考點(diǎn):分段函數(shù)的應(yīng)用
專題:函數(shù)的性質(zhì)及應(yīng)用,不等式的解法及應(yīng)用
分析:由題意可得函數(shù)f(x)為奇函數(shù),函數(shù)f(x)在R上是增函數(shù).令x2+x=12,求得x=3或x=-4(舍去).故由不等式f(x2-x+1)<12,可得 x2-x+1<3,由此求得x的范圍.
解答:
解:∵f(x)=
,
∴f(-x)=-f(x)恒成立,
∴函數(shù)f(x)為奇函數(shù),
再根據(jù)二次函數(shù)的圖象和性質(zhì)可得:
f(x)在(0,+∞)上是增函數(shù),f(0)=0,可得函數(shù)f(x)在R上是增函數(shù).
令x
2+x=12,求得x=3 或x=-4(舍去).
∴由不等式f(x
2-x+1)<12,可得 x
2-x+1<3,
即 (x+1)(x-2)<0,
解得-1<x<2,
故答案為:(-1,2).
點(diǎn)評(píng):本題主要考查分段函數(shù)的應(yīng)用,考查函數(shù)的單調(diào)性和奇偶性的應(yīng)用,屬于基礎(chǔ)題.