设函数f（x）=2的次方为|x+1|-|x-1|,求使f（x）＞=2根号2的取值范围
问题描述：
设函数f（x）=2的次方为|x+1|-|x-1|,求使f（x）＞=2根号2的取值范围
答
函数f（x）=2^(|x+1|-|x-1|)f（x）＞=2根号22^(|x+1|-|x-1|)＞=2根号22^(|x+1|-|x-1|)＞=2^1 * 2^(1/2)2^(|x+1|-|x-1|)＞=2^(3/2)|x+1|-|x-1|＞=3/2当x≤-1时-x-1-(1-x)＞=3/2-2＞=3/2不成立舍去当-1＜x＜1时x+1-(1-...为什么是2^(|x+1|-|x-1|)＞=2^(3/2)？3根号2是怎么出来的？谢谢！2根号2=2√2=2 * √2=(2^1) * 2^(1/2)=2^(1+1/2)=2^(3/2)