Question

已知f(x)=x^2－1/x,利用单调性的定义证明f(x)在(0,＋∞)上是增函数

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Answer 1

令0＜x1＜x2
f(x2)-f(x1) = (x2^2 - 1/x2) - (x1^2 - 1/x1)
= (x2^2-x1^2) - (1/x2 - 1/x1)
= (x2+x1)(x2-x1) - (x1-x2)/(x1x2)
= (x2+x1)(x2-x1) + (x2-x1)/(x1x2)
= (x2-x1) { (x2+x1)+1/(x1x2)}
∵0＜x1＜x2，∴x2-x1＞0
∵0＜x1＜x2，∴ (x2+x1)+1/(x1x2)＞0
∴f(x2)-f(x1) = (x2-x1) { (x2+x1)+1/(x1x2)} ＞0
∴f(x2)＞f(x1) ，f(x)在(0,＋∞)上是增函数