f(x)=2x∧2-ax+a在区间(负无穷,1)上不单调.则函数g(x)=f(x)/x在区间(2,正无穷)上一定

smile19822022-10-04 11:39:542条回答

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ido923 共回答了18个问题 | 采纳率88.9%
f'(x) = 4x - a = 0, x = a/4
f(x)在区间(-∞,1)上不单调, 则对称轴x = a/4 < 1, a < 4
g(x) = f(x)/x = 2x - a + a/x
g'(x) = 2 - a/x² = 0
(1) a = 0
g'(x) > 0, g(x)单调增
(2) 0 < a < 4
x² = a/2 < 2, x < 2
x > 2: g'(x) > 0, g(x)单调增
(3) a < 0
g'(x) > 0, g(x)单调增
g(x)在区间(2,∞)上一定单调增
1年前
liaojingyu225 共回答了1个问题 | 采纳率
13216574 鞍山
1年前

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