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数学
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函数y=根号下5-4x-x^2的递增区间
人气:138 ℃ 时间:2019-09-24 06:00:27
解答
定义域:
5-4x-x²≥0
x²+4x-5≤0
(x+5)(x-1)≤0
-5≤x≤1
这是一个复合函数,
y=√u在定义域内单调递增
u=5-4x-x²=-(x²+4x+4)+9=-(x+2)²+9
对称轴为x=-2 开口向下
u在[-5,-2]单调递增,在[-2,1]单调递减
所以y=根号下5-4x-x^2的递增区间为[-5,-2]
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