题目内容
18.已知函数f(x)=$\frac{(\frac{1}{2})^{x}-1}{(\frac{1}{2})^{x}+1}$.(1)判断f(x)的奇偶性;
(2)证明f(x)在定义域上单调递减.
分析 (1)根据函数奇偶性的定义即可判断f(x)的奇偶性;
(2)利用函数单调性的定义即可证明f(x)在定义域上单调递减.
解答 解:(1)函数的定义域为(-∞,+∞),
则f(-x)=$\frac{(\frac{1}{2})^{-x}-1}{(\frac{1}{2})^{-x}+1}$=$\frac{1-(\frac{1}{2})^{x}}{1+(\frac{1}{2})^{x}}$=-$\frac{(\frac{1}{2})^{x}-1}{(\frac{1}{2})^{x}+1}$=-f(x).
则f(x)为奇函数;
(2)证明f(x)在定义域上单调递减.
f(x)=$\frac{(\frac{1}{2})^{x}-1}{(\frac{1}{2})^{x}+1}$=$\frac{{(\frac{1}{2})}^{x}+1-2}{{(\frac{1}{2})}^{x}+1}$=1-$\frac{2}{(\frac{1}{2})^{x}+1}$,
设x1<x2,
则f(x1)-f(x2)=1-$\frac{2}{(\frac{1}{2})^{{x}_{1}}+1}$-1+$\frac{2}{(\frac{1}{2})^{{x}_{2}}+1}$=$\frac{2}{(\frac{1}{2})^{{x}_{2}}+1}$-$\frac{2}{(\frac{1}{2})^{{x}_{1}}+1}$=$\frac{2[(\frac{1}{2})^{{x}_{1}}-(\frac{1}{2})^{{x}_{2}}]}{[(\frac{1}{2})^{{x}_{1}}+1][(\frac{1}{2})^{{x}_{2}}+1]}$,
∵x1<x2,
∴$(\frac{1}{2})^{{x}_{1}}$>$(\frac{1}{2})^{{x}_{2}}$,即$(\frac{1}{2})^{{x}_{1}}$-$(\frac{1}{2})^{{x}_{2}}$>0,
则f(x1)>f(x2),
即f(x)在定义域上单调递减.
点评 本题主要考查函数奇偶性和单调性的判断,利用定义法是解决本题的关键.
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