What is the critical point of xe^x?

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Discussion Overview

The discussion revolves around finding the critical point of the function \( xe^x \). Participants explore the definition of critical points, the process of differentiation, and the implications of using calculators for such problems.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for the critical point of \( xe^x \) and mentions confusion with their calculator's output.
  • Another participant questions the definition of "critical point" and challenges the claim of an asymptote at \( x=1 \), prompting a clarification of the function's value at that point.
  • A mathematical derivation is presented, suggesting that the critical point occurs at \( x = -1 \), where the derivative \( \frac{\partial}{\partial x} xe^x = x e^x + e^x \) equals zero.
  • Some participants express confusion over the interpretation of values at specific points, particularly regarding the relationship between \( x=1 \) and the critical point.
  • There is a critique of reliance on calculators, suggesting that they complicate straightforward problems.
  • A participant acknowledges a misunderstanding in their previous statements about the critical point and reiterates that the critical point is indeed at \( x = -1 \).

Areas of Agreement / Disagreement

Participants do not reach consensus on the interpretation of the function's behavior at \( x=1 \) and the use of calculators. There are conflicting views on the clarity of the critical point definition and the implications of the derivative's value.

Contextual Notes

There are unresolved questions regarding the assumptions made about the function's behavior and the potential misinterpretation of calculator outputs. The discussion reflects varying levels of understanding of critical points and differentiation.

Mitchtwitchita
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can anybody tell me what the critical point of xe^x is? When I try putting it into my calculator, it just shows a line staring at zero with an asymptote at x=1.
 
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What is the definition of "critical point"?

And I can't help but wonder what you put into your calculator! y= xex does not have any asymptotes. What is the value of y= xex when x= 1?
 
[tex]\frac{\partial}{\partial x} xe^x = x*e^x + e^x[/tex]

so you need to find

[tex]0 = x*e^x + e^x[/tex]

so the critical point is

[tex]x = -1[/tex]

it is the place where the function is not locally a diffeomorphism, that is where the inverse function theorem don't apply, so for higher dimensions you need to calculate the jacobian.
 
Thanks for your help guys. I think I got it now. When x=1, the answer is -1. This is the critical point for the function. I don't what the deal was with my calculator. It could have been the scale. Anyhoo, thanks again.
 
not sure what you meen by:

When x=1, the answer is -1

it is not 'when x=1 or equal to anything then...'

you simply need to find where the derivative is equal to 0 or undefined, and for your function that is in x=-1.
 
another argument against calculators. a trivial problem made hard by using and believing a calculator.
 
Mitchtwitchita said:
Thanks for your help guys. I think I got it now. When x=1, the answer is -1. This is the critical point for the function. I don't what the deal was with my calculator. It could have been the scale. Anyhoo, thanks again.
Then you don't got it. If "when x= 1, the answer is -1" is in response to my question "What is the value of y= xex when x= 1?" (my point being that if it has a value, x= 1 cannot be an asyptote), then when x= 1, y= xex= 1(e1)= 1. I can't imagine how you would get a negative number for that. And you have already been told that the critical point is NOT at x= 1.
 
I apologize, I see that my answer has caused some clamor. I meant that when I differentiated and solved for x, the answer is -1...which was my initial query. Thanks again for the help guys!
 

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