Question

已知函数f(x)=log2(2"x-1)求函数f(x)的定义域；判断并证明函数f(x)单调性；

2"x-1是二x平方减一log2的2是log的不是2(2"x-1)

Answer 1

2x^2-1>0x^2>1/2x>√2/2或x<-√2/2f(x1)-f(x2)=log2 (2x1^2-1)-log2 (2x2^2-1)=log2 (2x1^2-1)/(2x2^2-1)=log2 (√2x1+1)(√2x1-1)/(√2x2+1)(√2x2-1)1）若x1<x2∈(√2/2,+∞)则(√2x2-1)>(√2x1-1)>0故0<(√2x1+1)(√2x1-1)/(√2x2+1)(√2x2-1)<1从而log2 (√2x1+1)(√2x1-1)/(√2x2+1)(√2x2-1)<0即f(x1)<f(x2),函数在(√2/2,+∞)上是增函数2）若x1<x2∈(-∞，-√2/2）则(√2x2-1)<(√2x1-1)<(√2x2+1)<(√2x1+1)<0故(√2x1+1)(√2x1-1)/(√2x2+1)(√2x2-1)>1从而log2 (√2x1+1)(√2x1-1)/(√2x2+1)(√2x2-1)>0即f(x1)>f(x2),函数在(-∞，-√2/2）上是减函数