Question

奇函数F（x）满足F（x+2）= —F（x）,且x属于（0,1）时,F（x）=2^x,求F（log1/2 18）的值

Answer 1

先化解log1/2 18如下：

log1/2 18 = - log2 18 = -(log2 (2*3^2))

= -(log2 2 + 2*log2 3 )

= -1 - 2* log2 3

于是有：

F(log1/2 18) = F(-1- 2* (log2 3) )

又由已知：

F（x+2) = -F(x)

F(x) = -F(x+2)

故

F(-1- 2*(log2 3)) = -F(2-1-2*(log2 3))

= F(2+2-1-2*(log2 3))

= F(3-2*(log2 3) )

= F(log2 8/9)

由于log2 8/9在区间(-1,0)

所以要求F(x) 在区间内的表达式

由其奇偶性可得

F(x) = -F(-x)

故当x 在(-1,0)时

F(x) = -2^(-x)

故F(log2 8/9) = -2^(log2 8/9) = -8/9