a |
x |
∴对任意的1≤x1<x2,有f(x1)<f(x2),
即log9(x1+8−
a |
x1 |
a |
x2 |
得x1+8−
a |
x1 |
a |
x2 |
a |
x1x2 |
∵x1-x2<0,∴1+
a |
x1x2 |
a |
x1x2 |
∵x2>x1≥1,∴要使a>-x1x2恒成立,只要a≥-1;
又∵函数f(x)=log9(x+8−
a |
x |
即a<9,综上a的取值范围为[-1,9).
另(用导数求解)令g(x)=x+8−
a |
x |
函数f(x)=log9(x+8−
a |
x |
∴g(x)=x+8−
a |
x |
a |
x2 |
∴1+8-a>0,且1+
a |
x2 |