Question

已知f（x）=3x+1/x+2，指出f（x）的单调区间

Answer 1

是f(x)=(3x+1)/(x+2)吧？下次把括号打一下。
f(x)=(3x+1)/(x+2)
=[3(x+2)-5]/(x+2)
=3 - [5/(x+2)]
①当x+2＞0，即x＞-2时，
y=1/(x+2)单调递减
y=5/(x+2)单调递减
y=-5/(x+2)单调递增
y=3-[5/(x+2)]单调递增

②当x+2＜0，即x＜-2时
y=1/(x+2)单调递减
y=5/(x+2)单调递减
y=-5/(x+2)单调递增
y=3-[5/(x+2)]单调递增

∴在(﹣∞，-2)和(-2，﹢∞)上单调递增！！！
注意：不能用“∪”

Answer 2

f(x)'=3-1/x^2=(3x^2-1)/x^2
x>根号3/3或x<负根号3/3,f(x)递增
负根号3/3<x<0 递减
0<x<负根号3/3 递减