题目内容
16.已知函数f(x)=1g$\frac{{x}^{2}-x+1}{{x}^{2}+1}$.(1)判断函数f(x)在区间[-1,1]上的单调性;
(2)当t∈R时.求证:1g$\frac{7}{10}$≤f(|t-$\frac{1}{6}$|-|t+$\frac{1}{6}$|)≤lg$\frac{13}{10}$.
分析 (1)容易求出该函数的定义域为R,根据单调性的定义,可设任意的x1,x2∈[-1,1],且x1<x2,通过作差,通分,及提取公因式便可以得出$\frac{{{x}_{1}}^{2}-x+1}{{{x}_{1}}^{2}+1}>\frac{{{x}_{2}}^{2}-{x}_{2}+1}{{{x}_{2}}^{2}+1}$,从而得出f(x1)>f(x2),这便得出函数f(x)在[-1,1]上为减函数;
(2)可去绝对值号便可求出|t$-\frac{1}{6}$|-$|t+\frac{1}{4}|$$∈[-\frac{1}{3},\frac{1}{3}]$,然后根据f(x)的单调性,便可得出$f(-\frac{1}{3})≤f(|t-\frac{1}{6}|-|t+\frac{1}{6}|)≤f(\frac{1}{3})$,这样便可看出求出$f(-\frac{1}{3})和f(\frac{1}{3})$,便可得出要证明的结论.
解答 解:(1)$\frac{{x}^{2}-x+1}{{x}^{2}+1}=\frac{(x-\frac{1}{2})^{2}+\frac{3}{4}}{{x}^{2}+1}>0$;
∴该函数定义域为R;
$\frac{{x}^{2}-x+1}{{x}^{2}+1}=1-\frac{x}{{x}^{2}+1}$,设x1,x2∈[-1,1],且x1<x2,则:
$\frac{{{x}_{1}}^{2}-{x}_{1}+1}{{{x}_{1}}^{2}+1}-\frac{{{x}_{2}}^{2}-{x}_{2}+1}{{{x}_{2}}^{2}+1}$=$\frac{{x}_{2}}{{{x}_{2}}^{2}+1}-\frac{{x}_{1}}{{{x}_{1}}^{2}+1}=\frac{({x}_{2}-{x}_{1})(1-{x}_{1}{x}_{2})}{({{x}_{2}}^{2}+1)({{x}_{1}}^{2}+1)}$;
∵x1,x2∈[-1,1],且x1<x2;
∴x2-x1>0,x1x2∈[-1,1];
∴1-x1x2>0;
∴$\frac{{{x}_{1}}^{2}-{x}_{1}+1}{{{x}_{1}}^{2}+1}>\frac{{{x}_{2}}^{2}-{x}_{2}+1}{{{x}_{2}}^{2}+1}$;
∴$lg\frac{{{x}_{1}}^{2}-{x}_{1}+1}{{{x}_{1}}^{2}+1}>lg\frac{{{x}_{2}}^{2}-{x}_{2}+1}{{{x}_{2}}^{2}+1}$;
即f(x1)>f(x2);
∴f(x)在[-1,1]上单调递减;
(2)证明:$|t-\frac{1}{6}|-|t+\frac{1}{6}|$=$\left\{\begin{array}{l}{\frac{1}{3}}&{t≤-\frac{1}{6}}\\{-2t}&{-\frac{1}{6}<t<\frac{1}{6}}\\{-\frac{1}{3}}&{t≥\frac{1}{6}}\end{array}\right.$;
∴$-\frac{1}{3}≤|t-\frac{1}{6}|-|t+\frac{1}{6}|≤\frac{1}{3}$;
∵f(x)在[-1,1]上为减函数;
∴$f(\frac{1}{3})≤f(|t-\frac{1}{6}|-|t+\frac{1}{6}|)≤f(-\frac{1}{3})$,f($\frac{1}{3}$)=$lg\frac{\frac{1}{9}-\frac{1}{3}+1}{\frac{1}{9}+1}=lg\frac{7}{10}$,$f(-\frac{1}{3})=lg\frac{\frac{1}{9}+\frac{1}{3}+1}{\frac{1}{9}+1}=lg\frac{13}{10}$;
∴$lg\frac{7}{10}≤f(|t-\frac{1}{6}|-|t+\frac{1}{6}|)≤lg\frac{13}{10}$.
点评 考查配方法出来二次式子,对数函数的单调性,单调性的定义,以及根据单调性定义判断一个函数单调性的方法和过程,作差法比较大小的应用,作差后分式的通分,能提取公因式的提取公因式,以及减函数定义的运用.
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