已知函數(shù)f(x)=log2(x+1),g(x+1)=log2(3x+2),求在g(x)≥f(x)成立的條件下,函數(shù)y=g(x)-f(x)的值域.
分析:先由g(x+1),求得g(x),再將“g(x)≥f(x)”轉(zhuǎn)化為“l(fā)og2(3x-1)≥log2(x+1)”,再利用對(duì)數(shù)函數(shù)的單調(diào)性求得x的取值范圍,即為新函數(shù)y=g(x)-f(x)的定義域,然后,利用函數(shù)的單調(diào)性求得新函數(shù)的值域.
解答:解:由題設(shè),g(x)=log
2(3x-1)--(2分)
由g(x)≥f(x)即:log
2(3x-1)≥log
2(x+1)得
?⇒x≥1∴使g(x)≥f(x)的x的取值范圍是
x≥1y=g(x)-f(x)=log
2(3x-1)-log
2(x+1)
=
log2=log2(3-)∵
x≥1∴1≤3-<3又∵y=log
2x在x∈(0,+∞)上單調(diào)遞增
∴當(dāng)
x≥1時(shí),log23>log2(3-)≥log21=0,
∴所求函數(shù)的值域?yàn)閇0,log
23)
點(diǎn)評(píng):本題主要考查對(duì)數(shù)的運(yùn)算法則及對(duì)數(shù)函數(shù)的定義域,單調(diào)性和值域,還考查了函數(shù)的構(gòu)造與轉(zhuǎn)化,體現(xiàn)了綜合性.