Question

求解题，谢谢。

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Answer 1

f(x) = x^2+(m-2)x +2
f(-x) = x^2 -(m-2)x +2 = f(x)
=>
m-2 =0
m=2
f(x) = x^2+2
g(x) =-x^2+(m+2)x-2
=-x^2+4x-2
=-(x-2)^2 +2
单调
减小 =[2, ∞)
增加 = (-∞,2]

(2)

f(x) domain {x|x≠0, x∈R } is odd
f(2)=0
f(x) 在 (0,+ ∞) is increasing
To find : solution set of f(x+1)<0

Solution
f(2) =0
f(x) 在 (0,+ ∞) is increasing
ie

x>2, f(x)>0
0<x<2 , f(x)<0
-2<x<0, f(x) >0
x<-2 ,f(x) <0

f(x+1) <0
=> 0<x+1 <2 or x+1<-2
=> -1<x<1 or x<-3

Answer 2

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