已知冪函數(shù)f(x)=xa和對數(shù)函數(shù)g(x)=logax,其中a為不等于1的正數(shù)
(1)若冪函數(shù)的圖象過點(27,3),求常數(shù)a的值,并說明冪函數(shù)f(x)的單調(diào)性;
(2)若0<a<1,且函數(shù)y=g(x+3)在區(qū)間[-2,-1]上總有|y|≤2,求a的取值范圍.
分析:(1)將點的坐標代入冪函數(shù)解析式求出α,據(jù)α>0,冪函數(shù)單調(diào)遞增.
(2)求出函數(shù)的解析式,根據(jù)0<a<1時,對數(shù)函數(shù)單調(diào)遞減,求出函數(shù)的最值,列出不等式求出a的范圍.
解答:解:(1)∵冪函數(shù)的圖象過點(27,3),
∴3=27
α∴
a=,
∴
f(x)=x故函數(shù)在(-∞,+∞)上是單調(diào)增函數(shù)
(2)y=g(x+3)=log
a(x+3)
∵0<a<1,
∴y=log
a(x+3)在區(qū)間[-2,-1]上單調(diào)遞減
所以當(dāng)x=-2時y取得最大值0,當(dāng)x=-1時y取得最小值log
a2
∵|y|≤2
∴-log
a2≤2
a∈(0,] 點評:本題考查利用待定系數(shù)法求函數(shù)的解析式、冪函數(shù)的性質(zhì)、對數(shù)函數(shù)的單調(diào)性及解對數(shù)不等式.