求单调区间fx=(|x|-3)(x-1)
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f(x)
=(|x|-3)(x-1)
=(-x-3)(x-1) ; x<0
=(x-3)(x-1) ; x≥0
case 1: x<0
f(x)
=(-x-3)(x-1)
=-x^2-2x+3
=-(x+1)^2 +4
单调
递增 =(-无穷, -1]
递减=[-1, 0)
case 2: x≥0
f(x)
=(x-3)(x-1)
=x^2-4x+3
=(x-2)^2 +2
单调
递增 =[2,-无穷)
递减=[0, 2]
//
所以
f(x) =(|x|-3)(x-1)
单调
递增 =(-无穷, -1] U [2,-无穷)
递减 =[-1, 2]
=(|x|-3)(x-1)
=(-x-3)(x-1) ; x<0
=(x-3)(x-1) ; x≥0
case 1: x<0
f(x)
=(-x-3)(x-1)
=-x^2-2x+3
=-(x+1)^2 +4
单调
递增 =(-无穷, -1]
递减=[-1, 0)
case 2: x≥0
f(x)
=(x-3)(x-1)
=x^2-4x+3
=(x-2)^2 +2
单调
递增 =[2,-无穷)
递减=[0, 2]
//
所以
f(x) =(|x|-3)(x-1)
单调
递增 =(-无穷, -1] U [2,-无穷)
递减 =[-1, 2]
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