# Thread: critical points of a multivariable function

1. ## critical points of a multivariable function

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.

2. ## Re: critical points of a multivariable function

Originally Posted by Dobby
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.
It doesn't look like it.

3. ## Re: critical points of a multivariable function

$\displaystyle f_x=3x^2-6x-9$
$\displaystyle f_y=-6x+6y$

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

4. ## Re: critical points of a multivariable function

Originally Posted by Dobby
$\displaystyle f_x=3x^2-6x-9$
$\displaystyle f_y=-6x+6y$

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
Go on.
What are the y values?

5. ## Re: critical points of a multivariable function

with $\displaystyle x=3, f_y=-6(3)+6(3)=0?$

same with x=-1

or should I be plugging it into x=y to get (3,3) and (-1,-1)

6. ## Re: critical points of a multivariable function

Originally Posted by Dobby
with $\displaystyle x=3, f_y=-6(3)+6(3)=0?$

same with x=-1

or should I be plugging it into x=y to get (3,3) and (-1,-1)
Yep!
That looks much better!

Note that you just filled in y=3 to get to $\displaystyle f_y=0$.

7. ## Re: critical points of a multivariable function

I don't get it can you please give me a little more help I've exerted my efforts maximally

Edit: I see thanks for your help ! very silly of me

8. ## Re: critical points of a multivariable function

Originally Posted by Dobby
I don't get it can you please give me a little more help I've exerted my efforts maximally
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.

9. ## Re: critical points of a multivariable function

Absolutely you are correct! It is very clear to me now! Thank your very much !

10. ## Re: critical points of a multivariable function

You are aware of the thanks button?

11. ## Re: critical points of a multivariable function

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

12. ## Re: critical points of a multivariable function

Originally Posted by Dobby
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.