关于导数求函数最值的问题！！数学高手帮帮忙！！！

y = x/(1 + x²)
y' = [(1 + x²) - (2x²)]/(x² + 1)² = (1 - x²)/(x² + 1)²
y' = 0 => 1 - x² = 0 => x = ±1
x<-1，x = -1，-1f'(x) ↓ 0 ↑ 0 ↓
f(x) 单减 极小值 单增 极大值 单减
所以y = f(x)在(-∞，+∞)上的最小值为f(-1) = -1/2
最大值为f(1) = 1/2
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y = √x • lnx
y' = 1/2√x • lnx + √x • 1/x = (lnx + 2)/(2√x)
y' = 0 => lnx + 2 = 0 => lnx = - 2 => x = 1/e²
y'' = - lnx/[4x^(3/2)]，y''|(1/e²) > 0，取得极小值f(1/e²) = - 2/e
f(1/4) = - ln2，f(1) = 0
最小值为- 2/e，最大值为0
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y = x²e^-x
y' = e^-x • 2x + x² • e^-x • (-1) = x(2 - x)e^-x
y' = 0 => x = 0 或 x = 2
x<0，x = 0，0f'(x) ↓ 0 ↑ 0 ↓
f(x) 单减 极小值 单增 极大值 单减
极小值f(0) = 0，极大值f(2) = 4/e²
f(-4) = 16e⁴，f(4) = 16/e⁴
最小值为0，最大值为16e⁴