学数答题160911-三次函数

题160911（14分）已知三次函数$f\left( x \right)$有三个零点${{x}_{1}},{{x}_{2}},{{x}_{3}}$，且在点$\left( {{x}_{i}},f\left( {{x}_{i}} \right) \right)$处切线的斜率为${{k}_{i}}$（$i=1,2,3$）．

证明：$\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{{{k}_{3}}}=0$．

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证：由题意不妨设$f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)$（$a\ne 0$），则

$f'\left( x \right)=a\left[ \left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)+\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)+\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right) \right]$

故$\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}}+\dfrac{1}{{{k}_{3}}}$

$=\dfrac{1}{f'\left( {{x}_{1}} \right)}+\dfrac{1}{f'\left( {{x}_{2}} \right)}+\dfrac{1}{f'\left( {{x}_{3}} \right)}$

$=\dfrac{1}{a\left( {{x}_{1}}-{{x}_{2}} \right)\left( {{x}_{1}}-{{x}_{3}} \right)}+\dfrac{1}{a\left( {{x}_{2}}-{{x}_{1}} \right)\left( {{x}_{2}}-{{x}_{3}} \right)}+\dfrac{1}{a\left( {{x}_{3}}-{{x}_{1}} \right)\left( {{x}_{3}}-{{x}_{2}} \right)}$

$=\dfrac{\left( {{x}_{2}}-{{x}_{3}} \right)+\left( {{x}_{3}}-{{x}_{1}} \right)+\left( {{x}_{1}}-{{x}_{2}} \right)}{a\left( {{x}_{1}}-{{x}_{2}} \right)\left( {{x}_{1}}-{{x}_{3}} \right)\left( {{x}_{2}}-{{x}_{3}} \right)}$

$=0$