高中数学必修 第一册（648题）

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知识点：第四章 指数函数与对数函数

参考答案：令\(t={2}^{x}-{2}^{-x}\)

易知 \(t={2}^{x}-{2}^{-x}\) 为增函数，

故由\(x\in \left [ {-1，1} \right ]\) ，

可得：\(t\in \left [ {-\frac {3} {2}，\frac {3} {2}} \right ]\)

对函数 \(g\left ( {x} \right )=h\left ( {t} \right )={t}^{2}-mt+2\)，\(t\in \left [ {-\frac {3} {2}，\frac {3} {2}} \right ]\)它的最小值为\(-2\)

等价于\(\left \{ \begin{gathered} {\frac {m} {2}<-\frac {3} {2}} \\ {h\left ( {-\frac {3} {2}} \right )=\frac {9} {4}+\frac {3} {2}m+2=-2} \end{gathered} \right .\) 或\(\left \{ \begin{gathered} {-\frac {3} {2}\leq \frac {m} {2}\leq \frac {3} {2}} \\ {h\left ( {\frac {m} {2}} \right )=\left ( {\frac {m} {2}} \right )^{2}-m\cdot \frac {m} {2}+2=-2} \end{gathered} \right .\) 或\(\left \{ \begin{gathered} {\frac {m} {2}>\frac {3} {2}} \\ {h\left ( {\frac {3} {2}} \right )=\frac {9} {4}-\frac {3} {2}m+2=-2} \end{gathered} \right .\) ，

解得：\(m=-\frac {25} {6}\) 或\(\frac {25} {6}\)