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anonymous
 one year ago
Is the value of (1exp(x))/(1+exp(x)) between (1,1)? How?
anonymous
 one year ago
Is the value of (1exp(x))/(1+exp(x)) between (1,1)? How?

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.3are you asking if that is the range?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=\frac{1\frac{1}{e^x}}{1+\frac{1}{e^x}}\] easier to see if you multiply by \(e^x\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3you get \[f(x)=\frac{e^x1}{e^x+1}\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3now 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
 one year ago
Best ResponseYou've already chosen the best response.0But the min value is not 1, right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3it 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
 one year ago
Best ResponseYou've already chosen the best response.0If xtends to inf, the value is 1 but if x tends to inf it is not 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The upper limit is not one, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But e to the power inf is infinity, so if x tends to inf, we have inf divided by inf which is not defined

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3you are talking about a limit right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3then 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
 one year ago
Best ResponseYou've already chosen the best response.0I m nr sure how the upper limit of the range is 1

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3what do you make of \[\lim_{x\to \infty}\frac{x^2+1}{x^21}\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The numerator goes to inf faster so the limit is inf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or Do we divide by x n see

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3oh no i see the confusion \[x^2+1\] goes to infinity just as fast as \(x^21\) the number out at the end makes no difference at all

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3you remember in some class, maybe called "pre calculus" finding the horizontal asymptote of a rational function?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3that 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
 one year ago
Best ResponseYou've already chosen the best response.0I m still not getting why the upper limit of my expression shd be 1, cn u pls explain that

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3of the expression \[\frac{e^x1}{e^x1}\]?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3ooops i made a typo there, should be \[\frac{e^x1}{e^x+1}\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3ok 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

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3and also since the numerator is less than the denominator, this is always smaller than 1, not larger right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3\(e^x\) grows very very quickly, much faster than \(x^2\) or in fact any power of \(x\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3already \(e^{10}\) is \(22026\) rounded so you would have \[\frac{22025}{22027}\] which is pretty close to 1, with \(x=10\) only

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3imagine what it would look like if \(x=100\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can u tell me what kind of function will have a range between 1 and infinity

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3i can make up one sure

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3how about \[f(x)=\sqrt{x}+1\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.3those would have range \([1,\infty)\) if you want \((1,\infty)\) open you could use for example \[f(x)=e^x+1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was looking for functions close to logit

perl
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{blue}{\text{Originally Posted by}}\) @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
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