計算下列各二次函數的極大值或極小值,並寫出所對應的x值
1. y=(2/3)x² -x+6
要用配方法計 (連步驟)
中四maths
2007-02-12 12:52 am
回答 (2)
2007-02-12 12:56 am
✔ 最佳答案
計算下列各二次函數的極大值或極小值,並寫出所對應的x值1. y=(2/3)x² -x+6
要用配方法計 (連步驟)
y
=(2/3)x² -x+6
=2/3[x^2-(3/2)x]+6
=2/3[(x-3/4)^2-9/16]+6
=2/3[(x-3/4)^2]-3/8+6
=2/3[(x-3/4)^2]+45/8
當x=3/4 時,y有極小值 45/8
2007-02-12 1:00 am
take derivative of the equation:
dy/dx = 2 (2/3) x - 1
= 4/3 x - 1
Max. or min appears at dy/dx = 0
i.e. 0 = 4/3 x - 1
gives : x = 3/4
Therefore, maximum y appears at x = 3/4.
dy/dx = 2 (2/3) x - 1
= 4/3 x - 1
Max. or min appears at dy/dx = 0
i.e. 0 = 4/3 x - 1
gives : x = 3/4
Therefore, maximum y appears at x = 3/4.
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