Local Minimum of y=x(e^1/x): Why Not (1,e)?

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Discussion Overview

The discussion revolves around the local minimum of the function y=x(e^1/x) at the point (1,e) and questions regarding the existence of an absolute minimum for this function. Participants explore the implications of the function's behavior across its domain.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant claims that the local minimum of y=x(e^1/x) is at (1,e) but questions why this point is not the absolute minimum.
  • Another participant asks for clarification on the domain of the function, suggesting that the domain excludes zero.
  • A further contribution discusses the behavior of the function as x approaches negative infinity, noting that e^(1/x) approaches 1, which may influence the function's minimum characteristics.

Areas of Agreement / Disagreement

Participants have not reached a consensus on whether (1,e) is the absolute minimum or why the function lacks an absolute minimum. Multiple viewpoints and questions remain unresolved.

Contextual Notes

The discussion includes assumptions about the function's domain and behavior at extreme values, which are not fully explored or resolved.

nickeisenberg
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I found that the local minimum of y=x(e^1/x) is (1,e) but can someone tell me why the absolute minimum is also not (1,e) and can someone tell me why this function does not have an absolute minimum? Thanks
 
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nickeisenberg said:
I found that the local minimum of y=x(e^1/x) is (1,e) but can someone tell me why the absolute minimum is also not (1,e) and can someone tell me why this function does not have an absolute minimum? Thanks

Hey nickeisenberg and welcome to the forums.

Just to clarify, what is the domain of your function?
 
all reals number except 0
 
y= x (e^1/x)

Think about the two components.
If x -> -∞
x -> -∞
e^(1/x) -> 1 (1/-∞) ~ 0, e^0 = 1
 

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