Question

14．設函數f（x）=$\left\{\begin{array}{l}{-{x}^{2}，x＞0}\\{{2}^{x}，x＜0}\end{array}\right.$，g（x）=$\left\{\begin{array}{l}{-\frac{1}{x}，x＞0}\\{x-1，x＜0}\end{array}\right.$則g（f（-1））的值為-2．

Answer 1

分析 先求出f（-1）=${2}^{-1}=\frac{1}{2}$，從而g（f（-1））=g（$\frac{1}{2}$）=-$\frac{1}{\frac{1}{2}}$=-2．由此能求出結果．

解答 解：∵函數f（x）=$\left\{\begin{array}{l}{-{x}^{2}，x＞0}\\{{2}^{x}，x＜0}\end{array}\right.$，
g（x）=$\left\{\begin{array}{l}{-\frac{1}{x}，x＞0}\\{x-1，x＜0}\end{array}\right.$
∴f（-1）=${2}^{-1}=\frac{1}{2}$，
g（f（-1））=g（$\frac{1}{2}$）=-$\frac{1}{\frac{1}{2}}$=-2．
故答案為：-2．

點評 本題考查函數值的求法，是基礎題，解題時要認真審題，注意函數性質的合理運用．