Question

4．若（a+2）${\;}^{-\frac{1}{3}}$＜（3-2a）${\;}^{-\frac{1}{3}}$，求a的取值范圍．

Answer 1

分析 借助于f（x）=x${\;}^{-\frac{1}{3}}$的單調(diào)性與單調(diào)區(qū)間分情況列出a+2和3-2a的大小關(guān)系解出．

解答 解：∵f（x）=x${\;}^{-\frac{1}{3}}$在（-∞，0）和（0，+∞）上是減函數(shù)，且當(dāng)x＞0時(shí)，f（x）＞0，當(dāng)x＜0時(shí)，f（x）＜0，
∵（a+2）${\;}^{-\frac{1}{3}}$＜（3-2a）${\;}^{-\frac{1}{3}}$，
∴$\left\{\begin{array}{l}{a+2＞0}\\{3-2a＞0}\\{a+2＞3-2a}\end{array}\right.$或$\left\{\begin{array}{l}{a+2＜0}\\{3-2a＜0}\\{a+2＞3-2a}\end{array}\right.$或$\left\{\begin{array}{l}{a+2＜0}\\{3-2a＞0}\end{array}\right.$，解得$\frac{1}{3}＜a$＜$\frac{3}{2}$或a＜-2．
∴a的取值范圍是（-∞，-2）∪（$\frac{1}{3}$，$\frac{3}{2}$）．

點(diǎn)評(píng) 本題考查了利用函數(shù)單調(diào)性解不等式，找到合理的函數(shù)模型是關(guān)鍵．