求函数y=﹣x²+2|x|+3的单调递增区间
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求函数y=﹣x²+2|x|+3的单调递增区间
求函数y=﹣x²+2|x|+3的单调递增区间
求函数y=﹣x²+2|x|+3的单调递增区间
数形结合发,先画出函数的图像,是偶函数
显然,单调递增区间(-∞,-,1),(0,1)
x>=0
y=-x^2+2x+3
y=-(x^2-2x-3)
y=-((x-1)^2-1-3)
y=-((x-1)^2-4)
y=-(x-1)^2+4
(-无穷,1]
[0,1]
2.x<0
y=-x^2-2x+3
y=-(x^2+2x-3)
y=-((x+1)^2-1-3)
y=-((x+1)^2-4)
y=-(x+1)^2+4
(-无穷,-1]
所以单调递增区间是[0,1]和(-无穷,-1]