Find all the critical points of $\displaystyle f(x,y)= x^3-6xy+3y^2-9x+7$
I get (3,0) and (-1,0) Did I do the question correctly? Please and thanks.
$\displaystyle f_x=3x^2-6x-9$
$\displaystyle f_y=-6x+6y$
now solving $\displaystyle f_y=0, x=y =>$ substitute that into $\displaystyle f_x $and solve and get $\displaystyle x=3$ and $\displaystyle x=-1$ then plug it back into $\displaystyle f_y$ to get the y values and thus the two C.P.'s
From $\displaystyle f_y=0$, you deduced that y=x.
Afterward you substituted that in $\displaystyle f_x=0$ to find x.
To find the corresponding y you have to use the equation you found for y.
From the fact that $\displaystyle f_y=0$, it is not right to conclude that y must be 0.
No, I was not but I just discovered it and will very well press. Also, please if I take this a step further,
Am I correct in my calculations that (3,3) is a local minimum and (-1,-1) is neither seeing as for (-1,-1) => (f_xx)(f_yy)-[f_xy]^2=0
Aha! You are taking it to the next level!
Not quite.
Yes, (3,3) is a local minimum, but as yet (-1,-1) is inconclusive.
See for instance: Second partial derivative test - Wikipedia, the free encyclopedia
This means that (-1,-1) could still be either a minimum, a maximum, or a saddle point.