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I need to find al the critical points of the following function

$f(x,y)=y^2-x^2y-3y+x^4-x^3$

Determine if they are local minima, local maxima, or saddle points, by looking at the Hessian matrices at the critical points.

$f_x$=$-2xy+4x^3-3x^2$

$f_y$=$2y-x^2-3$

To find the critical points I will solve both equations by setting them equal to $0$.

Equation $1$:

$-2xy+4x^3-3x^2$=0

$x(2y+4x^2-3x)=0$

Equation $2$:

$2y-x^2-3 =0$

$2y=x^2+3$

$y=\frac{x^2+3}{2}$ I'm going to plug this into my $f_x$ equation.

I will get $x(5x^2-3x+3)=0$

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  • $\begingroup$ Could you please explain how the critical points were found in the example in the link you attached? I think that's where I'm getting stuck. $\endgroup$ – user2553807 Jan 30 '14 at 18:46
  • $\begingroup$ I added a little more to my question, am I on the right track to find the critical points? $\endgroup$ – user2553807 Jan 30 '14 at 18:58
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You will find the points where $f_x=f_y=0$, then look at the Hessian for each point. This is pretty straight forward for this equation, but ensuring you have found all the points where this is true requires careful bookkeeping:

$$\begin{align}f_x=0&=-2xy+4x^3-3x^2\\ f_y=0&=2y-x^2-3\end{align}$$

To answer the question directly: No, you don't need a different method just because you will look at the Hessian after finding each stationary point.

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