已知函數(shù)f(x)對(duì)任意實(shí)數(shù)x,y恒有f(x+y)=f(x)+f(y)且當(dāng)x>0,f(x)<0.又f(1)=-2.
(1)判斷函數(shù)f(x)的奇偶性;
(2)求函數(shù)f(x)在區(qū)間[-3,3]上的最大值;
(3)解關(guān)于x的不等式f(ax2)-2f(x)<f(ax)+4.
分析:(1)先求f(0)=0,再取y=-x,則f(-x)=-f(x)對(duì)任意x∈R恒成立,故可得函數(shù)為奇函數(shù);
(2)先判斷函數(shù)在(-∞,+∞)上是減函數(shù),再求f(-3)=-f(3)=6,從而可求函數(shù)的最大值;
(3)利用函數(shù)為奇函數(shù),可整理得f(ax2-2x)<f(ax-2),利用f(x)在(-∞,+∞)上是減函數(shù),可得ax2-2x>ax-2,故問題轉(zhuǎn)化為解不等式.
解答:解:(1)取x=y=0,則f(0+0)=2f(0),∴f(0)=0…1′
取y=-x,則f(x-x)=f(x)+f(-x)∴f(-x)=-f(x)對(duì)任意x∈R恒成立∴f(x)為奇函數(shù).…3′
(2)任取x
1,x
2∈(-∞,+∞)且x
1<x
2,則x
2-x
1>0,∴f(x
2)+f(-x
1)=f(x
2-x
1)<0,…4′
∴f(x
2)<-f(-x
1),
又f(x)為奇函數(shù)∴f(x
1)>f(x
2)
∴f(x)在(-∞,+∞)上是減函數(shù).∴對(duì)任意x∈[-3,3],恒有f(x)≤f(-3)…6′
而f(3)=f(2+1)=f(2)+f(1)=3f(1)=-2×3=-6,
∴f(-3)=-f(3)=6,∴f(x)在[-3,3]上的最大值為6…8′
(3)∵f(x)為奇函數(shù),∴整理原式得 f(ax
2)+f(-2x)<f(ax)+f(-2),
進(jìn)一步得f(ax
2-2x)<f(ax-2),
而f(x)在(-∞,+∞)上是減函數(shù),
∴ax
2-2x>ax-2…10′∴(ax-2)(x-1)>0.
∴當(dāng)a=0時(shí),x∈(-∞,1)
當(dāng)a=2時(shí),x∈{x|x≠1且x∈R}
當(dāng)a<0時(shí),
x∈{x|<x<1}當(dāng)0<a<2時(shí),
x∈{x|x>或x<1}當(dāng)a>2時(shí),
x∈{x|x<或x>1}…12′
點(diǎn)評(píng):本題考查抽象函數(shù)的性質(zhì),賦值法事常用方法,同時(shí)借助于函數(shù)的單調(diào)性,抽象函數(shù)的不等式問題可以轉(zhuǎn)化為具體函數(shù)求解.