f(x)=x+根号下x^2+2 证明函数f(x)在R上单调递增

Question

请用定义法求得 请告之详细解题过程 谢谢

Answer 1

f(x)=x+√(x²+2)
设 x1<x2，
则 f(x1)-f(x2)=x1-x2+√(x1²+2) -√(x2²+2)
=x1-x2 +[√(x1²+2) -√(x2²+2)]·[√(x1²+2) +√(x2²+2)] /[√(x1²+2) +√(x2²+2)]
=(x1-x2) +[(x1²+2)-(x2²+2)]/[√(x1²+2) +√(x2²+2)]
=(x1-x2) +(x1-x2)(x1+x2)/[√(x1²+2) +√(x2²+2)]
=(x1-x2)[√(x1²+2) +√(x2²+2)+(x1+x2) ]/[√(x1²+2) +√(x2²+2)]
因为 √(x1²+2) +√(x2²+2)+(x1+x2)> |x1|+|x2|+x1+x2>0
从而 上式<0，即 f(x1)<f(x2)，f(x)在R上是增函数