题目内容
4.设函数f(x)=$\left\{\begin{array}{l}{{2}^{-x}(x<1)}\\{lo{g}_{4}x(x≥1)}\end{array}\right.$.(1)求f(0),f(2),f(f(3))的值;
(2)求不等式f(x)≤2的解集.
分析 (1)由分段函数,代入数值,计算即可得到所求,注意运用对数的性质和恒等式;
(2)由题意可得,$\left\{\begin{array}{l}{{2}^{-x}≤2}\\{x<1}\end{array}\right.$或$\left\{\begin{array}{l}{lo{g}_{4}x≤2}\\{x≥1}\end{array}\right.$,运用指数函数和对数函数的单调性,解出它们,再求交集即可得到所求不等式的解集.
解答 解:(1)函数f(x)=$\left\{\begin{array}{l}{{2}^{-x}(x<1)}\\{lo{g}_{4}x(x≥1)}\end{array}\right.$,
可得f(0)=20=1,f(2)=log42=$\frac{1}{2}$,
f(3)=log43<1,f(f(3))=2-log43=${4}^{\frac{1}{2}lo{g}_{4}\frac{1}{3}}$=${4}^{lo{g}_{4}\frac{\sqrt{3}}{3}}$=$\frac{\sqrt{3}}{3}$;
(2)由题意可得$\left\{\begin{array}{l}{{2}^{-x}≤2}\\{x<1}\end{array}\right.$或$\left\{\begin{array}{l}{lo{g}_{4}x≤2}\\{x≥1}\end{array}\right.$,
即为$\left\{\begin{array}{l}{-x≤1}\\{x<1}\end{array}\right.$或$\left\{\begin{array}{l}{0<x≤16}\\{x≥1}\end{array}\right.$,
即有-1≤x<1或1≤x≤16,
可得-1≤x≤16,
则不等式的解集为[-1,16].
点评 本题考查分段函数的运用:求函数值和解不等式,考查运算能力,属于中档题.
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