已知函数=+有如下性质:如果常数>0,那么该函数在0,上是减函

已知函数有如下性质:如果常数0,那么该函数在0上是减函数,在,+∞上是增函数.

1)如果函数0)的值域为6,+∞,求的值;

2)研究函数(常数0)在定义域内的单调性,并说明理由;

3)对函数(常数0)作出推广,使它们都是你所推广的函数的特例.研究推广后的函数的单调性(只须写出结论,不必证明)。

答案

解:(1)函数y=x+(x>0)的最小值是2,则2=6, b=log29.

     (2)  0<x1<x2,y2y1=.

 <x1<x2, y2>y1, 函数y=[,+∞)上是增函数;当0<x1<x2<y2<y1, 函数y=(0,]上是减函数.y=是偶函数,于是,该函数在(∞,]上是减函数, [,0)上是增函数;

     (3) 可以把函数推广为y=(常数a>0),其中n是正整数.

        n是奇数时,函数y=(0,]上是减函数,[,+∞) 上是增函数,-

  (∞,]上是增函数, [,0)上是减函数

        n是偶数时,函数y=(0,]上是减函数,[,+∞) 上是增函数,

(∞,]上是减函数, [,0)上是增函数.

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