Question

9．已知函數(shù)f（x）=$\left\{\begin{array}{l}{{x}^{2}，x＜0}\\{x-1，x≥0}\end{array}\right.$，若f（x）≤1，則x的取值范圍是[-1，2]．

Answer 1

分析 首先由f（x）≤1結(jié)合分段函數(shù)轉(zhuǎn)化為不等式組，然后分別解不等式組取并集即可．

解答 解：由已知，f（x）≤1，即$\left\{\begin{array}{l}{{x}^{2}≤1}\\{x＜0}\end{array}\right.$和$\left\{\begin{array}{l}{x-1≤1}\\{x≥0}\end{array}\right.$，
分別解這兩個不等式組得-1≤x＜0和0≤x≤2，所以f（x）≤1的解集為[-1，2]．
故答案為：[-1，2]．

點評 本題考查了分段函數(shù)與不等式組的解法；將f（x）≤1轉(zhuǎn)化為具體的不等式組是關(guān)鍵．