求y=3x-x3的绝对值在[-2,2]上的最大值及最小值

来源：百度知道 编辑：UC知道 时间：2024/09/27 05:55:55
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y = |3x - x^3| = |x||3 - x^2|

-2 <= x < -3^(1/2), y = -x(x^2 - 3) = 3x - x^3, y' = 3 - 3x^2 = 3(1-x^2) < 0, 函数单调递减。y(-2) = -6 + 8 = 2 >= y >= y(-3^(1/2)) = -3*3^(1/2) + 3*3^(1/2) = 0.

-3^(1/2) <= x < 0, y = -x(3 - x^2) = x^3 - 3x, y' = 3x^2 - 3 = 3(x^2 - 1)

-3^(1/2) <= x < -1, y'>0, 函数单调递增。y(-3^(1/2)) = 0 <= y <= y(-1) = -1 + 3 = 2.

-1 <= x < 0, y'<0, 函数单调递减。y(-1) = 2 >= y >= y(0) = 0.

0 <= x < 3^(1/2), y = x(3-x^2) = 3x - x^3, y'=3-3x^2 = 3(1-x^2).

0 <= x < 1, y'>0,函数单调递增。y(0) = 0 <= y <= y(1) = 3 - 1 = 2.

1 <= x < 3^(1/2), y'<0,函数单调递减。y(1) = 2 >= y >= y(3^(1/2)) = 0.

3^(1/2) <= x <= 2, y = x(x^2 - 3)=x^3-3x, y'=3x^2-3=3(x^2-1)>0.函数单调递增。y(3^(1/2)) = 0 <= y <= y(2) = 8 - 6 = 2.

所以，
y=3x-x3的绝对值在[-2,2]上的最大值及最小值分别为2和0.