X 1, x2 >;; 0 has [f (x1)-f (x2)]/(x1-x2) > 0, available:
F(x) is increasing function on (0, positive infinity), so f(x) is also increasing function on (-infinity, 0).
f(2)=0,f(-2)=0
On (0, positive infinity) 0 < x & lt2f (x) < =0; x & gt=2,f(x)>0
On (-infinity,0)x < =-2 f(x)& lt; =0; -2 & lt; x & lt0,f(x)>0
Odd function x=0 f(0)=0
The solution set of inequality xf(x)< 0 is _ (-2,0) ∩ (0,2).
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