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20.已知函数f(x)=x²-2|x|.

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20.已知函数f(x)=x²-2|x|.
(1)判断并证明函数的奇偶性,并作函数图像(2)判断函数f(x)在(-1,0)上的单调性并加以证明.
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 x10
So f(x1)-f(x2)=(x1²-x2²)+2(x2-x1)>0
In this way we can prove that the function increases in the section (-1,0).