Logistic function with a slope but no asymptotes?












6












$begingroup$


The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve







share|cite|improve this question











$endgroup$












  • $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    14 hours ago










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    14 hours ago










  • $begingroup$
    I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
    $endgroup$
    – jld
    14 hours ago






  • 4




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    14 hours ago








  • 2




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    9 hours ago


















6












$begingroup$


The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve







share|cite|improve this question











$endgroup$












  • $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    14 hours ago










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    14 hours ago










  • $begingroup$
    I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
    $endgroup$
    – jld
    14 hours ago






  • 4




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    14 hours ago








  • 2




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    9 hours ago
















6












6








6





$begingroup$


The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve







share|cite|improve this question











$endgroup$




The logistic function has an output range 0 to 1, and asymptotic slope is zero on both sides.



What is an alternative to a logistic function that doesn't flatten out completely at its ends? Whose asymptotic slopes are approaching zero but not zero, and the range is infinite?






sigmoid-curve




sigmoid-curve






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 22 mins ago









Neil G

9,79012970




9,79012970










asked 15 hours ago









AksakalAksakal

39k452120




39k452120












  • $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    14 hours ago










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    14 hours ago










  • $begingroup$
    I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
    $endgroup$
    – jld
    14 hours ago






  • 4




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    14 hours ago








  • 2




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    9 hours ago




















  • $begingroup$
    The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
    $endgroup$
    – jld
    14 hours ago










  • $begingroup$
    Basically I want a function that looks like sigmoid but has a slope
    $endgroup$
    – Aksakal
    14 hours ago










  • $begingroup$
    I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
    $endgroup$
    – jld
    14 hours ago






  • 4




    $begingroup$
    $operatorname{sign}(x)log(1 + |x|)$?
    $endgroup$
    – steveo'america
    14 hours ago








  • 2




    $begingroup$
    Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
    $endgroup$
    – usεr11852
    9 hours ago


















$begingroup$
The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
$endgroup$
– jld
14 hours ago




$begingroup$
The title seems to disagree with how i read your question -- is this new function required to have asymptotes or not?
$endgroup$
– jld
14 hours ago












$begingroup$
Basically I want a function that looks like sigmoid but has a slope
$endgroup$
– Aksakal
14 hours ago




$begingroup$
Basically I want a function that looks like sigmoid but has a slope
$endgroup$
– Aksakal
14 hours ago












$begingroup$
I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
$endgroup$
– jld
14 hours ago




$begingroup$
I just updated -- is that more what you mean? I'm still not sure what you mean by a "slope". Do you mean a sigmoid shape but with $lim_{xtopminfty}f(x) = pm infty$, i.e. it doesn't flatten out into horizontal asymptotes at $|x|$ grows?
$endgroup$
– jld
14 hours ago




4




4




$begingroup$
$operatorname{sign}(x)log(1 + |x|)$?
$endgroup$
– steveo'america
14 hours ago






$begingroup$
$operatorname{sign}(x)log(1 + |x|)$?
$endgroup$
– steveo'america
14 hours ago






2




2




$begingroup$
Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
$endgroup$
– usεr11852
9 hours ago






$begingroup$
Beginning of the decade called, it wants its neural network activation functions back. (Sorry bad joke, but realistically this is why people moved to ReLUs) (+1 though, relevant question)
$endgroup$
– usεr11852
9 hours ago












3 Answers
3






active

oldest

votes


















8












$begingroup$

Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
$$
text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
$$



This is unbounded but grows like $log$ for large $|x|$ and looks like
asinh



I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




Original answer



$newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
$$
lim_{xtopm infty} f(x) = 0.
$$



Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
$$
exists x_1 : x < x_1 implies |f(x)| < e
$$
and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



This means that any such function can't be continuous. Would something like
$$
f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
$$ work?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
    $endgroup$
    – Sycorax
    12 hours ago












  • $begingroup$
    @Sycorax thanks, i was wondering about that
    $endgroup$
    – jld
    10 hours ago










  • $begingroup$
    My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
    $endgroup$
    – Ingolifs
    7 hours ago



















6












$begingroup$

You could just add a term to a logistic function:



$$
f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
$$



The asymptotes will have slopes $d$.



Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



Sigmoid






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
    $endgroup$
    – user1717828
    5 hours ago



















5












$begingroup$

I will go ahead and turn the comment into an answer. I suggest
$$
f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
$$
which has slope tending towards zero, but is unbounded.






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      12 hours ago












    • $begingroup$
      @Sycorax thanks, i was wondering about that
      $endgroup$
      – jld
      10 hours ago










    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      7 hours ago
















    8












    $begingroup$

    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      12 hours ago












    • $begingroup$
      @Sycorax thanks, i was wondering about that
      $endgroup$
      – jld
      10 hours ago










    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      7 hours ago














    8












    8








    8





    $begingroup$

    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?






    share|cite|improve this answer











    $endgroup$



    Initially I was thinking you did want the horizontal asymptotes at $0$ still; I moved my original answer to the end. If you instead want $lim_{xtopm infty} f(x) = pminfty$ then would something like the inverse hyperbolic sine work?
    $$
    text{asinh}(x) = logleft(x + sqrt{1 + x^2}right)
    $$



    This is unbounded but grows like $log$ for large $|x|$ and looks like
    asinh



    I like this function a lot as a data transformation when I've got heavy tails but possibly zeros or negative values.



    Another nice thing about this function is that $text{asinh}'(x) = frac{1}{sqrt{1+x^2}}$ so it has a nice simple derivative.




    Original answer



    $newcommand{e}{varepsilon}$Let $f : mathbb Rtomathbb R$ be our function and we'll assume
    $$
    lim_{xtopm infty} f(x) = 0.
    $$



    Suppose $f$ is continuous. Fix $e > 0$. From the asymptotes we have
    $$
    exists x_1 : x < x_1 implies |f(x)| < e
    $$
    and analogously there's an $x_2$ such that $x > x_2 implies |f(x)| < e$. Therefore outside of $[x_1,x_2]$ $f$ is within $(-e, e)$. And $[x_1,x_2]$ is a compact interval so by continuity $f$ is bounded on it.



    This means that any such function can't be continuous. Would something like
    $$
    f(x) = begin{cases} x^{-1} & xneq 0 \ 0 & x = 0end{cases}
    $$     work?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 12 hours ago

























    answered 14 hours ago









    jldjld

    12.3k23352




    12.3k23352








    • 1




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      12 hours ago












    • $begingroup$
      @Sycorax thanks, i was wondering about that
      $endgroup$
      – jld
      10 hours ago










    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      7 hours ago














    • 1




      $begingroup$
      The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
      $endgroup$
      – Sycorax
      12 hours ago












    • $begingroup$
      @Sycorax thanks, i was wondering about that
      $endgroup$
      – jld
      10 hours ago










    • $begingroup$
      My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
      $endgroup$
      – Ingolifs
      7 hours ago








    1




    1




    $begingroup$
    The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
    $endgroup$
    – Sycorax
    12 hours ago






    $begingroup$
    The "Related" threads include this unanswered question, in case anyone else has asked themselves the natural followup "what happens if you use asinh in a neural network?" stats.stackexchange.com/questions/359245/…
    $endgroup$
    – Sycorax
    12 hours ago














    $begingroup$
    @Sycorax thanks, i was wondering about that
    $endgroup$
    – jld
    10 hours ago




    $begingroup$
    @Sycorax thanks, i was wondering about that
    $endgroup$
    – jld
    10 hours ago












    $begingroup$
    My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
    $endgroup$
    – Ingolifs
    7 hours ago




    $begingroup$
    My ears did indeed prick up. I have in the past found asinh() useful when you want to 'do log stuff' to both positive and negative numbers. It also gets around the quandry you can get in, where you need to do a log transform on data with zeros and have to judge an appropriate value of $a$ for $log(x + a)$
    $endgroup$
    – Ingolifs
    7 hours ago













    6












    $begingroup$

    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      5 hours ago
















    6












    $begingroup$

    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      5 hours ago














    6












    6








    6





    $begingroup$

    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid






    share|cite|improve this answer









    $endgroup$



    You could just add a term to a logistic function:



    $$
    f(x; a, b, c, d, e)=frac{a}{1+bexp(-cx)} + dx + e
    $$



    The asymptotes will have slopes $d$.



    Here is an example with $a=10, b = 1, c = 2, d = frac{1}{20}, e = -5$:



    Sigmoid







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 13 hours ago









    COOLSerdashCOOLSerdash

    16.4k75293




    16.4k75293












    • $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      5 hours ago


















    • $begingroup$
      I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
      $endgroup$
      – user1717828
      5 hours ago
















    $begingroup$
    I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
    $endgroup$
    – user1717828
    5 hours ago




    $begingroup$
    I think this answer is the best because if you zoom out far enough it's just a straight line with a little wiggle in the middle. Gives the most intuitive behavior at large x but retains the sigmoid shape.
    $endgroup$
    – user1717828
    5 hours ago











    5












    $begingroup$

    I will go ahead and turn the comment into an answer. I suggest
    $$
    f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
    $$
    which has slope tending towards zero, but is unbounded.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      I will go ahead and turn the comment into an answer. I suggest
      $$
      f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
      $$
      which has slope tending towards zero, but is unbounded.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        I will go ahead and turn the comment into an answer. I suggest
        $$
        f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
        $$
        which has slope tending towards zero, but is unbounded.






        share|cite|improve this answer









        $endgroup$



        I will go ahead and turn the comment into an answer. I suggest
        $$
        f(x) = operatorname{sign}(x)log{left(1 + |x|right)},
        $$
        which has slope tending towards zero, but is unbounded.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        steveo'americasteveo'america

        23318




        23318






























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