Question

一道函数题

已知定义在R上的偶函数f(x)在[0,正无穷）上是增函数，且f(1/3)=0，则满足f(log1/8x)>0(1/8是底数）的x取值范围

要具体过程

Answer 1

偶函数，则有f(x)=f(|x|)
故f(log1/8 x)>0
即f(|log1/8 x|)>f(1/3)
增函数得：|log1/8x|>1/3.
即log1/8x>1/3 或log1/8x<-1/3
得0<x<1/2或x>2

Answer 2

f在[0,正无穷）上是增函数
=>f在（负无穷，0]上是减函数

f(1/3) =0
for x> 1/3
f(x) > f(1/3) =0
log(1/8) x > 1/3
x > (1/8) ^(1/3)
=> x > 1/2

or
for x < -1/3
f(x) > f(-1/3) =0
=> log(1/8) x < -1/3
x < (1/8) ^(-1/3)
=> x < 2

ie 1/2 < x < 2 #