$f(x) = frac{x^2 - 1}{x-1} $ is different from $f(x) = x + 1$, in spite of the fact that rules of algebra is...












3












$begingroup$


Suppose there be a function, $$ f(x) = frac {x^2-1}{x-1} $$



For $x=1$, the function becomes un defined. But, in Algebra we know it is allowed to Cancel Denominator and numerator by the common factor and this would result in the same expression which is equivalent to the first one. But,



$$require{cancel} f(x) = frac{(x+1)cancel{(x-1)}}{cancel{(x-1)}} =y$$
$$Rightarrow y = x + 1 $$



But, here when I see that Graph of both the equation they look same however the first one is undefined for x = 1 which is not with the case of the second one. But, just canceling a common term in numerator and denominator or multiplying them, changes the whole function why?



So, rules of algebra don't work? or there is a problem with my understanding?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - frac1x - 1$ instead of $frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by.
    $endgroup$
    – Arthur
    Sep 17 '18 at 13:43












  • $begingroup$
    @Arthur Okay let me add them, nice idea, btw :)
    $endgroup$
    – Abhas Kumar Sinha
    Sep 17 '18 at 13:44










  • $begingroup$
    @AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:45








  • 1




    $begingroup$
    @AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $mathbb{R}-{1}$
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:53








  • 2




    $begingroup$
    @AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case.
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 14:00


















3












$begingroup$


Suppose there be a function, $$ f(x) = frac {x^2-1}{x-1} $$



For $x=1$, the function becomes un defined. But, in Algebra we know it is allowed to Cancel Denominator and numerator by the common factor and this would result in the same expression which is equivalent to the first one. But,



$$require{cancel} f(x) = frac{(x+1)cancel{(x-1)}}{cancel{(x-1)}} =y$$
$$Rightarrow y = x + 1 $$



But, here when I see that Graph of both the equation they look same however the first one is undefined for x = 1 which is not with the case of the second one. But, just canceling a common term in numerator and denominator or multiplying them, changes the whole function why?



So, rules of algebra don't work? or there is a problem with my understanding?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - frac1x - 1$ instead of $frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by.
    $endgroup$
    – Arthur
    Sep 17 '18 at 13:43












  • $begingroup$
    @Arthur Okay let me add them, nice idea, btw :)
    $endgroup$
    – Abhas Kumar Sinha
    Sep 17 '18 at 13:44










  • $begingroup$
    @AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:45








  • 1




    $begingroup$
    @AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $mathbb{R}-{1}$
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:53








  • 2




    $begingroup$
    @AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case.
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 14:00
















3












3








3





$begingroup$


Suppose there be a function, $$ f(x) = frac {x^2-1}{x-1} $$



For $x=1$, the function becomes un defined. But, in Algebra we know it is allowed to Cancel Denominator and numerator by the common factor and this would result in the same expression which is equivalent to the first one. But,



$$require{cancel} f(x) = frac{(x+1)cancel{(x-1)}}{cancel{(x-1)}} =y$$
$$Rightarrow y = x + 1 $$



But, here when I see that Graph of both the equation they look same however the first one is undefined for x = 1 which is not with the case of the second one. But, just canceling a common term in numerator and denominator or multiplying them, changes the whole function why?



So, rules of algebra don't work? or there is a problem with my understanding?










share|cite|improve this question











$endgroup$




Suppose there be a function, $$ f(x) = frac {x^2-1}{x-1} $$



For $x=1$, the function becomes un defined. But, in Algebra we know it is allowed to Cancel Denominator and numerator by the common factor and this would result in the same expression which is equivalent to the first one. But,



$$require{cancel} f(x) = frac{(x+1)cancel{(x-1)}}{cancel{(x-1)}} =y$$
$$Rightarrow y = x + 1 $$



But, here when I see that Graph of both the equation they look same however the first one is undefined for x = 1 which is not with the case of the second one. But, just canceling a common term in numerator and denominator or multiplying them, changes the whole function why?



So, rules of algebra don't work? or there is a problem with my understanding?







algebra-precalculus functions






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 22:16









N. F. Taussig

45k103358




45k103358










asked Sep 17 '18 at 13:40









Abhas Kumar SinhaAbhas Kumar Sinha

304115




304115








  • 1




    $begingroup$
    Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - frac1x - 1$ instead of $frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by.
    $endgroup$
    – Arthur
    Sep 17 '18 at 13:43












  • $begingroup$
    @Arthur Okay let me add them, nice idea, btw :)
    $endgroup$
    – Abhas Kumar Sinha
    Sep 17 '18 at 13:44










  • $begingroup$
    @AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:45








  • 1




    $begingroup$
    @AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $mathbb{R}-{1}$
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:53








  • 2




    $begingroup$
    @AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case.
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 14:00
















  • 1




    $begingroup$
    Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - frac1x - 1$ instead of $frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by.
    $endgroup$
    – Arthur
    Sep 17 '18 at 13:43












  • $begingroup$
    @Arthur Okay let me add them, nice idea, btw :)
    $endgroup$
    – Abhas Kumar Sinha
    Sep 17 '18 at 13:44










  • $begingroup$
    @AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:45








  • 1




    $begingroup$
    @AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $mathbb{R}-{1}$
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 13:53








  • 2




    $begingroup$
    @AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case.
    $endgroup$
    – Piotr Wasilewicz
    Sep 17 '18 at 14:00










1




1




$begingroup$
Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - frac1x - 1$ instead of $frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by.
$endgroup$
– Arthur
Sep 17 '18 at 13:43






$begingroup$
Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - frac1x - 1$ instead of $frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by.
$endgroup$
– Arthur
Sep 17 '18 at 13:43














$begingroup$
@Arthur Okay let me add them, nice idea, btw :)
$endgroup$
– Abhas Kumar Sinha
Sep 17 '18 at 13:44




$begingroup$
@Arthur Okay let me add them, nice idea, btw :)
$endgroup$
– Abhas Kumar Sinha
Sep 17 '18 at 13:44












$begingroup$
@AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1
$endgroup$
– Piotr Wasilewicz
Sep 17 '18 at 13:45






$begingroup$
@AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1
$endgroup$
– Piotr Wasilewicz
Sep 17 '18 at 13:45






1




1




$begingroup$
@AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $mathbb{R}-{1}$
$endgroup$
– Piotr Wasilewicz
Sep 17 '18 at 13:53






$begingroup$
@AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $mathbb{R}-{1}$
$endgroup$
– Piotr Wasilewicz
Sep 17 '18 at 13:53






2




2




$begingroup$
@AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case.
$endgroup$
– Piotr Wasilewicz
Sep 17 '18 at 14:00






$begingroup$
@AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case.
$endgroup$
– Piotr Wasilewicz
Sep 17 '18 at 14:00












4 Answers
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active

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$begingroup$

The functions are different because they have different domains.



A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.



The implied domain of the function defined by
$$f(x) = frac{x^2 - 1}{x - 1}$$
is the largest set of real numbers that do not make the denominator equal to zero, which is
$${x in mathbb{R} mid x neq 1} = mathbb{R} - {1}$$

Thus, we should write that $f$ is the function $f: mathbb{R} - {1} to mathbb{R}$ defined by
$$f(x) = frac{x^2 - 1}{x - 1}$$
Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since
$$lim_{x to 1} f(x) = lim_{x to 1} = frac{x^2 - 1}{x + 1} = lim_{x to 1} (x + 1) = 1 + 1 = 2$$



Its graph is the punctured line shown below.



graph_of_punctured_line



Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain
$$f(x) = frac{x^2 - 1}{x - 1} = frac{(x + 1)(x - 1)}{x - 1} = x + 1$$
for each $x in text{Dom}_f = mathbb{R} - {1}$.



The function $g: mathbb{R} to mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.



graph_of_line



While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    We can prove that both of these functions have the same graph for domain $D=mathbb{R}-{1}$:



    let $a in D$ be any number from domain. Then we have:



    $f(a) = a + 1 = a + 1 cdot frac{a-1}{a-1} = frac{a^{2}-1}{a-1} = f(a)$



    So the graph is the same for $D=mathbb{R}-{1}$: but in general these two functions are different because of Domain.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      There are no two functions given. To speak about a function we have to know the image and domain.
      $endgroup$
      – Cornman
      Sep 17 '18 at 13:55










    • $begingroup$
      @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
      $endgroup$
      – Piotr Wasilewicz
      Sep 17 '18 at 13:58










    • $begingroup$
      In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
      $endgroup$
      – Cornman
      Sep 17 '18 at 14:03










    • $begingroup$
      @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
      $endgroup$
      – Piotr Wasilewicz
      Sep 17 '18 at 14:05










    • $begingroup$
      But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
      $endgroup$
      – Piotr Wasilewicz
      Sep 17 '18 at 14:06



















    1












    $begingroup$

    The problem is, that $f(x)=frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:



    $f:mathbb{R}setminus{1}tomathbb{R}$.



    One might consider $mathbb{R}setminus{1}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero).
    So this is a pole.



    Or is it?



    It is completly fine to define $f:mathbb{R}tomathbb{R}$, since the pole is "liftable".
    A fake pole, if you please.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The only difference is that $frac{(x^2-1)}{(x-1)} $ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same.






      share|cite|improve this answer









      $endgroup$














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        4 Answers
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        4 Answers
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        active

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        $begingroup$

        The functions are different because they have different domains.



        A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.



        The implied domain of the function defined by
        $$f(x) = frac{x^2 - 1}{x - 1}$$
        is the largest set of real numbers that do not make the denominator equal to zero, which is
        $${x in mathbb{R} mid x neq 1} = mathbb{R} - {1}$$

        Thus, we should write that $f$ is the function $f: mathbb{R} - {1} to mathbb{R}$ defined by
        $$f(x) = frac{x^2 - 1}{x - 1}$$
        Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since
        $$lim_{x to 1} f(x) = lim_{x to 1} = frac{x^2 - 1}{x + 1} = lim_{x to 1} (x + 1) = 1 + 1 = 2$$



        Its graph is the punctured line shown below.



        graph_of_punctured_line



        Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain
        $$f(x) = frac{x^2 - 1}{x - 1} = frac{(x + 1)(x - 1)}{x - 1} = x + 1$$
        for each $x in text{Dom}_f = mathbb{R} - {1}$.



        The function $g: mathbb{R} to mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.



        graph_of_line



        While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          The functions are different because they have different domains.



          A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.



          The implied domain of the function defined by
          $$f(x) = frac{x^2 - 1}{x - 1}$$
          is the largest set of real numbers that do not make the denominator equal to zero, which is
          $${x in mathbb{R} mid x neq 1} = mathbb{R} - {1}$$

          Thus, we should write that $f$ is the function $f: mathbb{R} - {1} to mathbb{R}$ defined by
          $$f(x) = frac{x^2 - 1}{x - 1}$$
          Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since
          $$lim_{x to 1} f(x) = lim_{x to 1} = frac{x^2 - 1}{x + 1} = lim_{x to 1} (x + 1) = 1 + 1 = 2$$



          Its graph is the punctured line shown below.



          graph_of_punctured_line



          Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain
          $$f(x) = frac{x^2 - 1}{x - 1} = frac{(x + 1)(x - 1)}{x - 1} = x + 1$$
          for each $x in text{Dom}_f = mathbb{R} - {1}$.



          The function $g: mathbb{R} to mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.



          graph_of_line



          While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            The functions are different because they have different domains.



            A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.



            The implied domain of the function defined by
            $$f(x) = frac{x^2 - 1}{x - 1}$$
            is the largest set of real numbers that do not make the denominator equal to zero, which is
            $${x in mathbb{R} mid x neq 1} = mathbb{R} - {1}$$

            Thus, we should write that $f$ is the function $f: mathbb{R} - {1} to mathbb{R}$ defined by
            $$f(x) = frac{x^2 - 1}{x - 1}$$
            Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since
            $$lim_{x to 1} f(x) = lim_{x to 1} = frac{x^2 - 1}{x + 1} = lim_{x to 1} (x + 1) = 1 + 1 = 2$$



            Its graph is the punctured line shown below.



            graph_of_punctured_line



            Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain
            $$f(x) = frac{x^2 - 1}{x - 1} = frac{(x + 1)(x - 1)}{x - 1} = x + 1$$
            for each $x in text{Dom}_f = mathbb{R} - {1}$.



            The function $g: mathbb{R} to mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.



            graph_of_line



            While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.






            share|cite|improve this answer









            $endgroup$



            The functions are different because they have different domains.



            A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.



            The implied domain of the function defined by
            $$f(x) = frac{x^2 - 1}{x - 1}$$
            is the largest set of real numbers that do not make the denominator equal to zero, which is
            $${x in mathbb{R} mid x neq 1} = mathbb{R} - {1}$$

            Thus, we should write that $f$ is the function $f: mathbb{R} - {1} to mathbb{R}$ defined by
            $$f(x) = frac{x^2 - 1}{x - 1}$$
            Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since
            $$lim_{x to 1} f(x) = lim_{x to 1} = frac{x^2 - 1}{x + 1} = lim_{x to 1} (x + 1) = 1 + 1 = 2$$



            Its graph is the punctured line shown below.



            graph_of_punctured_line



            Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain
            $$f(x) = frac{x^2 - 1}{x - 1} = frac{(x + 1)(x - 1)}{x - 1} = x + 1$$
            for each $x in text{Dom}_f = mathbb{R} - {1}$.



            The function $g: mathbb{R} to mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.



            graph_of_line



            While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 17 '18 at 14:29









            N. F. TaussigN. F. Taussig

            45k103358




            45k103358























                2












                $begingroup$

                We can prove that both of these functions have the same graph for domain $D=mathbb{R}-{1}$:



                let $a in D$ be any number from domain. Then we have:



                $f(a) = a + 1 = a + 1 cdot frac{a-1}{a-1} = frac{a^{2}-1}{a-1} = f(a)$



                So the graph is the same for $D=mathbb{R}-{1}$: but in general these two functions are different because of Domain.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  There are no two functions given. To speak about a function we have to know the image and domain.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 13:55










                • $begingroup$
                  @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 13:58










                • $begingroup$
                  In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 14:03










                • $begingroup$
                  @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:05










                • $begingroup$
                  But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:06
















                2












                $begingroup$

                We can prove that both of these functions have the same graph for domain $D=mathbb{R}-{1}$:



                let $a in D$ be any number from domain. Then we have:



                $f(a) = a + 1 = a + 1 cdot frac{a-1}{a-1} = frac{a^{2}-1}{a-1} = f(a)$



                So the graph is the same for $D=mathbb{R}-{1}$: but in general these two functions are different because of Domain.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  There are no two functions given. To speak about a function we have to know the image and domain.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 13:55










                • $begingroup$
                  @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 13:58










                • $begingroup$
                  In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 14:03










                • $begingroup$
                  @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:05










                • $begingroup$
                  But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:06














                2












                2








                2





                $begingroup$

                We can prove that both of these functions have the same graph for domain $D=mathbb{R}-{1}$:



                let $a in D$ be any number from domain. Then we have:



                $f(a) = a + 1 = a + 1 cdot frac{a-1}{a-1} = frac{a^{2}-1}{a-1} = f(a)$



                So the graph is the same for $D=mathbb{R}-{1}$: but in general these two functions are different because of Domain.






                share|cite|improve this answer









                $endgroup$



                We can prove that both of these functions have the same graph for domain $D=mathbb{R}-{1}$:



                let $a in D$ be any number from domain. Then we have:



                $f(a) = a + 1 = a + 1 cdot frac{a-1}{a-1} = frac{a^{2}-1}{a-1} = f(a)$



                So the graph is the same for $D=mathbb{R}-{1}$: but in general these two functions are different because of Domain.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 17 '18 at 13:52









                Piotr WasilewiczPiotr Wasilewicz

                25819




                25819












                • $begingroup$
                  There are no two functions given. To speak about a function we have to know the image and domain.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 13:55










                • $begingroup$
                  @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 13:58










                • $begingroup$
                  In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 14:03










                • $begingroup$
                  @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:05










                • $begingroup$
                  But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:06


















                • $begingroup$
                  There are no two functions given. To speak about a function we have to know the image and domain.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 13:55










                • $begingroup$
                  @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 13:58










                • $begingroup$
                  In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
                  $endgroup$
                  – Cornman
                  Sep 17 '18 at 14:03










                • $begingroup$
                  @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:05










                • $begingroup$
                  But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
                  $endgroup$
                  – Piotr Wasilewicz
                  Sep 17 '18 at 14:06
















                $begingroup$
                There are no two functions given. To speak about a function we have to know the image and domain.
                $endgroup$
                – Cornman
                Sep 17 '18 at 13:55




                $begingroup$
                There are no two functions given. To speak about a function we have to know the image and domain.
                $endgroup$
                – Cornman
                Sep 17 '18 at 13:55












                $begingroup$
                @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
                $endgroup$
                – Piotr Wasilewicz
                Sep 17 '18 at 13:58




                $begingroup$
                @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right?
                $endgroup$
                – Piotr Wasilewicz
                Sep 17 '18 at 13:58












                $begingroup$
                In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
                $endgroup$
                – Cornman
                Sep 17 '18 at 14:03




                $begingroup$
                In my opinion the widest domain which fits $f(x)=frac{x^2-1}{x-1}$ is indeed $mathbb{R}$, since the pole can be "fixed". It is $lim_{xto 1} f(x)=2$.
                $endgroup$
                – Cornman
                Sep 17 '18 at 14:03












                $begingroup$
                @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
                $endgroup$
                – Piotr Wasilewicz
                Sep 17 '18 at 14:05




                $begingroup$
                @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$
                $endgroup$
                – Piotr Wasilewicz
                Sep 17 '18 at 14:05












                $begingroup$
                But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
                $endgroup$
                – Piotr Wasilewicz
                Sep 17 '18 at 14:06




                $begingroup$
                But you can put $f(x) = frac{x^{2}-1}{x-1}$ for $x neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different.
                $endgroup$
                – Piotr Wasilewicz
                Sep 17 '18 at 14:06











                1












                $begingroup$

                The problem is, that $f(x)=frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:



                $f:mathbb{R}setminus{1}tomathbb{R}$.



                One might consider $mathbb{R}setminus{1}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero).
                So this is a pole.



                Or is it?



                It is completly fine to define $f:mathbb{R}tomathbb{R}$, since the pole is "liftable".
                A fake pole, if you please.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  The problem is, that $f(x)=frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:



                  $f:mathbb{R}setminus{1}tomathbb{R}$.



                  One might consider $mathbb{R}setminus{1}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero).
                  So this is a pole.



                  Or is it?



                  It is completly fine to define $f:mathbb{R}tomathbb{R}$, since the pole is "liftable".
                  A fake pole, if you please.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The problem is, that $f(x)=frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:



                    $f:mathbb{R}setminus{1}tomathbb{R}$.



                    One might consider $mathbb{R}setminus{1}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero).
                    So this is a pole.



                    Or is it?



                    It is completly fine to define $f:mathbb{R}tomathbb{R}$, since the pole is "liftable".
                    A fake pole, if you please.






                    share|cite|improve this answer









                    $endgroup$



                    The problem is, that $f(x)=frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:



                    $f:mathbb{R}setminus{1}tomathbb{R}$.



                    One might consider $mathbb{R}setminus{1}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero).
                    So this is a pole.



                    Or is it?



                    It is completly fine to define $f:mathbb{R}tomathbb{R}$, since the pole is "liftable".
                    A fake pole, if you please.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 17 '18 at 13:47









                    CornmanCornman

                    3,38421229




                    3,38421229























                        0












                        $begingroup$

                        The only difference is that $frac{(x^2-1)}{(x-1)} $ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The only difference is that $frac{(x^2-1)}{(x-1)} $ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The only difference is that $frac{(x^2-1)}{(x-1)} $ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same.






                            share|cite|improve this answer









                            $endgroup$



                            The only difference is that $frac{(x^2-1)}{(x-1)} $ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Sep 17 '18 at 13:54









                            Anirudh RoyAnirudh Roy

                            111




                            111






























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