anonymous
  • anonymous
Is the value of (1-exp(-x))/(1+exp(-x)) between (-1,1)? How?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
are you asking if that is the range?
anonymous
  • anonymous
Yes

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misty1212
  • misty1212
\[f(x)=\frac{1-\frac{1}{e^x}}{1+\frac{1}{e^x}}\] easier to see if you multiply by \(e^x\)
misty1212
  • misty1212
you get \[f(x)=\frac{e^x-1}{e^x+1}\]
misty1212
  • misty1212
now it is pretty clearly always less than 1, since \(e^x>0\) for all \(x\) and the numerator is one less than the denominator
anonymous
  • anonymous
But the min value is not -1, right?
misty1212
  • misty1212
it cannot be \(-1\) but that is the lower limit, which is why the range is the open interval \((-1,1)\) not the closed one \([-1,1]\)
anonymous
  • anonymous
If xtends to -inf, the value is -1 but if x tends to inf it is not 1
misty1212
  • misty1212
why not?
anonymous
  • anonymous
The upper limit is not one, right?
misty1212
  • misty1212
sure it is
anonymous
  • anonymous
But e to the power inf is infinity, so if x tends to inf, we have inf divided by inf which is not defined
misty1212
  • misty1212
?
misty1212
  • misty1212
you are talking about a limit right?
anonymous
  • anonymous
Ya
misty1212
  • misty1212
then lets go slow, because the statement "if x tends to inf, we have inf divided by inf which is not defined" makes no sense in terms of limits
anonymous
  • anonymous
I m nr sure how the upper limit of the range is 1
misty1212
  • misty1212
what do you make of \[\lim_{x\to \infty}\frac{x^2+1}{x^2-1}\]?
anonymous
  • anonymous
The numerator goes to inf faster so the limit is inf
anonymous
  • anonymous
Or Do we divide by x n see
misty1212
  • misty1212
oh no i see the confusion \[x^2+1\] goes to infinity just as fast as \(x^2-1\) the number out at the end makes no difference at all
anonymous
  • anonymous
So it tends to 1
misty1212
  • misty1212
you remember in some class, maybe called "pre calculus" finding the horizontal asymptote of a rational function?
anonymous
  • anonymous
Ya
misty1212
  • misty1212
that is what you are doing exactly if the degrees are the same, it is the ratio of the leading coefficients you do not say "it is infinity over infinity therefore undefined"
anonymous
  • anonymous
I m still not getting why the upper limit of my expression shd be 1, cn u pls explain that
misty1212
  • misty1212
of the expression \[\frac{e^x-1}{e^x-1}\]?
anonymous
  • anonymous
Ya
misty1212
  • misty1212
ooops i made a typo there, should be \[\frac{e^x-1}{e^x+1}\]
misty1212
  • misty1212
ok we agree that the numerator is one less then the denominator right? and therefore it can never be one, because a fraction is only one if the numerator and denominator are the same
anonymous
  • anonymous
Ya
misty1212
  • misty1212
and also since the numerator is less than the denominator, this is always smaller than 1, not larger right?
anonymous
  • anonymous
Right
misty1212
  • misty1212
so let me ask you a question it can never be one it is always smaller than one what do you think the limit might be? \(\frac{1}{2}\)?
misty1212
  • misty1212
\(e^x\) grows very very quickly, much faster than \(x^2\) or in fact any power of \(x\)
anonymous
  • anonymous
Ok
misty1212
  • misty1212
already \(e^{10}\) is \(22026\) rounded so you would have \[\frac{22025}{22027}\] which is pretty close to 1, with \(x=10\) only
anonymous
  • anonymous
I see
misty1212
  • misty1212
imagine what it would look like if \(x=100\)
anonymous
  • anonymous
Can u tell me what kind of function will have a range between 1 and infinity
misty1212
  • misty1212
i can make up one sure
misty1212
  • misty1212
how about \[f(x)=\sqrt{x}+1\]
misty1212
  • misty1212
or \[f(x)=x^2+1\]
anonymous
  • anonymous
I see that
misty1212
  • misty1212
those would have range \([1,\infty)\) if you want \((1,\infty)\) open you could use for example \[f(x)=e^x+1\]
anonymous
  • anonymous
I was looking for functions close to logit
misty1212
  • misty1212
ok wise guy
perl
  • perl
\(\color{blue}{\text{Originally Posted by}}\) @misty1212 now it is pretty clearly always less than 1, since \(e^x>0\) for all \(x\) and the numerator is one less than the denominator \(\color{blue}{\text{End of Quote}}\) The numerator* is 2 less than the denominator
misty1212
  • misty1212
you must be bored
anonymous
  • anonymous
No way
anonymous
  • anonymous
U explain so well
anonymous
  • anonymous
R u prof of math?

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