Question

y=x+根号[x(2-x)]求值域,用逆求法

Answer 1

√[x(2-x)]:x(2-x) ≥ 0,x(x-2) ≤ 0

x ≥ 0 且x ≤ 2,即 0 ≤ x ≤ 2

或x ≥ 2 且x ≤ 0,无解

x(2-x) = -x^2 +2x -1 + 1 = -(x-1)^2 +1,x=1时,x(2-x)及√[x(2-x)]取最大值1,此时x+√[x(2-x)]取最大值2.

显然x+√[x(2-x)]在[0,2]中,最小值为0.

值域[0,2]