Question

f(x)=logax+b/x-b(a>0,b>0,a≠1） 求单调性,不要复合函数单调性,要定义证明,

Answer 1

f(x) = log a (x+b)/(x-b) = log a (x-b+2b)/(x-b) = log a { 1+2b/(x+b) } = log a { 1+2/(x/b+1) }

x∈（-∞,-b)时,x/b单调减, 1+2/(x/b+1)单调增,

其中当0＜a＜1时,f(x)单调减；当a＞1时,f(x)单调增

x∈（b,+∞)时,x/b单调增, 1+2/(x/b+1)单调减,

其中当0＜a＜1时,f(x)单调增；当a＞1时,f(x)单调减