Question

12．已知函數(shù)$f（x）=\left\{\begin{array}{l}{3^{x+1}}，x≤0\\{log_{\frac{1}{2}}}x，x＞0\end{array}\right.$則不等式f（x）＞1的解集為$（-1，\frac{1}{2}）$．

Answer 1

分析 根據(jù)題意，由f（x）＞1，變形可得$\left\{\begin{array}{l}{{3}^{x+1}＞1}\\{x≤0}\end{array}\right.$①或$\left\{\begin{array}{l}{lo{g}_{\frac{1}{2}}x＞1}\\{x＜0}\end{array}\right.$②，解①②再取并集可得x的取值范圍，即可得答案．

解答 解：根據(jù)題意，函數(shù)的解析式為$f（x）=\left\{\begin{array}{l}{3^{x+1}}，x≤0\\{log_{\frac{1}{2}}}x，x＞0\end{array}\right.$，
若不等式f（x）＞1，$\left\{\begin{array}{l}{{3}^{x+1}＞1}\\{x≤0}\end{array}\right.$①或$\left\{\begin{array}{l}{lo{g}_{\frac{1}{2}}x＞1}\\{x＞0}\end{array}\right.$②，
解①可得：-1＜x≤0，
解②可得：0＜x＜$\frac{1}{2}$，
綜合可得：x的取值范圍：-1＜x＜$\frac{1}{2}$，
即（x）＞1的解集為（-1，$\frac{1}{2}$）；
故答案為：（-1，$\frac{1}{2}$）．

點(diǎn)評(píng) 本題考查分段函數(shù)的應(yīng)用，涉及指數(shù)、對(duì)數(shù)不等式的解法，注意轉(zhuǎn)化為兩個(gè)不等式組進(jìn)行分開求解．