Question

判断函数f(x)=x^2-2|x|..x∈R,在(-1,0)上的单调性并加以证明

Answer 1

令,0>X2>X1,有X2-X1>0,

f(x2)-f(x1)=x2^2-2|x2|-x1^+2|x1|

=(x2^2-x1^2)-2(|x2|-|x1|)

=(x2-x1)(x2+x1)-2(|x2|-|x1|).

∵x∈R,在(-1,0),∴|x|0,x∈R,在(-1,0),(X2+X1+2)>0,

有(X2-X1)(X2+X1+2)>0.

f(x2)-f(x1)>0,

f(x2)>f(x1).

∴f(x)在(-1,0)上的单调递增.