Question

Trying to take a function and identify all of its critical points, prompt posted below momentarily.

Answer 1

\[\text{Find and identify all critical points of the function.}\]

Answer 2

\[y^3-3yx^2-3y^2-3x^2\]

Answer 3

Alright, if I remember correctly, first thing I should do in this is take the gradient of the given function.

Answer 4

\[\triangledown f=(-6yx-6x)i+(3y^2-3x^2-6y)j\]

Answer 5

And now I need to do something like set it equal to zero, something or other.

Answer 6

Am I on the right track? Or how should I proceed?

Answer 7

Yes, set the components of the gradient to zero and solve for points \((x,y)\) that satisfy.

Answer 8

You're spot on. Set the partial derivatives equal to x and solve simultaneously for your critical points.

Answer 9

Alright, moving forward momentarily. \[-6yx-6x=0;\]\[3y^2-3x^2-6y=0.\]

Answer 10

\[6yx=6x; \ \ \ 6y=6; \ \ \ y=1. \]

Answer 11

Be careful, you've divided by x which can be 0 so you also need to consider the case x=0.

Answer 12

(This is the part that always gets me, and the part that I always mess up on, so help is appreciated particularly on solving for what *could* be potential critical points, same thing with using Lagrange multipliers.)

Answer 13

Alright, so are you saying that another possibility is that x could be zero? So like, if I divided across by a given variable (that was entirely removed from the equation as a result), I could say that whatever it was, x^2, y^3, x, could be zero? Or how do I think about that?)

Answer 14

e.g., every time I divide across and wholly eliminate a variable, do I need to consider that variable at zero to be a possible solution, or what???

Answer 15

Yes, because if that variable is 0 then you're dividing by 0 which, as you know, is a big no no. Another way to look at it is by factoring:\[6yx=6x\]\[6yx-6x=0\]\[x(y-1)=0\]Hench x=0 or y=1. Do you see where this magic 0 comes from now?

Answer 16

*Hence, lol

Answer 17

(One second, getting water, be right back)

Answer 18

Back.

Answer 19

Yeah, I see it that way and also thought of it that way; so we are not allowed to (at least for the sake of being proper) divide by what could be zero, we just have to represent what will give us the same potential points in a different way.

Answer 20

Now I just need to find the other critical points.

Answer 21

The second equation is something I'm having trouble dealing with; do you have any suggestions as how to algebraically deal with it?

Answer 22

You can divide by what can potentially be zero, you just have to account for the possibility like we have here, by knowing that x=0 is also a solution. You simply lose a root when you divide by 0 in this case.

Answer 23

Bear with me, I'm working on the second one.

Answer 24

No wonder I can't get this right, I was using wrong data to start off with - and so were you.\[-6yx-6x=0\]\[6x=-6yx\]\[x(1+y)=0\]
x=0, y=-1 (not 1!) :) Now sub these values into the other equation. Let's start with x:

\[y^2-0^2-2y=0\]\[y^2-2y=0\]\[(y-1)^2=1\]\[y=1\pm1\]
So far we have (0,0), (0,2). Now for y:
\[(-1)^2-x^2-2(-1)=0\]\[1-x^2+2=0\]\[x=\pm\sqrt{3}\]

The critical points:
\[(0,0), (0,2),(\sqrt3,-1),(-\sqrt3,-1)\]

Answer 25

Alright, I just saw this, taking a look at it now.

Answer 26

Awesome. Thank you very much. I'm going to take a few more questions like this up.

Answer 27

Sorry for the late reply, had a really busy week at uni and I moved back home over the weekend. No problem! Feel free to give me a shout if you get stuck with any others.