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May 5th, 2015, 02:59 PM   #1
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Critical Points, Partial Derivatives

I'm partially deriving the equation $\displaystyle f(x,y)=x^3+x^2y+y^2-4y+2$ and want to find critical points. I'm at somewhat of a dilemma for my solutions of $x$

$\displaystyle f_{x}=0$, $\displaystyle x$ solutions are:

$\displaystyle x_1=0 $
$\displaystyle x_2=4 $
$\displaystyle x_3=-1$

$\displaystyle f_{y}=0$ $\displaystyle , x$ solutions are:

$\displaystyle x_1=4$
$\displaystyle x_2=-1$

I'm unsure as whether or not $\displaystyle x=0$ is ignored in this instance.

When set in terms of $\displaystyle y$,

$\displaystyle f_x=0$ when $\displaystyle x=0$

$\displaystyle f_y=2$ when $\displaystyle x=0$

My first order derivatives are correct, I have checked via second order derivatives and $\displaystyle f_{xy}=f_{yx}=2x$

Last edited by hyperbola; May 5th, 2015 at 03:06 PM.
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May 5th, 2015, 07:23 PM   #2
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Quote:
Originally Posted by hyperbola View Post
I'm partially deriving the equation $\displaystyle f(x,y)=x^3+x^2y+y^2-4y+2$ and want to find critical points. I'm at somewhat of a dilemma for my solutions of $x$

$\displaystyle f_{x}=0$, $\displaystyle x$ solutions are:

$\displaystyle x_1=0 $
$\displaystyle x_2=4 $
$\displaystyle x_3=-1$
$\displaystyle f_x= 3x^2+ 2xy$.
f_x(0, y)= 0
f_x(4, y)= 48+ 8y. Whether that is 0 or not depends on y.
f_x(-1, y)= 3- 2y. Whether that is 0 or not depends on y.

Quote:
$\displaystyle f_{y}=0$ $\displaystyle , x$ solutions are:

$\displaystyle x_1=4$
$\displaystyle x_2=-1$
$\displaystyle f_y= x^2- 2y+ 4$
$\displaystyle f_y(4, y)= 20- 2y$
whether that is 0 or not depends on y
$\displaystyle f_y(-1, y)= 5- 2y$
whether that is 0 or not depends on y.

Quote:
I'm unsure as whether or not $\displaystyle x=0$ is ignored in this instance.

When set in terms of $\displaystyle y$,

$\displaystyle f_x=0$ when $\displaystyle x=0$

$\displaystyle f_y=2$ when $\displaystyle x=0$

My first order derivatives are correct, I have checked via second order derivatives and $\displaystyle f_{xy}=f_{yx}=2x$
Frankly, you don't seem to have any idea what you are doing! You are apparently finding values of x without giving any thought to y.

Yes, $\displaystyle f_x= 3x^2+ 2xy$ and $\displaystyle f_y= x^2- 2y+ 4$
set those equal to 0, $\displaystyle 3x^2+ 2xy= 0$ and $\displaystyle x^2- 2y+ 4= 0$ and find paired values of x and y that satisfy both equations.

From the second equation, $\displaystyle y= \frac{x^2+ 4}{2}$. Putting that into the first equation gives a single 3 degree equation for x. Then find the corresponding value of x.
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May 5th, 2015, 08:07 PM   #3
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I get:

$\displaystyle f_x(x,y)=x(3x+2y)\tag{1}$

$\displaystyle f_y(x,y)=x^2+2y-4\tag{2}$

Equating both to zero, from (2), we find:

$\displaystyle 2y=4-x^2$

and then from (1), we obtain:

$\displaystyle x\left(x^2-3x-4\right)=0$

Factor, and then you will have 3 values for $x$, and then use:

$\displaystyle y=2-\frac{1}{2}x^2$

to get the corresponding $y$-values and you now have 3 critical points.
Thanks from hyperbola
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May 5th, 2015, 11:38 PM   #4
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Quote:
Originally Posted by Country Boy View Post
$\displaystyle f_x= 3x^2+ 2xy$.

Frankly, you don't seem to have any idea what you are doing!
10/10 for positivity...

Thanks MarkFL, your replies are to the point with no extra rubbish on the side.
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May 6th, 2015, 01:06 AM   #5
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Quote:
Originally Posted by Country Boy View Post
$\displaystyle f_x= 3x^2+ 2xy$.
Putting that into the first equation gives a single 3 degree equation for x. Then find the corresponding value of x.
Naturally, I would've thought that both equations would give me an equal number of x values. Instead, I ended up with a 2nd degree (2 $x$ values) and a 3rd degree (3 $x$ values) polynomial. I've never before encountered this when solving simultaneous equations, hence the uncertainty.

Last edited by hyperbola; May 6th, 2015 at 01:08 AM.
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