(選修4-5:不等式選講)
已知函數(shù)f(x)=|x-4|+|x-1|.
(1)求f(x)的最小值;
(2)解不等式|x-4|+|x-1|≤5.
解:(1)函數(shù)f(x)=|x-4|+|x-1|=
,故當(dāng)1≤x≤4時(shí),f(x)
min=3.
(2)由于|x-4|+|x-1|表示數(shù)軸上的x對(duì)應(yīng)點(diǎn)到4和1對(duì)應(yīng)點(diǎn)的距離之和,而0和5 對(duì)應(yīng)點(diǎn)到4和1對(duì)應(yīng)點(diǎn)的距離之和正好等于5,
故不等式|x-4|+|x-1|≤5的解集為{x|0≤x≤5}.
分析:(1)根據(jù)函數(shù)f(x)=
,可得當(dāng)1≤x≤4時(shí),f(x)有最小值為3.
(2)由于|x-4|+|x-1|表示數(shù)軸上的x對(duì)應(yīng)點(diǎn)到4和1對(duì)應(yīng)點(diǎn)的距離之和,而0和5 對(duì)應(yīng)點(diǎn)到4和1對(duì)應(yīng)點(diǎn)的距離之和正好等于5,由此求得不等式|x-4|+|x-1|≤5的解集.
點(diǎn)評(píng):本題主要考查絕對(duì)值的意義,絕對(duì)值不等式的解法,屬于中檔題.