Intermideate value therem question

[nhrock3]

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?

?

 

[LCKurtz]

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

 

[nhrock3]

close to zero from minus
its minus infinity

when x goes to minus infinity its 0

how it helps me?

 

[nhrock3]

close to sero is 0

 

[LCKurtz]

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

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?

 

[nhrock3]

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?

 

[estro]

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.