求a的取值范围

显示全部楼层 · 2008-11-2 13:29:21

已知函数f（x）=（ax+1）/（x+2）在区间（-2，正无穷）上为增函数，则实数a的取值范围为？（要过程）

千问 · 2008-11-2 13:29:21

f(x)=(ax+2a-2a+1)/(x+2) =[a(x+2)-(2a-1)]/(x+2) =a-(2a-1)/(x+2) 在(-2，+∞)上单调递增所以-(2a-1)/(x+2)在(-2，+∞)上单调递增所以(2a-1)/(x+2)在(-2，+∞)上单调递减反比例函数递减则系数大于0 (2a-1)/(x+2)就是把反比例函数(2a-1)/x向左移2个单位所以不影响系数的符号所以2a-1>0 a>1/2...

千问 · 2008-11-2 13:29:21

令X2＞X1＞-2，则 f(X2)-f(X1)=(aX2+1)/(X2+2)-(aX1+1)/(X1+2) =[(aX2+1)(X1+2)-(aX1+1)(X2+2)]/[(X2+1)(X1+1)] =[2aX2-2aX1-(X2-X1)]/[(X2+1)(X1+1)] =[(X2-X1)(2a-1)]/[(X2+1)(X1+1)] 因为是增...