函数f(x)=ax-1/(x+2)在区间（-无穷,-2）上单调递增,则实数a的取值范围是,

Question

函数f(x)=ax-1/(x+2)在区间（-无穷,-2）上单调递增,则实数a的取值范围是___,

Answer 1

设x1＜x2＜－2,
因为f(x)=ax-1/(x+2)在区间（-无穷,-2）上单调递增
所以f(x2)－f(x1)＞0即ax2-1/(x2+2)－[ax1-1/(x1+2)]＞0
[(ax2-1)(x1+2)－(ax1-1)(x2+2)]／[(x2+2)(x1+2)]＞0
[(x2－x1)(1＋2a)]／[(x2+2)(x1+2)]＞0
因为x1＜x2＜－2,即x2－x1＞0
所以(x2+2)＜0,(x1+2)＜0则(x2+2)(x1+2)＞0
要使[(x2－x1)(1＋2a)]／[(x2+2)(x1+2)]＞0成立,则1＋2a＞0
所以a＞－1/2