对勾函数

以下是对勾函数${\displaystyle f(x)=x+{\frac {1}{x}}}$ 的图像

• a、b同正，在${\displaystyle \left(-\infty ,-{\sqrt {\frac {b}{a}}}\right]}$ 单调递增，在${\displaystyle \left[-{\sqrt {\frac {b}{a}}},0\right)}$ 单调递减，在${\displaystyle \left(0,{\sqrt {\frac {b}{a}}}\right]}$ 单调递减，在${\displaystyle \left[{\sqrt {\frac {b}{a}}},+\infty \right)}$ 单调递增。
• a、b同负，在${\displaystyle \left(-\infty ,-{\sqrt {\frac {b}{a}}}\right]}$ 单调递减，在${\displaystyle \left[-{\sqrt {\frac {b}{a}}},0\right)}$ 单调递增，在${\displaystyle \left(0,{\sqrt {\frac {b}{a}}}\right]}$ 单调递增，在${\displaystyle \left[{\sqrt {\frac {b}{a}}},+\infty \right)}$ 单调递减。

函数单调性的证明

对${\displaystyle f(x)=x+{\frac {a}{x}}(a>0)}$ ，任取${\displaystyle 0<x_{1}<x_{2}\leq {\sqrt {a}}}$ ，则有${\displaystyle {\begin{cases}x_{1}-x_{2}<0\\x_{1}\cdot x_{2}>0\\0<x_{1}\cdot x_{2}<a\\\end{cases}}}$ 
${\displaystyle \therefore f(x_{1})-f(x_{2})=x_{1}+{\frac {a}{x_{1}}}-x_{2}-{\frac {a}{x_{2}}}={\frac {(x_{1}-x_{2})(x_{1}{\cdot }x_{2}-a)}{x_{1}{\cdot }x_{2}}}>0}$ ，即${\displaystyle f(x_{1})>f(x_{2})}$ 
${\displaystyle \therefore f(x)}$ 在${\displaystyle (0,{\sqrt {a}}]}$ 上单调递减。同理，${\displaystyle f(x)}$ 在${\displaystyle [{\sqrt {a}},+\infty )}$ 上单调递增；在${\displaystyle (-\infty ,-{\sqrt {a}}]}$ 上单调递增；在${\displaystyle \left[-{\sqrt {a}},0\right)}$ 上单调递减。