Critical Point of Multivariable Function

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SUMMARY

The critical point of the function e^-(x^(2)+y^(2)+2x) is determined by setting the partial derivatives Fx and Fy to zero. The derivatives are given by Fx(x,y) = (-2x-2)e^(-x^(2)-y^(2}-2x) and Fy = -2ye^(-x^(2)-y^(2}-2x). The critical point occurs when Fx(x,y) = 0, which simplifies to solving for x in the equation 2x - 2 = 0, leading to the critical point at x = 1. The logarithmic approach suggested in the discussion is incorrect as the logarithm of zero is undefined.

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  • Knowledge of exponential functions
  • Ability to solve equations simultaneously
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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
 
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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.
 
ah thank you.
 

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