)=a(x2-x1)(x2-x3)=p/(x3-x1)k3=f'(x3)=a(x3-x1)(x3-x2)=p/(x...非常感谢您细致的解答
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