# Intermideate value therem question

1. Sep 11, 2011

### 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?

?

2. Sep 11, 2011

### LCKurtz

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

3. Sep 11, 2011

### nhrock3

close to zero from minus
its minus infinity

when x goes to minus infinity its 0

how it helps me?

4. Sep 11, 2011

### nhrock3

close to sero is 0

5. Sep 11, 2011

### LCKurtz

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?

6. Sep 12, 2011

### 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?

7. Sep 12, 2011

### 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.

Last edited: Sep 12, 2011