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高中数学必修 第一册（648题）
证明：函数
知识点：第三章 函数的概念与性质
参考答案：设\({x}_{1}，{x}_{2}\)是R上的两个任意实数，且\({x}_{1}<{x}_{2}\)，则\(f({x}_{1})-f({x}_{2})=({x}^{3}_{1}+{x}_{1})-({x}^{3}_{2}+{x}_{2})\)
\(=({x}^{3}_{1}-{x}^{3}_{2})+({x}_{1}-{x}_{2})=({x}_{1}-{x}_{2})({x}^{2}_{1}+{x}_{1}{x}_{2}+{x}^{2}_{2})+({x}_{1}-{x}_{2})=({x}_{1}-{x}_{2})({x}^{2}_{1}+{x}_{1}{x}_{2}+{x}^{2}_{2}+1)\)
\(=\left ( {{x}_{1}-{x}_{2}} \right )\left [ {\left ( {{x}_{1}+\frac {{x}_{2}} {2}} \right )^{2}+\frac {3} {4}{x}^{2}_{2}+1} \right ]\)，\(∵{x}_{1}<{x}_{2}，∴{x}_{1}-{x}_{2}<0\)而\(\left ( {{x}_{1}+\frac {{x}_{2}} {2}} \right )^{2}+\frac {3} {4}{x}^{2}_{2}+1>0\)，\(\therefore f({x}_{1})-f({x}_{2})<0\)，
即\(f({x}_{1})<f({x}_{2})\)．\(\therefore f\left ( {x} \right )={x}^{3}+x\)在R上是增函数．