Question

已知函数f(x)=|x+1|.(1)求不等式f(x)<|2x+1|-1的解集M;(2)设a,b∈M,证明:f(ab)>f(a)-f(-b).

Answer 1

• 令g(x)=f(x)-|2x+1|+1=|x+1|-|2x+1|+1,分段：

  g(x)=x+1  x≤-1

  f(x)=3x+3 -1<x<-½ 

  f(x)=-x+1  x≥-½

  

1. 解集M为|x|<1

2. f(x)=-x-1 x<-1 单调递减

   f(x)=x+1 x>1   单调递增

   显然①f(|x|)=f(-|x|)+2 ②|ab|>|a|、|ab|>|b|  (a,b∈M)

   a<-1 b<-1时 ①

   f(a)=-a-1 f(-b)=f(b)+2=-b-1+2=1-b

   f(a)-f(-b)=-a-1+b-1=b-a-2

   f(ab)=ab+1>-a+1>-a-2>b-a-2=f(a)-f(-b)

   a>1 b>1时    ②
   

   f(a)=a+1 f(-b)=b+1-2=b-1

   f(a)-f(-b)=a+1-b+1=a-b+2＜a-1+2=a+1

   f(ab)=ab+1>a+1>f(a)-f(-b)

   a>1 b<-1        ③

   f(a)=a+1 f(-b)=f(b)+2=-b-1+2=1-b

   f(a)-f(-b)=a+1+b-1=a+b<a-1

   f(ab)=f(-ab)-2=-ab-1>a-1>f(a)-f(-b)

   a<-1 b>1        ④

   f(a)=-a-1 f(-b)=f(b)-2=b+1-2=b-1

   f(a)-f(-b)=-a-1-b+1=-a-b<-a-1

   f(ab)=f(-ab)-2=-ab-1>-a-1>f(a)-f(-b)

   得证。

Answer 2

引用善言而不辩的回答：
令g(x)=f(x)-|2x+1|+1=|x+1|-|2x+1|+1,分段：
g(x)=x+1 x≤-1
f(x)=3x+3 -1<x<-½
f(x)=-x+1 x≥-½

解集M为|x|<1
f(x)=-x-1 x<-1 单调递减
f(x)=x+1 x>1 单调递增
显然①f(|x|)=f(-|x|)+2 ②|ab|>|a|、|ab|>|b| (a,b∈M)
a<-1 b<-1时 ①
f(a)=-a-1 f(-b)=f(b)+2=-b-1+2=1-b
f(a)-f(-b)=-a-1+b-1=b-a-2
f(ab)=ab+1>-a+1>-a-2>b-a-2=f(a)-f(-b)
a>1 b>1时 ②
f(a)=a+1 f(-b)=b+1-2=b-1
f(a)-f(-b)=a+1-b+1=a-b+2＜a-1+2=a+1
f(ab)=ab+1>a+1>f(a)-f(-b)
a>1 b<-1 ③
f(a)=a+1 f(-b)=f(b)+2=-b-1+2=1-b
f(a)-f(-b)=a+1+b-1=a+b<a-1
f(ab)=f(-ab)-2=-ab-1>a-1>f(a)-f(-b)
a<-1 b>1 ④
f(a)=-a-1 f(-b)=f(b)-2=b+1-2=b-1
f(a)-f(-b)=-a-1-b+1=-a-b<-a-1
f(ab)=f(-ab)-2=-ab-1>-a-1>f(a)-f(-b)
得证。

第一问是x＜-1或x＞1