Question

已知函数f(x)=2^x-1/[2^（绝对值x）]

Answer 1

若x0,

2^x-1/2^x=2

(2^x)^2-2*2^x-1=0

2^x=1+√2或2^x=1-√2

又x>0使得2^x>1,所以2^x=1+√2

x=log(2)(1+√2)

2^t(2^(2t)-1/2^(2t))+m(2^t-1/2^t)≥0

(2^t)^4+m*(2^t)^2-m-1≥0

(2^2t)^2+m*(2^2t)-m-1≥0

t∈[1,2],2^2t∈[4,16]

g(x)=x^2+mx-m-1=0

(x+m+1)(x-1)=0

则-m-1-5