Can anyone help me?
The equation is x^3+y^3=3xy
It is said that find the stationary values and determine whether they are maxima or minima.
I differentiate it fist and get dy/dx=(y-x^2)/(y^2-x)
and then get dy/dx=0 and y-x^2=0. So y=x^2
I have already find out the stationary points (0,0) and (cube root of 2, cube root of 4) .
(cube root of 2, cube root of 4 )is a maximum point if I substitute is into the second derivative.
However, At point (0,0) the second derivative of the function =0 then how do I know whether it is a maxima or minima?
Thanks a lot!!
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pepperzealot
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- 29-01-2015 10:22
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- 29-01-2015 10:28
(Original post by pepperzealot)
Can anyone help me?
The equation is x^3+y^3=3xy
It is said that find the stationary values and determine whether they are maxima or minima.
I differentiate it fist and get dy/dx=(y-x^2)/(y^2-x)
and then get dy/dx=0 and y-x^2=0. So y=x^2
I have already find out the stationary points (0,0) and (cube root of 2, cube root of 4) .
(cube root of 2, cube root of 4 )is a maximum point if I substitute is into the second derivative.
However, At point (0,0) the second derivative of the function =0 then how do I know whether it is a maxima or minima?
Thanks a lot!!
f'(-0.1) and f'(0.1) -
davros
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- 29-01-2015 13:40
(Original post by pepperzealot)
Can anyone help me?
The equation is x^3+y^3=3xy
It is said that find the stationary values and determine whether they are maxima or minima.
I differentiate it fist and get dy/dx=(y-x^2)/(y^2-x)
and then get dy/dx=0 and y-x^2=0. So y=x^2
I have already find out the stationary points (0,0) and (cube root of 2, cube root of 4) .
(cube root of 2, cube root of 4 )is a maximum point if I substitute is into the second derivative.
However, At point (0,0) the second derivative of the function =0 then how do I know whether it is a maxima or minima?
Thanks a lot!!
I'm not saying you're wrong, but it looks from what you've calculated that dy/dx is undefined at (0,0) so it may need special consideration! -
pepperzealot
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- 29-01-2015 13:42
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pepperzealot
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- 29-01-2015 13:46
(Original post by davros)
Are you sure that (0,0) is a stationary point?
I'm not saying you're wrong, but it looks from what you've calculated that dy/dx is undefined at (0,0) so it may need special consideration!
Yeah I think it is a correct value.... Actually there are two parts of this question. The first part is just find the x coordinates of the stationary points. I have just checked with the answer at the back of the book. It is said that their x coordinates are 0 and cube root2.. But the second part of the question there is no explanationT T -
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- 29-01-2015 13:49
(Original post by pepperzealot)
I have tried. But the value of y is also needed to find dy/dx. If I substitute x=0.1to the original equation to get the value of y, there will be three values of y for x=0.1..... It is really complicated!
this curve is the folium of Descartes and there is a node at the origin which complicates things.
Are you an A level student or undergrad before I continue? -
the bear
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- 29-01-2015 13:53
both the first and second derivatives are undefined at (0,0)
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pepperzealot
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- 29-01-2015 13:58
(Original post by TeeEm)
let me tell you about this question because I did not pay much attention the first time
this curve is the folium of Descartes and there is a node at the origin which complicates things.
Are you an A level student or undergrad before I continue?I am an AS student...
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- 29-01-2015 14:02
I remember a recent question on MEI C3 paper where this came up, they drew the graph to show you you were looking for a local max, so you could ignore the origin
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