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：证明：由(1)知 \(f(x) = \frac{{1 - {2^x}}}{{1 + {2^x}}}\) .任取 \({x_1},{x_2} \in R\)，且 \({x_1} < {x_2}\), 则 \(\begin{gathered}
f({x_1}) - f({x_2}) \\
= \frac{{1 - {2^{{x_1}}}}}{{1 + {2^{{x_1}}}}} - \frac{{1 - {2^{{x_2}}}}}{{1 + {2^{{x_2}}}}} \\
= \frac{{\left( {1 - {2^{{x_1}}}} \right)\left( {1 + {2^{{x_2}}}} \right) - \left( {1 - {2^{{x_2}}}} \right)\left( {1 + {2^{{x_1}}}} \right)}}{{\left( {1 + {2^{{x_1}}}} \right)\left( {1 + {2^{{x_2}}}} \right)}} \\
= \frac{{2\left( {{2^{{x_2}}} - {2^{{x_1}}}} \right)}}{{\left( {1 + {2^{{x_1}}}} \right)\left( {1 + {2^{{x_2}}}} \right)}} \\
\end{gathered} \) \(\begin{gathered}
\because {x_1} < {x_2} \\
\therefore {2^{{x_2}}} - {2^{{x_1}}} > 0 \\
\end{gathered} \)又 \(\left( {{2^{{x_1}}} + 1} \right)\left( {{2^{{x_2}}} + 1} \right) > 0\)故 \(f\left( {{x_1}} \right) - f\left( {{x_2}} \right) > 0\) ，即 \(f\left( {{x_1}} \right) > f\left( {{x_2}} \right)\) ，所以 \(f(x)\) 为 \(R\) 上的减函数．