Finding the Critical Point(negative square root)

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SUMMARY

The discussion focuses on finding the critical points of the function g(x, y, z) = x³ + xy² + x² + y² + 3z². The user derived three equations from the partial derivatives set to zero: (1) 3x² + y² + 2x = 0, (2) 2xy + 2y = 0, and (3) 6z = 0. The analysis revealed that z must equal 0, and while x can be -1, substituting this value into equation (1) leads to an impossible condition for y², indicating that y cannot be isolated as a real number. The discussion raises the question of whether complex numbers are necessary for further analysis.

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Luscinia
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Homework Statement


Find the critical points for the function g(x, y, z) = x3+xy2+x2+y2+3z2.

Homework Equations

The Attempt at a Solution


I've come up with the following 3 equations (derivatives set so that they are equal to 0)
(1) 3x2+y2+2x=0
(2) 2xy+2y=0
(3) 6z=0

From (3),
z=0​
From (2),
2y(x)+2y(1)=0
x+1=0
x=-1​
From (1) using what I have obtained from (2),
3(-1)2+y2+2(-1)=0
3-2+y2=0
1=-y2
y2=-1​
I do not know how I am supposed to isolate y in this case since square roots are supposed to be positive. Do I need to get i involved? If so, how?
 
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From (2) you have two possibilities: 0 = 2xy + 2y = 2y(x + 1) so either y = 0 or x = -1 or both.

You must also satisfy (1). You've shown that if x = -1 then (1) requires that y^2 = -1, so that doesn't give you a critical point. There remains the y = 0 possibility.
 
Luscinia said:

Homework Statement


Find the critical points for the function g(x, y, z) = x3+xy2+x2+y2+3z2.

Homework Equations

The Attempt at a Solution


I've come up with the following 3 equations (derivatives set so that they are equal to 0)
(1) 3x2+y2+2x=0
(2) 2xy+2y=0
(3) 6z=0

From (3),
z=0​
From (2),
2y(x)+2y(1)=0
x+1=0
x=-1​
From (1) using what I have obtained from (2),
3(-1)2+y2+2(-1)=0
3-2+y2=0
1=-y2
y2=-1​
I do not know how I am supposed to isolate y in this case since square roots are supposed to be positive. Do I need to get i involved? If so, how?

##(2) \Longrightarrow 2y(x+1) = 0 \Longrightarrow y = 0 \; \text{or }\; x+1 = 0##.
 

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