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May 5th, 2015, 02:59 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 $\displaystyle f(x,y)=x^3+x^2y+y^24y+2$ 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. 
May 5th, 2015, 07:23 PM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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:
$\displaystyle f_y(4, y)= 20 2y$ 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:
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, 08:07 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
I get: $\displaystyle f_x(x,y)=x(3x+2y)\tag{1}$ $\displaystyle f_y(x,y)=x^2+2y4\tag{2}$ Equating both to zero, from (2), we find: $\displaystyle 2y=4x^2$ and then from (1), we obtain: $\displaystyle x\left(x^23x4\right)=0$ 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. 
May 5th, 2015, 11:38 PM  #4 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  
May 6th, 2015, 01:06 AM  #5 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  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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