Question

求函数y=根号下{log以1/3为底（3+2x-x²)的对数}的定义域

Answer 1

y = √ { log (1/3) [(3+2x-x²)]}

首先,零和负数无对数,所以3+2x-x²＞0,(x+1)(x-3)＜0,-1＜x＜3

第二,根号下无负数,∴ log (1/3) [(3+2x-x²)]＞0,∴3+2x-x²＜1,x²-2x+1＜3,（x-1)²＜3,1-√3＜x＜1+√3

∴1-√3＜x＜1+√3