# Thread: mean value theory infinity proble

1. ## mean value theory infinity proble

pprove that for alll $\displaystyle 0<c<\frac{{1}}{2}$ there is x>0 so $\displaystyle \frac{1}{x}-\frac{1}{e^{x}-1}=c$

i have tried $\displaystyle f(x)=\frac{1}{x}-\frac{1}{e^{x}-1}$
limit of f(x) x->infinity = zero
limit of f(x) x->0 = pi/2

so in similar problems i have defined

g(x)= 0.5 when x=0
g(x)= f(x) when 0<x<infinity
g(x)=0 when x=infinity

and then i have to point one is 0 the other is 1/2
and i do mean value theorem

but its not legat to definne such function
infinity is not a point,what to do?

2. ## Re: mean value theory infinity proble

Originally Posted by transgalactic
pprove that for alll $\displaystyle 0<c<\frac{{1}}{2}$ there is x>0 so $\displaystyle \frac{1}{x}-\frac{1}{e^{x}-1}=c$
$\displaystyle f(x)=\frac{1}{x}-\frac{1}{e^{x}-1}$
Please review the statement of this question.
As written it is not true.
$\displaystyle f$ is decreasing and on $\displaystyle (0,0.5]$ the minimum is $\displaystyle c.0.4585$

3. ## Re: mean value theory infinity proble

Originally Posted by transgalactic
limit of f(x) x->infinity = zero
limit of f(x) x->0 = pi/2
The first limit is correct; the second one should be 1/2. Thus, for each 0 < c < 1/2, you can find x1 > 0 such that c <= f(x1) < 1/2 and x2 > 0 such that f(x2) <= c. The function is continuous for x > 0, so by the intermediate value theorem, there exists an x1 <= x <= x2 such that f(x) = c.

4. ## Re: mean value theory infinity proble

limit of f(x) x->infinity = zero
limit of f(x) x->0 = 1/2

my problem is how to explain that there is such x1 c <= f(x1) < 1/2
from the limit?

same thing goes for
f(x2) <= c

i dont know how to get to these to innequalitties from limit definition

from the first i got
for every x>M -epsilon<f(x)<epsilon

from the second limit definition i get
for every |x-0|<delta 1/2-epsilon<f(x)<epsilon+1/2

i dont know how from these limit definition i get these innequlities
?

5. ## Re: mean value theory infinity proble

For the limit x -> infinity, take epsilon = c. For the limit x -> 0, take epsilon = 1/2 - c.

6. ## Re: mean value theory infinity proble

1/2-epsilon<f(x)<epsilon+1/2
take epsilon = 1/2 - c
so i get

1+c<f(x)<1-c

-epsilon<f(x)<epsilon
epsilon=c
so i get

-c<f(x)<c
so from the second i get f(x2)<c

how i get c <= f(x1) < 1/2
from
1+c<f(x)<1-c
why its smaller then 1/2
?

7. ## Re: mean value theory infinity proble

Originally Posted by transgalactic
1/2-epsilon<f(x)<epsilon+1/2
take epsilon = 1/2 - c
so i get

1+c<f(x)<1-c
Check the last line again. Also, it is helpful to think not about algebraic expressions but about distances on the numerical line. Imagine 1/2, then c located below it and some point x whose distance to 1/2 is less than the distance between c and 1/2.

8. ## Re: mean value theory infinity proble

c<f(x)<1-c

and it is given that 0<c<1/2

but stil it doesnt prove f(x)<1/2

f(x)<t

t is allways bigger then 1/2

if c was 1/2 we would get f(x)<1/2

but its not
c is bellow 1/2

?

9. ## Re: mean value theory infinity proble

Originally Posted by transgalactic
c<f(x)<1-c

and it is given that 0<c<1/2

but stil it doesnt prove f(x)<1/2
I guess you are concerned about f(x) < 1/2 because I said:
Originally Posted by emakarov
Thus, for each 0 < c < 1/2, you can find x1 > 0 such that c <= f(x1) < 1/2
According to the graph, f(x) is monotonically decreasing, so f(x) < 1/2. However, it is not important for the current problem. The important thing is that one can find x1 such that f(x1) > c and x2 such that f(x2) < c.

10. ## Re: mean value theory infinity proble

oohh i get it thanks

11. ## Re: mean value theory infinity proble

never mind the question
i had a question and i solved it