Is the value of (1-exp(-x))/(1+exp(-x)) between (-1,1)? How?

- anonymous

Is the value of (1-exp(-x))/(1+exp(-x)) between (-1,1)? How?

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- schrodinger

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- misty1212

HI!!

- misty1212

are you asking if that is the range?

- anonymous

Yes

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## More answers

- misty1212

\[f(x)=\frac{1-\frac{1}{e^x}}{1+\frac{1}{e^x}}\] easier to see if you multiply by \(e^x\)

- misty1212

you get
\[f(x)=\frac{e^x-1}{e^x+1}\]

- 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

But the min value is not -1, right?

- 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

If xtends to -inf, the value is -1 but if x tends to inf it is not 1

- misty1212

why not?

- anonymous

The upper limit is not one, right?

- misty1212

sure it is

- 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

you are talking about a limit right?

- anonymous

Ya

- 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

I m nr sure how the upper limit of the range is 1

- misty1212

what do you make of
\[\lim_{x\to \infty}\frac{x^2+1}{x^2-1}\]?

- anonymous

The numerator goes to inf faster so the limit is inf

- anonymous

Or Do we divide by x n see

- 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

So it tends to 1

- misty1212

you remember in some class, maybe called "pre calculus" finding the horizontal asymptote of a rational function?

- anonymous

Ya

- 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

I m still not getting why the upper limit of my expression shd be 1, cn u pls explain that

- misty1212

of the expression
\[\frac{e^x-1}{e^x-1}\]?

- anonymous

Ya

- misty1212

ooops i made a typo there, should be
\[\frac{e^x-1}{e^x+1}\]

- 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

Ya

- misty1212

and also since the numerator is less than the denominator, this is always smaller than 1, not larger right?

- anonymous

Right

- 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

\(e^x\) grows very very quickly, much faster than \(x^2\) or in fact any power of \(x\)

- anonymous

Ok

- 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

I see

- misty1212

imagine what it would look like if \(x=100\)

- anonymous

Can u tell me what kind of function will have a range between 1 and infinity

- misty1212

i can make up one sure

- misty1212

how about
\[f(x)=\sqrt{x}+1\]

- misty1212

or
\[f(x)=x^2+1\]

- anonymous

I see that

- misty1212

those would have range \([1,\infty)\) if you want \((1,\infty)\) open you could use for example
\[f(x)=e^x+1\]

- anonymous

I was looking for functions close to logit

- misty1212

ok wise guy

- 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

you must be bored

- anonymous

No way

- anonymous

U explain so well

- anonymous

R u prof of math?

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