問數,關於函數取值問題~

y = x^2(1-3x)
y = x^2 - 3x^3
y' = 2x - 9x^2
y' = x(2 - 9x)
For y = 0, x = 0, 0, 1/3
For y' = 0, x = 0, 2/9
for 0 < x < 1/3
0 < x^2 < 1/9
0 > -3x > -1
1 > 1-3x > 0
0 < 1-3x < 1
so 0 < x^2(1-3x) < 1/9
thus for 0 < x < 1/3 the max value for y is at x=2/9, and the min value for y is at x=0 and x=1/3
The max value for y is (2/9)^2(1-3*2/9) = 4/243
The min value for y is (0)^2(1-3*0) or (1/3)^2(1-3*1/3)
= 0 or 0
The min value for y is 0
I hope this can help your understanding. :)

如果可以解釋一下constraint optimization要考慮boundary就更容易理解了。