Question

9．設(shè)函數(shù)f（x）=$\left\{\begin{array}{l}{{e}^{x-1}，x＜1}\\{{x}^{\frac{1}{3}}，x≥1}\end{array}\right.$則使得f（x）≤e成立的x的取值范圍是（-∞，e³]．

Answer 1

分析 原題等價于$\left\{\begin{array}{l}{{e}^{x-1}≤e}\\{x＜1}\end{array}\right.$或$\left\{\begin{array}{l}{{x}^{\frac{1}{3}}≤e}\\{x≥1}\end{array}\right.$，由此能求出x的取值范圍．

解答 解：∵f（x）=$\left\{\begin{array}{l}{{e}^{x-1}，x＜1}\\{{x}^{\frac{1}{3}}，x≥1}\end{array}\right.$，f（x）≤e，
∴$\left\{\begin{array}{l}{{e}^{x-1}≤e}\\{x＜1}\end{array}\right.$或$\left\{\begin{array}{l}{{x}^{\frac{1}{3}}≤e}\\{x≥1}\end{array}\right.$，
解得x＜1或1≤x≤e³，
∴x的取值范圍是（-∞，e³]．
故答案為：（-∞，e³]．

點評 本題考查實數(shù)的取值范圍的求法，是基礎(chǔ)題，解題時要認真審題，注意等價轉(zhuǎn)化思想的合理運用．