Question

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

Answer 1

HI!!

Answer 2

are you asking if that is the range?

Answer 3

Yes

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

But the min value is not -1, right?

Answer 8

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 \([-1,1]\)

Answer 9

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

Answer 10

why not?

Answer 11

The upper limit is not one, right?

Answer 12

sure it is

Answer 13

But e to the power inf is infinity, so if x tends to inf, we have inf divided by inf which is not defined

Answer 14

?

Answer 15

you are talking about a limit right?

Answer 16

Ya

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

The numerator goes to inf faster so the limit is inf

Answer 21

Or Do we divide by x n see

Answer 22

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

Answer 23

So it tends to 1

Answer 24

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

Answer 25

Ya

Answer 26

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"

Answer 27

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

Answer 28

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

Answer 29

Ya

Answer 30

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

Answer 31

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

Answer 32

Ya

Answer 33

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

Answer 34

Right

Answer 35

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? \(\frac{1}{2}\)?

Answer 36

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

Answer 37

Ok

Answer 38

already \(e^{10}\) is \(22026\) rounded so you would have \[\frac{22025}{22027}\] which is pretty close to 1, with \(x=10\) only

Answer 39

I see

Answer 40

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

Answer 41

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

Answer 42

i can make up one sure

Answer 43

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

Answer 44

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

Answer 45

I see that

Answer 46

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

Answer 47

I was looking for functions close to logit

Answer 48

ok wise guy

Answer 49

\(\color{blue}{\text{Originally Posted by}}\) @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

Answer 50

you must be bored

Answer 51

No way

Answer 52

U explain so well

Answer 53

R u prof of math?