已知函数f(x)对一切实数x,y
1)令x=y=0
f(0+0)=f(0)×f(0)
==>f(0) =[f(0)]²
f(x)≠0
==> f(0)=1
当x0
f(x+ -x) =f(x))×f(-x)
即f(0)= 1 =f(x)×f(-x)
==> f(-x) =1/f(x)
f(x)>1
====>00时，0＜f(x)＜1
2）设y>0
f(x+y)-f(x) =f(x)×f(y) -f(x)
=f(x)[f(y-1)]
（1）已经证明了f(x)>0 ，
而0＜f(y)＜1
所以 f(x)[f(y-1)]＜0
====>f(x)在R上递减
3）f(-1)=2
f[-1 +(-1)]=f(-1))×f(-...全部
1)令x=y=0
f(0+0)=f(0)×f(0)
==>f(0) =[f(0)]²
f(x)≠0
==> f(0)=1
当x0
f(x+ -x) =f(x))×f(-x)
即f(0)= 1 =f(x)×f(-x)
==> f(-x) =1/f(x)
f(x)>1
====>00时，0＜f(x)＜1
2）设y>0
f(x+y)-f(x) =f(x)×f(y) -f(x)
=f(x)[f(y-1)]
（1）已经证明了f(x)>0 ，
而0＜f(y)＜1
所以 f(x)[f(y-1)]＜0
====>f(x)在R上递减
3）f(-1)=2
f[-1 +(-1)]=f(-1))×f(-1)
==> f(-2)=4
f(x²/2+x-1)>1/4
==> 4f(x²/2+x-1)>1
即f(-2)f((x²/2+x-1)>1
即f(-2)f(x²/2+x-1) =f[-2 +(x²/2+x-1) =f(x²/2+x-3)>1
因为f(0)=1 ，f(x)在R上递减
==〉x²/2+x-3 >0
x²+2x-6 >0
x∈(-∞ ,-1 -√7)∪(-1+√7 。
+∞ )
。收起
