For any point (x,y) on the certain curve,
http://i30.tinypic.com/654l79.jpg
and y has a local minimum of 7 when x=1.
a) Find the equation of the curve
b) Find the maximun point of y
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2008-05-30 8:34 am
回答 (2)
2008-05-30 10:23 am
✔ 最佳答案
d^2y/dx^2=6x-4dy/dx=3x^2-4x+C
sub x=1,dy/dx=0
3-4+C=0
C=1
dy/dx=3x^2-4x+1=(x-1)(3x-1)
The equation of the curve is
x^3-2x^2+x+K where K is a constant
sub x=1, y=7
1-2+1+K=7
K=7
So x^3-2x^2+x+7
Let dy/dx=0,then x=1 or x=1/3
So the maximun point of y is (1/3)^3-2(1/3)^2+(1/3)+7
=1/27-1/18+1/3+7
2008-05-30 6:35 pm
d²y/dx²=6x-4
dy/dx=3x²-4x+C, where C is a constant
When x=1, y has a local minimum
∴3(1)²-4(1)+C=0
C=1
∴dy/dx=3x²-4x+1
∴y=x³-2x²+x+K, where K is a constant
When x=1, y=7
∴7=1-2+1+K
K=7
∴y=x³-2x²+x+7
dy/dx=0, 3x²-4x+1=0
(3x-1)(x-1)=0
x=1 or 1/3
d²y/dx²=6(1/3)-4=-2 < 0
∴x=1/3 is a max. pt.
∴Max. pt. of y=1/27-2/9+1/3+7=193/27
dy/dx=3x²-4x+C, where C is a constant
When x=1, y has a local minimum
∴3(1)²-4(1)+C=0
C=1
∴dy/dx=3x²-4x+1
∴y=x³-2x²+x+K, where K is a constant
When x=1, y=7
∴7=1-2+1+K
K=7
∴y=x³-2x²+x+7
dy/dx=0, 3x²-4x+1=0
(3x-1)(x-1)=0
x=1 or 1/3
d²y/dx²=6(1/3)-4=-2 < 0
∴x=1/3 is a max. pt.
∴Max. pt. of y=1/27-2/9+1/3+7=193/27
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