已知函数f（x）=x3次方＋x,求证f（x）在R上是增函数,

任取x1＜x2,x1、x2∈R
则f(x2)-f(x1)
=(x2)³+x2-(x1)³-x1
=(x2-x1)[(x2)²+x1x2+(x1)²]+(x2-x1)
=(x2-x1)[(x2)²+x1x2+(x1)²+1]
因为x2＞x1
所以x2-x1＞0
又(x2)²+x1x2+(x1)²+1
=(x2)²/2+(x1)²/2+[(x1)+(x2)]²/2+1
＞0
所以f(x2)-f(x1)＞0
故f(x)在R上是增函数.