若函數(shù)f(x)=4x-2x+1+3的定義域為[-1,1],則f(x)值域為________.
[2,3]
分析:令2
x=t,t∈[

,2],則函數(shù)f(x)=4
x-2
x+1+3可轉(zhuǎn)化為g(t)=t
2-2t+3=(t-1)
2+2,然后根據(jù)二次函數(shù)的性質(zhì)可求出函數(shù)的值域.
解答:令2
x=t,t∈[

,2]
則g(t)=t
2-2t+3=(t-1)
2+2
當(dāng)t=1時即x=0時,函數(shù)取最小值2;
當(dāng)t=2時即x=1時,函數(shù)取最大值3;
故f(x)值域為[2,3]
故答案為:[2,3]
點評:本題主要考查了指數(shù)型復(fù)合函數(shù)的性質(zhì)及應(yīng)用,以及二次函數(shù)在閉區(qū)間上的最值,同時考查了轉(zhuǎn)化的思想,屬于基礎(chǔ)題.