application of differentiation

2008-10-02 3:06 am
Consider the curve y = x^3 - 9x^2 + 15x + 9
a. Find the range of x such that dy/dx < 0
b. Sketch the shape of the curve : y^2 = x^3 - 9x^2 + 15x + 9
更新1:

yes, thank eelyw very much. the question ask us to sketch the shape of the curve : y^2 = x^3 - 9x^2 + 15x + 9 not y = x^3 - 9x^2 + 15x + 9

更新2:

thank eelyw again. will the curve cut x-axis at x=3-2 sqrt3 and x=3+2 sqrt3?

更新3:

I mean the curve of y^2 = x^3 - 9x^2 + 15x + 9, I think it cuts the x-axis at x=3-2 sqrt3 and x=3+2 sqrt3, but the solution is that: http://good-times.webshots.com/photo/2328090110103642988KGWjNg

更新4:

I don't understand. Would you show me the shape of the curve required, please?

回答 (1)

2008-10-02 3:42 am
✔ 最佳答案
a)y= x^3 - 9x^2 + 15x + 9
dy/dx = 3x^2 - 18x + 15 .
So 3x^2 - 18x + 15 < = 0
x^2 - 6x + 5 < = 0
(x -1)(x -5) < = 0
So the range is 1 < = x < = 5.
b)
d^2y/dx^2 = 6x - 18.
For x = 1, d^2y/dx^2 is negative, so max. is at x = 1, y = 1 - 9 + 15 + 9 = 16.
For x = 5, d^2y/dx^2 is positive, so min. is at x = 5, y = 125 - 225 + 75 + 9 = -16.
When x = 3, y = 0, therefore, the curve cuts the x -axis at (3,0).
y = (x -3)(x^2 -6x - 3).
So the curve also cuts the x-axis at another 2 points, namely, [6 +/- sqrt(36 + 12)]/2 or (3 + 2sqrt3) and (3 - 2sqrt3).
When x tends to positive infinity, y tends to positive infinity.
When x tends to negative infinity, y also tends to negative infinity.
With the above data, you can sketch the curve.

2008-10-02 11:29:17 補充:
Sorry, I overlooked the y^2 term. Let y^2 = k. When x < (3-2 sqrt3), k is -ve, so y is unreal, so no curve. When (3 - 2sqrt3)< x < 3, k is +ve, so y = +/- sqrt k with max. point at (1, 4) and min. point at (1, -4).When 3 < x < 5, k is again -ve, so no curve. When 5< x, k is +ve, so y = +/- sqrt k.

2008-10-02 15:15:18 補充:
1. Yes, the curve will cuts the x-axis at (3-2sqrt3), 3 and (3+2sqrt3) because no matter it is y or y^2 or even y^3, the polynomial x^3 - 9x^ + 15x + 9 equals to zero at x = these 3 values.

2008-10-02 15:18:01 補充:
2. Lets take x = 10, so y^2 = 1000 - 900 + 150 + 9 = 259. That is y = +/- sqrt(259), but you don't see anything beyond x = 3 in the curve at good-times. So the curve should not be correct.

2008-10-02 15:43:53 補充:
3. By some calculation, you can find that the polynomial x^3 - 9x^2 + 15x + 9 can be re-written as (x-3)^3 - 12(x-3) = y^2. By the principle of translation, this is in fact the curve y^2 = x^3 - 12x but shifted to the right by 3 units.

2008-10-02 15:46:41 補充:
Fig.9 at staff.jccc.net/swilson/planecurves/cubics.htm shows the full shape of y^2 = x^3 - 3x. The curve at good-times is a section of the curve of the full curve.

2008-10-02 15:55:30 補充:
Correction: The curve at good-times is a section of the full curve with 1< x < 3.


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