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 May 5th, 2015, 03:59 PM #1 Senior Member   Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty 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 04:06 PM. May 5th, 2015, 08:23 PM   #2
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Quote:
 Originally Posted by hyperbola 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. May 5th, 2015, 09:07 PM #3 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs 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 May 6th, 2015, 12:38 AM   #4
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Quote:
 Originally Posted by Country Boy $\displaystyle f_x= 3x^2+ 2xy$. 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. May 6th, 2015, 02:06 AM   #5
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Quote:
 Originally Posted by Country Boy $\displaystyle f_x= 3x^2+ 2xy$. 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 02:08 AM. Tags critical, derivatives, partial, points ### critical points of a hyperbola

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