Question

判断式子|1-x|-√x^2-8x+16有没有最小值.若有,求出此值,并求出此时x的取值范围

Answer 1

√x^2-8x+16=|x-4|

当x≥4时,原式=x-1-(x-4)=3

当x≤1时,原式=1-x-(4-x)=-3

当1