Question

已知函数f(x)=2的x次方+2的负x次方a(常数a属于R) 若a小于等于4,求证；函数f（x）在区间【1,正无穷】上是

Answer 1

f(x) = 2^x+2^(-x) *ax∈【1,+∞）令1≤x1＜x2f(x2)-f(x1) = 【2^x2+2^(-x2) *a】-【2^x1+2^(-x1) *a】= (2^x2 - 2^x1) + a{ 1/2^x2 - 1/2^x1 }= (2^x2 - 2^x1) + a ( 2^x1-2^x2)/(2^x2 * 2^x1)= (2^x2 - 2^x1) - a ...