Intermideate value therem question

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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.
 
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