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How to show logistic function is continuous monotonic increasing?

$$\frac{1}{1+e^{-ax}}$$

Thanks in advanced..

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    $\begingroup$ Hint: show the derivative is positive. $\endgroup$
    – vadim123
    Oct 11, 2014 at 15:00
  • $\begingroup$ I did it vadim, but I am curious only to know is there a different way to show it is monotonic increasing? $\endgroup$
    – valerie
    Oct 11, 2014 at 15:20

1 Answer 1

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If you look at $e^{-ax}$, can you say something about the monotony of that? If so, what does that mean for $1+e^{-ax}$?

Now what does it tell you if your denominator is monotonous, and the numerator is constant?

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  • $\begingroup$ Hi, is that enough to show it through the derivative is always positive? I did it, but I want to know if there is another way to show that..or through the limit with $x->0$ = 1 also enough to claim the function is monotonic increasing? $\endgroup$
    – valerie
    Oct 11, 2014 at 15:18
  • $\begingroup$ The derivative being positive is equivalent to your function being strictly increasing. My approach would have been to say that $e^{-ax}=\frac{1}{e^{ax}}=(\frac{1}{e^x})^a$, and I know that $e^x$ is strictly increasing - but then, again, wanting to prove the latter I can't think of a quicker method than to derive the $e^x$. I don't know what you mean by 'through the limit', since monotony is a local property I don't see how the limit would tell you about monotony elsewhere? $\endgroup$ Oct 11, 2014 at 16:02
  • $\begingroup$ I mean that through the limit of the logistic function will bring you that for $x->\infty=1$ while for $x->-\infty=0 $ $\endgroup$
    – valerie
    Oct 11, 2014 at 16:19
  • $\begingroup$ That the limit for $x\to\infty$ is greater than for $x\to -\infty$ is a necessary requirement for the function to be increasing. It is, however, not a proof: take a look at $f(x)=x^3-x$, for instance: we have $\lim_{x\to\infty}{f(x)}=+\infty$, $\lim_{x\to -\infty}{f(x)}=-\infty$, but in $[-0.5,0.5]$ we are strictly decreasing. $\endgroup$ Oct 11, 2014 at 16:54

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