20.已知函数f(x)=x²-2|x|.

(1)f(-x)=(-x)²-2|-x|=x²-2|x|=f(x)So the function is an even one。(2)Let us suppose that x1 and x2 are two numbers in (-1,0) and x10So f(x1)-f(x2)=(x1²-x2²)+2(x2-x1)>0In this way we can prove that the function increases in the section (-1,0)。