Intermideate value therem question

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Homework Help Overview

The discussion revolves around proving the existence and uniqueness of a solution to the equation \( xe^{\frac{1}{x}} = c \) for \( c < 0 \). Participants are exploring the behavior of the function and its limits as \( x \) approaches different values.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants suggest analyzing the function's behavior near zero and as \( x \) approaches negative infinity. There are questions about how to select appropriate limits and values to demonstrate the existence of a solution. Some participants discuss the implications of the function approaching zero and negative infinity.

Discussion Status

The discussion is ongoing, with participants sharing insights and hints about the function's behavior in different intervals. Some guidance has been provided regarding the limits and the implications for the existence of solutions, but no consensus has been reached yet.

Contextual Notes

Participants are working under the constraints of proving uniqueness and existence for \( c < 0 \) and are questioning the assumptions and definitions related to the function's behavior in the specified intervals.

nhrock3
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prove that for c<0

there is only one solution to

xe^{\frac{1}{x}}=c



??



for x=1 we have f(1)>0



the limit as x->-infinity is -infinity



what to do?

?
 
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Begin by drawing a graph to get an idea. In particular, what if x is less than 0 but close to 0?
 
close to zero from minus
its minus infinity

when x goes to minus infinity its 0

how it helps me?
 
close to sero is 0
 
LCKurtz said:
Begin by drawing a graph to get an idea. In particular, what if x is less than 0 but close to 0?

nhrock3 said:
close to zero from minus
its minus infinity

when x goes to minus infinity its 0

how it helps me?

nhrock3 said:
close to sero is 0

If c < 0 and you know your function approaches 0 as x → 0- and approaches -∞ a x → -∞, can you conclude your function = c for some x?
 
i need to show that there is x1 f(x1)<c
f(x2)>c


from the limit when x goes to sero we get zero -e<f(x)<e
from the limit when x goes minus infinity f(x)<-N

what e to chhose?
what N to choose?
 
LCKurtz provided great hint, I'll try to help as well.

To prove that the solution is single:
If c<0 what can you conclude about the existence of the solution (to your function) in [0,\infty).
In addition what can you tell about yours function behavior in (-\infty,0), how this helps you?

To prove solution existence:
See LKurtzs hint.
 
Last edited:

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