求x^2+x*(4-x^2)^(1/2)最大值
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y = x^2 + x√(4-x^2) , 定义域 x ∈[-2, 2]
y' = 2x +√(4-x^2) - x^2/√(4-x^2)
= 2[x√(4-x^2)+2-x^2]/√(4-x^2)
令 y' = 0, 得 x√(4-x^2)+2-x^2 = 0
x√(4-x^2) = x^2 - 2， x^2(4-x^2) = x^4 - 4x^2 + 4 ，
x^4 - 4x^2 + 2 = 0 , x^2 = [4±√(16-8)]/2 = 2±√2
x^2 = 2-√2, 或 x^2 = 2+√2
y(-2) = 4, y[-√(2+√2)] = 2+√2 - √(2+√2)√(2-√2) = 2,
y[-√(2-√2)] = 2-√2 - √(2-√2)√(2+√2) = 2-2√2
y[√(2-√2)] = 2-√2 + √(2-√2)√(2+√2) = 2
y[√(2+√2)] = 2+√2 + √(2+√2)√(2-√2) = 2+2√2,
y(2) = 4,
则得最大值 y[√(2+√2)] = 2+2√2， 最小值 y[-√(2-√2)] = 2-2√2
y' = 2x +√(4-x^2) - x^2/√(4-x^2)
= 2[x√(4-x^2)+2-x^2]/√(4-x^2)
令 y' = 0, 得 x√(4-x^2)+2-x^2 = 0
x√(4-x^2) = x^2 - 2， x^2(4-x^2) = x^4 - 4x^2 + 4 ，
x^4 - 4x^2 + 2 = 0 , x^2 = [4±√(16-8)]/2 = 2±√2
x^2 = 2-√2, 或 x^2 = 2+√2
y(-2) = 4, y[-√(2+√2)] = 2+√2 - √(2+√2)√(2-√2) = 2,
y[-√(2-√2)] = 2-√2 - √(2-√2)√(2+√2) = 2-2√2
y[√(2-√2)] = 2-√2 + √(2-√2)√(2+√2) = 2
y[√(2+√2)] = 2+√2 + √(2+√2)√(2-√2) = 2+2√2,
y(2) = 4,
则得最大值 y[√(2+√2)] = 2+2√2， 最小值 y[-√(2-√2)] = 2-2√2
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