Question

已知函数f(x)=ax+1/x+2在区间（-2，正无穷）上是增函数，求a的取值范围

Answer 1

f(x)=(ax+2a-2a+1)/(x+2)
=a(x+2)/(x+2)+(-2a+1)/(x+2)
=a+(-2a+1)/(x+2)

反比例函数在x>0是增函数则系数小于0
所以这里有-2a+1<0
a>1/2

Answer 2

f(x)=(ax+2a+1-2a)/(x+2)=a+[(1-2a)/(x+2)]
因为f(x)在（-2，正无穷）上是增函数，
而g(x)=1/(x+2)在（-2，正无穷）上是减函数则必有
1-2a<0即a>1/2

Answer 3

解f(x)=(ax+2a+1-2a)/(x+2)=a+[(1-2a)/(x+2)]
因为f(x)在（-2，正无穷）上是增函数，
而g(x)=1/(x+2)在（-2，正无穷）上是减函数则必有
1-2a<0即a>1/2

Answer 4

f(x)=(ax+1)/(x+2)
不妨设x1＞x2＞-2
因为f(x)在(-2,+∞)上为增函数
则，f(x1)-f(x2)=(ax1+1)/(x1+2)-(ax2+1)/(x2+2)
=[(ax1+1)(x2+2)-(ax2+1)(x1+2)]/[(x1+2)(x2+2)]
=[(ax1x2+2ax1+x2+2)-(ax1x2+x1+2ax2+2)]/[(x1+2)(x2+2)]
=[(2a-1)(x1-x2)]/[(x1+2)(x2+2)]
＞0
上式中，x1-x2＞0，(x1+2)(x2+2)＞0
所以，2a-1＞0
所以，a＞1/2

Answer 5

f(x)=(ax+2a+1-2a)/(x+2)
=a+(1-2a)/(x+2)
因为函数在（-2，正无穷）上递增
1-2a<0
a>1/2