【设值】任取x1、x2∈R,且x1＜x2
【作差】 f(x2)-f(x1)
【变形】=(x2)³+x2-(x1)³-x1
=(x2)³-(x1)³+(x2-x1)
=(x2-x1)[(x2)²+x1x2+(x1)²]+(x2-x1)
=(x2-x1)[(x2)²+x1x2+(x1)²+1]
【判符】因为x2＞x1
所以x2-x1＞0
(x2)²+x1x2+(x1)²+1=1/2[(x2)²+2x1x2+(x1)²]+(x2)²/2+(x1)²/2+1＞1
故f(x2)-f(x1)＞0
即f(x2)＞f(x1)
【定论】故函数f(x)=x^3 +x在区间(-∞,+∞)上是增函数