The equation of a curve is y=x^3-4x^2-16x+15.
(a) Find the points of intersection of the curve and the x-axis.
(b) Find the points on the curve such that the values of y are maximum or minimum.
(c)Sketch the curve for values of x from -3 to 6.
F.5 A.maths--Graphic differentiation
2006-12-19 8:59 am
回答 (1)
2006-12-19 1:20 pm
✔ 最佳答案
(a)let y=0
x^3-4x^2-16x+15=0
(x+3)(x^2-7x+5)=0
x=-3 or x=1/2[7-√29] or 1/2[7+√29]
points of intersection (-3,0) (1/2[7-√29],0) (1/2[7+√29],0)
(b)
dy/dx=3x^2-4x-16
let dy/dx=0
3x^2-8x-16=0
(3x+4)(x-4)=0
x=4 or x=-4/3
d^2y/dx^2=6x-4
when x=4, d^2y/dx^2>0
when x=-4/3, d^2y/dx^2<0
so (4,-49) is a minimum point (-4/3, 725/27) is a maximum point
(c)
see http://hk.geocities.com/myisland8132/doc2.pub
收錄日期: 2021-04-12 19:13:13
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