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$3x^3 +9x^2 -72x+2y^3-12y^2-126y+19$

I want to find critical points and their nature of this equation. When I equated $f_x$ and $f_y$ to 0, I m getting $x=2,x=-4$ for $f_x=0$ and $y=-3,y=7$ for $f_y=0$.

now, when I substitute x=2 in f(x,y) I m getting $2y^3-12y^2-126y=65$ which is a cubic equation. I am stuck here and don't know how to solve this...can somebody help?

https://www.geekandnerd.org/criticalsaddle-point-calculator-for-fxy/ - it shows four critical points by combining above ones.can we always do this?whats the logic behind doing the same?

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https://www.geekandnerd.org/criticalsaddle-point-calculator-for-fxy/ - it shows four critical points by combining above ones.can we always do this?whats the logic behind doing the same?

The logic is that critical points are the points satisfying $f_x = 0$ and $f_y=0$, it is a system of two equations. Your calculations show that $x=2$ and $x=-4$ satisfy the first equation and $y=-3$ and $y=7$ satisfy the second, so there are four points which satisfy both equations (simulnateously): $$\left(2,-3\right) \;,\; \left(2,7\right)\;,\; \left(-4,-3\right)\;,\; \left(-4,7\right)$$ Now you still need to check the nature of these critical points, e.g. with the second derivative test. Can you take it from there?


Addition after comment. The system you need to solve is: $$\left\{\begin{array}{l} f_x = 0 \\ f_y = 0 \end{array}\right. \iff \left\{\begin{array}{l} 9 (x - 2) (x + 4) = 0 \\ 6 (y - 7) (y + 3) = 0 \end{array}\right.$$ A solution to this system of two equations is not an $x$-value (or a $y$-value) but a pair $(x,y)$ satisfying both equations. Does that help?

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  • $\begingroup$ But wait – how can the cubic equation $2y^3-12y^2-126y=65$ have just 2 solutions, and how can $y=-3$ and $y=7$ be solutions? $\endgroup$ – Gerry Myerson May 4 '17 at 7:31
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    $\begingroup$ Gerry, that equation we get after substituing x=2 in f(x,y). y=-3 and y=7 are the solution we get when we equate partial derivetive of y to 0 $\endgroup$ – Kashmira Zambad May 4 '17 at 7:33
  • $\begingroup$ @GerryMyerson You never arrive at the cubic, the solutions follow from the system $\left\{f_x=0,f_y=0\right\}$. $\endgroup$ – StackTD May 4 '17 at 7:34
  • $\begingroup$ @GerryMyerson I added a bit too my answer. $\endgroup$ – StackTD May 4 '17 at 7:37
  • $\begingroup$ Yes! Thanks, I can take it forward! $\endgroup$ – Kashmira Zambad May 4 '17 at 7:49

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