關於Differentiation的簡單A.Maths.問題

2009-08-24 11:33 pm
For each of the following functions:
(a) determine the range of values of x that the function is
(i) increasing,
(ii) decreasing;
(b) find the maximum and minimum points.

1. y = x^6 – 3x^2

dy / dx = 6x^5 – 6x
= 6x(x^2 + 1)(x + 1)(x – 1)

when dy / dx = 0,
6x(x^2 + 1)(x + 1)(x – 1) = 0
x = 0, 1 or -1





X


X < -1


X = -1


-1 <x<0


X = 0


0 < x <1


X =1


X > 1



Dy / dx


-


0


+


0


-


0


+


(a) (i) increasing: [-1, 0], [1, ∞]←此方括是為圓括。
(ii) decreasing: (-∞, -1) ←此圓括是為方括。, [0, 1]

(b) maximum point: (0, 0)
minimum point: (1, -2), (-1, -2)

問題一:「6x(x^2 + 1)(x + 1)(x – 1) = 0」如何得出x = 0, 1 or -1?請列明計算步驟。
問題二:minimum point為何會有兩個,minimum不是最少的那個嗎?

回答 (1)

2009-08-25 7:00 am
✔ 最佳答案
問題一:「6x(x2 + 1)(x + 1)(x – 1) = 0」如何得出x = 0, 1 or -1?請列明計算步驟。6x(x2 + 1)(x + 1)(x – 1)是四組數相乘,他們是6x, (x2 + 1), (x + 1)及(x – 1)
他們的積是零,則可能性是他們其中一個是零,但x2 + 1不可能是零,因x2不會是負數。那麼只餘下三個可能性:
6x = 0 => x = 0或
x + 1 = 0 => x = -1或
x – 1 = 0 => x = 1
問題二:minimum point為何會有兩個,minimum不是最少的那個嗎?
在微分的概念裏,minimum可以是絶對(absolute)的或是local的。就像地勢一樣,可以有很多山峰(local maximum),也有許多山谷(local minimum),Mount Everest就是所有maximum points中的absolute maximum。
圖為這道題的Graph可見一山峰及二山谷,其中二山谷的y值剛好相同。但留意那山峰不是絶對的高峰。
http://img89.imageshack.us/img89/6397/curve.jpg

圖片參考:http://img89.imageshack.us/img89/6397/curve.jpg


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