# Thread: indermideate value therem question

1. ## intermidiate value therem question

prove that for c<0
there is only one solution to
$xe^{\frac{1}{x}}=c$

??
i tookk the left side as f(x)
for x=1 we have f(1)>0

the limit as x->-infinity is -infinity

what to do?
?

2. ## Re: intermidiate value therem question

Originally Posted by transgalactic
prove that for c<0
there is only one solution to
$xe^{\frac{1}{x}}=c$
If $f(x)=xe^{x^{-1}}$ can you show that if $x<0$
then $f '(x)>0$.
If so then $f$ is increasing and continuous on $(-\infty,0).$

3. ## Re: intermidiate value therem question

ok even if its encreasing
i need t show function vules bellow and above c
to use the intermidiete theorem

could you guide me regardnig how do i get thos values

because saying that its increasing is not sufficient

4. ## Re: intermidiate value therem question

Originally Posted by transgalactic
ok even if its encreasing
i need t show function vules bellow and above c
to use the intermidiete theorem
could you guide me regardnig how do i get thos values
$\lim _{x \to - \infty } f(x) = - \infty$

5. ## Re: intermidiate value therem question

ok so from the limit definition
for x<-M f(x)<-N

i can choose N to be C

but its not continues in zero
i cant take x=1 as the other point of intermidiete value theorem

6. ## Re: intermidiate value therem question

Originally Posted by transgalactic
ok so from the limit definition
for x<-M f(x)<-N
but its not continues in zero
Zero has absolutely nothing to do with this question.
Look at the OP.
Originally Posted by transgalactic
prove that for c<0
there is only one solution to

$xe^{\frac{1}{x}}=c$
?
$\lim _{x \to 0^ - } f(x) = 0$

$c<0$

7. ## Re: intermidiate value therem question

yes so by limit definition
|x|<delta
-epsilon<f(x)<epsilon

what epsilon to chose to get a value smaller then c
?

8. ## Re: intermidiate value therem question

Originally Posted by transgalactic
yes so by limit definition
|x|<delta-epsilon<f(x)<epsilon what epsilon to chose to get a value smaller then c
?
At this point, do you even know what you are talking about?
$\delta/\varepsilon$ has absolutely nothing to do with it, as far as I can tell.

If $c<0$ then $2c then because $f$ is increasing $f(2c).

9. ## Re: intermidiate value therem question

Originally Posted by Plato
At this point, do you even know what you are talking about?
$\delta/\varepsilon$ has absolutely nothing to do with it, as far as I can tell.

If $c<0$ then $2c then because $f$ is increasing $f(2c).
ok i see it now .
if its decreasing we take smaller values and they get a bigger function value

so from the decreasing property we can directly write
$f(2c)

from here its straight forward indermidiate value theorem

so why we need the limit
when x goes to zero the limit is sero

?

10. ## Re: intermidiate value therem question

Originally Posted by transgalactic
from here its straight forward indermidiate value theorem
so why we need the limit
when x goes to zero the limit is zero?
We want to know that $f$ to map $(-\infty,0)$ one-to-one and onto $(-\infty,0)$.

11. ## Re: intermidiate value therem question

ok so we need to havea value at some point then say that its decreasing asnd we get one to one

but here we need to prove the solution is c too
how
?

12. ## Re: intermidiate value therem question

Originally Posted by transgalactic
ok so we need to havea value at some point then say that its decreasing asnd we get one to one

but here we need to prove the solution is c too
how
?
I simply give up. Good luck.

13. ## Re: intermidiate value therem question

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 soby limit definition -e<f(x)<e
from the limit when x goes minus infinity f(x)<-N
what e to chhose?
what N to choose?

14. ## Re: intermidiate value therem question

Originally Posted by transgalactic
prove that for c<0
there is only one solution to
$xe^{\frac{1}{x}}=c$
As stated, the problem is to prove that the equation $xe^{1/x}=c$ has 0 or 1 solution for each c < 0. That's what "only one" means: there cannot be more than one. If the problem said "there exists one solution to this equation," that would mean "at least one, but possibly more." To stress that there exists exactly one solution, people sometimes say, "there exists one and only one solution."

Now, this interpretation is arguable and possibly not every everybody would agree that "only one" means <= 1 and "one" means ">= 1." I would therefore avoid statements containing "only one." However, since it is given, the default interpretation of this problem for me is to prove that there cannot be two solutions.

So, what exactly do you need to prove here? You may need to contact your instructor for clarification.