Critical Point of Multivariable Function

In summary, to find the critical point of the function e^-(x^(2)+y^(2)+2x), differentiate the function and set it to zero for both fx and fy. This gives the equations ln(-2x-2)-x^(2)-y^(2)-2x = 0 and ln(-2y)-x^(2)-y^(2)-2x = 0. However, since e^a is never equal to 0, the only way Fx(x,y) could be 0 is if (2x-2) is 0. Therefore, the critical point is when x=1 and y=0.
  • #1
danny_manny
42
0

Homework Statement



Locate the critical point of the function,

e^-(x^(2)+y^(2)+2x)

Homework Equations



none

The Attempt at a Solution


Ok first step differentiate the function and set it to zero for both fx and fy,

Fx(x,y) = (-2x-2)e^(-x^(2)-y^(2)-2x)
Fy = -2ye^(-x^(2)-y^(2)-2x)

Now I need to solve both equations simultaneously,

I get

ln(-2x-2)-x^(2)-y^(2)-2x = 0

and

ln(-2y)-x^(2)-y^(2)-2x = 0

and this is where I am stuck. :( any advice would be welcomed :D

thanks,
Dan
 
Physics news on Phys.org
  • #2
You need to solve Fx(x,y)=0. You don't want to take the log of that. log(0) is undefined, it's NOT 0. The way to do this is to notice e^a is NEVER 0. So e^(-x^(2)-y^(2)-2x) is NEVER 0. The only way Fx(x,y) could be 0 is if (2x-2) is 0.
 
  • #3
ah thank you.
 

Related to Critical Point of Multivariable Function

What is a critical point in multivariable calculus?

A critical point in multivariable calculus is a point on a function's domain where the gradient (or derivative) of the function is equal to zero. This means that the function is neither increasing nor decreasing at that point, and can indicate local extrema or saddle points.

How do you find critical points in multivariable calculus?

To find critical points in multivariable calculus, you must calculate the partial derivatives of the function with respect to each variable and set them equal to zero. Then, solve the resulting system of equations to find the values of the variables at the critical points.

What is the significance of critical points in multivariable calculus?

Critical points are important in multivariable calculus because they can indicate where a function has local extrema (maximum or minimum values) or saddle points. They can also be used to find the global maximum or minimum of a function over a given domain.

What is the difference between a local and global extremum?

A local extremum is a point where a function has the highest or lowest value within a small neighborhood, but it may not be the highest or lowest value overall. A global extremum is the absolute highest or lowest value of a function over its entire domain.

How do you determine the nature of a critical point in multivariable calculus?

The nature of a critical point can be determined by examining the second derivative of the function at that point. If the second derivative is positive, the point is a local minimum. If it is negative, the point is a local maximum. If the second derivative is zero, further analysis is needed to determine the nature of the point.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
285
  • Calculus and Beyond Homework Help
Replies
1
Views
521
  • Calculus and Beyond Homework Help
Replies
2
Views
616
  • Calculus and Beyond Homework Help
Replies
6
Views
902
  • Calculus and Beyond Homework Help
Replies
6
Views
927
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Replies
5
Views
1K
Back
Top