How to detect sharp corners in graphs

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I have drawn a graph for this function



$$logleft(1+|x|right)$$



It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".







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  • Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
    – tmaths
    Aug 16 at 9:09










  • How can I be sure that a sharp corner exists? Do I need to check the derivates?
    – Cesare
    Aug 16 at 9:10















up vote
0
down vote

favorite












I have drawn a graph for this function



$$logleft(1+|x|right)$$



It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".







share|cite|improve this question




















  • Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
    – tmaths
    Aug 16 at 9:09










  • How can I be sure that a sharp corner exists? Do I need to check the derivates?
    – Cesare
    Aug 16 at 9:10













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have drawn a graph for this function



$$logleft(1+|x|right)$$



It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".







share|cite|improve this question












I have drawn a graph for this function



$$logleft(1+|x|right)$$



It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 16 at 9:04









Cesare

53129




53129











  • Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
    – tmaths
    Aug 16 at 9:09










  • How can I be sure that a sharp corner exists? Do I need to check the derivates?
    – Cesare
    Aug 16 at 9:10

















  • Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
    – tmaths
    Aug 16 at 9:09










  • How can I be sure that a sharp corner exists? Do I need to check the derivates?
    – Cesare
    Aug 16 at 9:10
















Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
– tmaths
Aug 16 at 9:09




Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
– tmaths
Aug 16 at 9:09












How can I be sure that a sharp corner exists? Do I need to check the derivates?
– Cesare
Aug 16 at 9:10





How can I be sure that a sharp corner exists? Do I need to check the derivates?
– Cesare
Aug 16 at 9:10











4 Answers
4






active

oldest

votes

















up vote
2
down vote



accepted










You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :



$$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$



So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :



enter image description here






share|cite|improve this answer



























    up vote
    2
    down vote













    Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.



    Then:



    $ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$



    and



    $ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.






    share|cite|improve this answer



























      up vote
      1
      down vote













      On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point






      share|cite|improve this answer




















      • Thanks, does my function has a sharp corner?
        – Cesare
        Aug 16 at 9:14










      • Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
        – Davide Morgante
        Aug 16 at 9:22










      • Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
        – Davide Morgante
        Aug 16 at 9:32

















      up vote
      0
      down vote













      Basic properties of functions:



      • Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.


      • Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.


      • Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.


      For more information, read about Smoothness.






      share|cite|improve this answer




















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote



        accepted










        You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :



        $$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$



        So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :



        enter image description here






        share|cite|improve this answer
























          up vote
          2
          down vote



          accepted










          You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :



          $$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$



          So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :



          enter image description here






          share|cite|improve this answer






















            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :



            $$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$



            So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :



            enter image description here






            share|cite|improve this answer












            You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :



            $$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$



            So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 16 at 9:21









            tmaths

            1,332113




            1,332113




















                up vote
                2
                down vote













                Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.



                Then:



                $ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$



                and



                $ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.



                  Then:



                  $ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$



                  and



                  $ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.



                    Then:



                    $ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$



                    and



                    $ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.






                    share|cite|improve this answer












                    Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.



                    Then:



                    $ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$



                    and



                    $ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 16 at 9:28









                    Fred

                    38k1238




                    38k1238




















                        up vote
                        1
                        down vote













                        On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point






                        share|cite|improve this answer




















                        • Thanks, does my function has a sharp corner?
                          – Cesare
                          Aug 16 at 9:14










                        • Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
                          – Davide Morgante
                          Aug 16 at 9:22










                        • Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
                          – Davide Morgante
                          Aug 16 at 9:32














                        up vote
                        1
                        down vote













                        On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point






                        share|cite|improve this answer




















                        • Thanks, does my function has a sharp corner?
                          – Cesare
                          Aug 16 at 9:14










                        • Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
                          – Davide Morgante
                          Aug 16 at 9:22










                        • Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
                          – Davide Morgante
                          Aug 16 at 9:32












                        up vote
                        1
                        down vote










                        up vote
                        1
                        down vote









                        On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point






                        share|cite|improve this answer












                        On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Aug 16 at 9:13









                        Davide Morgante

                        2,357322




                        2,357322











                        • Thanks, does my function has a sharp corner?
                          – Cesare
                          Aug 16 at 9:14










                        • Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
                          – Davide Morgante
                          Aug 16 at 9:22










                        • Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
                          – Davide Morgante
                          Aug 16 at 9:32
















                        • Thanks, does my function has a sharp corner?
                          – Cesare
                          Aug 16 at 9:14










                        • Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
                          – Davide Morgante
                          Aug 16 at 9:22










                        • Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
                          – Davide Morgante
                          Aug 16 at 9:32















                        Thanks, does my function has a sharp corner?
                        – Cesare
                        Aug 16 at 9:14




                        Thanks, does my function has a sharp corner?
                        – Cesare
                        Aug 16 at 9:14












                        Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
                        – Davide Morgante
                        Aug 16 at 9:22




                        Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
                        – Davide Morgante
                        Aug 16 at 9:22












                        Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
                        – Davide Morgante
                        Aug 16 at 9:32




                        Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età! Keep up the good work
                        – Davide Morgante
                        Aug 16 at 9:32










                        up vote
                        0
                        down vote













                        Basic properties of functions:



                        • Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.


                        • Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.


                        • Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.


                        For more information, read about Smoothness.






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          Basic properties of functions:



                          • Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.


                          • Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.


                          • Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.


                          For more information, read about Smoothness.






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Basic properties of functions:



                            • Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.


                            • Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.


                            • Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.


                            For more information, read about Smoothness.






                            share|cite|improve this answer












                            Basic properties of functions:



                            • Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.


                            • Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.


                            • Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.


                            For more information, read about Smoothness.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 16 at 9:38









                            Phil H

                            9531513




                            9531513






















                                 

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