How is $frac {x^2 + x - 6}{x - 2}$ different from $(x+3)$? [duplicate]











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  • Why does factoring eliminate a hole in the limit?

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Let's consider the function:



$$ f(x) = frac {x^2 + x - 6}{x - 2} $$



At x = 2, the value fo the function is undefined, because $ x - 2 $ = 0. But if factor the expression on the numerator, we get $ frac {(x + 3) (x-2)}{x-2} $, which on simplifying yields $ x + 3 $. At $ x = 2 $, the function gives 5, which is not what we get before simplifying.



If all we do too the initial function is simple algebraic manipulation, shouldn't both functions yield the same value?










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marked as duplicate by Martin R, Hans Lundmark, Community Nov 29 at 14:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











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    Related: Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same?
    – StackTD
    Nov 29 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 at 12:48















up vote
1
down vote

favorite













This question already has an answer here:




  • Why does factoring eliminate a hole in the limit?

    13 answers



  • Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same? [duplicate]

    3 answers




Let's consider the function:



$$ f(x) = frac {x^2 + x - 6}{x - 2} $$



At x = 2, the value fo the function is undefined, because $ x - 2 $ = 0. But if factor the expression on the numerator, we get $ frac {(x + 3) (x-2)}{x-2} $, which on simplifying yields $ x + 3 $. At $ x = 2 $, the function gives 5, which is not what we get before simplifying.



If all we do too the initial function is simple algebraic manipulation, shouldn't both functions yield the same value?










share|cite|improve this question















marked as duplicate by Martin R, Hans Lundmark, Community Nov 29 at 14:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 2




    Related: Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same?
    – StackTD
    Nov 29 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 at 12:48













up vote
1
down vote

favorite









up vote
1
down vote

favorite












This question already has an answer here:




  • Why does factoring eliminate a hole in the limit?

    13 answers



  • Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same? [duplicate]

    3 answers




Let's consider the function:



$$ f(x) = frac {x^2 + x - 6}{x - 2} $$



At x = 2, the value fo the function is undefined, because $ x - 2 $ = 0. But if factor the expression on the numerator, we get $ frac {(x + 3) (x-2)}{x-2} $, which on simplifying yields $ x + 3 $. At $ x = 2 $, the function gives 5, which is not what we get before simplifying.



If all we do too the initial function is simple algebraic manipulation, shouldn't both functions yield the same value?










share|cite|improve this question
















This question already has an answer here:




  • Why does factoring eliminate a hole in the limit?

    13 answers



  • Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same? [duplicate]

    3 answers




Let's consider the function:



$$ f(x) = frac {x^2 + x - 6}{x - 2} $$



At x = 2, the value fo the function is undefined, because $ x - 2 $ = 0. But if factor the expression on the numerator, we get $ frac {(x + 3) (x-2)}{x-2} $, which on simplifying yields $ x + 3 $. At $ x = 2 $, the function gives 5, which is not what we get before simplifying.



If all we do too the initial function is simple algebraic manipulation, shouldn't both functions yield the same value?





This question already has an answer here:




  • Why does factoring eliminate a hole in the limit?

    13 answers



  • Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same? [duplicate]

    3 answers








algebra-precalculus functions arithmetic






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edited Nov 29 at 13:48









user21820

38.1k541150




38.1k541150










asked Nov 29 at 12:26









WorldGov

2029




2029




marked as duplicate by Martin R, Hans Lundmark, Community Nov 29 at 14:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Martin R, Hans Lundmark, Community Nov 29 at 14:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    Related: Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same?
    – StackTD
    Nov 29 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 at 12:48














  • 2




    Related: Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same?
    – StackTD
    Nov 29 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 at 12:48








2




2




Related: Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same?
– StackTD
Nov 29 at 12:47






Related: Why aren't the functions $f(x) = frac{x-1}{x-1}$ and $f(x) = 1$ the same?
– StackTD
Nov 29 at 12:47














See also: Why can we resolve indeterminate forms?
– StackTD
Nov 29 at 12:48




See also: Why can we resolve indeterminate forms?
– StackTD
Nov 29 at 12:48










5 Answers
5






active

oldest

votes

















up vote
8
down vote



accepted










The simplification is only valid for $xneq 2$, so you are not allowed to put $x=2$ into the resulting simplified expression. $f(2)$ is still not defined.



In other words, it is not true that “$f(x)=x+3$”. What is true is that “$f(x)=x+3$ provided that $xneq 2$”.



It is critical to retain any provisos you implicitly introduce when simplifying. To fail to do this is sloppy and result in erroneous conclusions.






share|cite|improve this answer




























    up vote
    5
    down vote













    They do agree. But remember that $frac{x^2+x-6}{x-2}$ is not defined at $x = 2$, so technically neither is the $x+3$ you get from simplifying it. Check any other value for $x$, and you will see that they agree.






    share|cite|improve this answer




























      up vote
      5
      down vote













      Note, that when a function is defined, it's not only about a formula but also about its domain.



      For the function given



      $$f(x) = frac {x^2 + x - 6}{x - 2}$$



      if it is to be expressed in that way, it must clearly be $xneq 2$ and thus its domain is $D_f = mathbb R setminus {2}$. This means, that even after the simplification, the value $x=2$ does not belong to the domain of the given function and thus cannot be substituted in.






      share|cite|improve this answer




























        up vote
        4
        down vote













        This is known as a removable discontinuity. If you check the graph, you would see that the two functions produce the same output at every point except $x = 2$ where the original function is undefined. By simplifying, you removed the discontinuity at that point.



        Take this as another example:



        $$f_1(x) = frac{x(x+1)}{x+1} quad f_2(x) = x$$



        At first, it might seem the two functions are equivalent. Yet, you have to notice $f_2(x)$ is defined for all $x in mathbb{R}$ whereas $f_1(x)$ is defined for $x neq -1$, so the two functions aren't the "same" as their domains differ. They are equivalent at every other point except $x = -1$, where the first function is undefined. (You can think of it as a gap in $f_1(x)$. Otherwise, the graphs look exactly the same.) The exact same idea applies to your example.






        share|cite|improve this answer






























          up vote
          2
          down vote













          Surely not because you are manipulating the definition of continuity. For example$$f(x)=begin{cases}1quad,quad xne 0end{cases}\g(x)=1$$ are clearly different. One is continuous on $Bbb R$ and one is not but we have $$f(x)={xover x}ne 1=g(x)$$You can see the difference in the illustration of the both function. $f(x)$ contains a gap in $x=0$ while $g(x)$ doesn't.






          share|cite|improve this answer




























            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            8
            down vote



            accepted










            The simplification is only valid for $xneq 2$, so you are not allowed to put $x=2$ into the resulting simplified expression. $f(2)$ is still not defined.



            In other words, it is not true that “$f(x)=x+3$”. What is true is that “$f(x)=x+3$ provided that $xneq 2$”.



            It is critical to retain any provisos you implicitly introduce when simplifying. To fail to do this is sloppy and result in erroneous conclusions.






            share|cite|improve this answer

























              up vote
              8
              down vote



              accepted










              The simplification is only valid for $xneq 2$, so you are not allowed to put $x=2$ into the resulting simplified expression. $f(2)$ is still not defined.



              In other words, it is not true that “$f(x)=x+3$”. What is true is that “$f(x)=x+3$ provided that $xneq 2$”.



              It is critical to retain any provisos you implicitly introduce when simplifying. To fail to do this is sloppy and result in erroneous conclusions.






              share|cite|improve this answer























                up vote
                8
                down vote



                accepted







                up vote
                8
                down vote



                accepted






                The simplification is only valid for $xneq 2$, so you are not allowed to put $x=2$ into the resulting simplified expression. $f(2)$ is still not defined.



                In other words, it is not true that “$f(x)=x+3$”. What is true is that “$f(x)=x+3$ provided that $xneq 2$”.



                It is critical to retain any provisos you implicitly introduce when simplifying. To fail to do this is sloppy and result in erroneous conclusions.






                share|cite|improve this answer












                The simplification is only valid for $xneq 2$, so you are not allowed to put $x=2$ into the resulting simplified expression. $f(2)$ is still not defined.



                In other words, it is not true that “$f(x)=x+3$”. What is true is that “$f(x)=x+3$ provided that $xneq 2$”.



                It is critical to retain any provisos you implicitly introduce when simplifying. To fail to do this is sloppy and result in erroneous conclusions.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 at 12:29









                MPW

                29.7k11956




                29.7k11956






















                    up vote
                    5
                    down vote













                    They do agree. But remember that $frac{x^2+x-6}{x-2}$ is not defined at $x = 2$, so technically neither is the $x+3$ you get from simplifying it. Check any other value for $x$, and you will see that they agree.






                    share|cite|improve this answer

























                      up vote
                      5
                      down vote













                      They do agree. But remember that $frac{x^2+x-6}{x-2}$ is not defined at $x = 2$, so technically neither is the $x+3$ you get from simplifying it. Check any other value for $x$, and you will see that they agree.






                      share|cite|improve this answer























                        up vote
                        5
                        down vote










                        up vote
                        5
                        down vote









                        They do agree. But remember that $frac{x^2+x-6}{x-2}$ is not defined at $x = 2$, so technically neither is the $x+3$ you get from simplifying it. Check any other value for $x$, and you will see that they agree.






                        share|cite|improve this answer












                        They do agree. But remember that $frac{x^2+x-6}{x-2}$ is not defined at $x = 2$, so technically neither is the $x+3$ you get from simplifying it. Check any other value for $x$, and you will see that they agree.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 29 at 12:30









                        Arthur

                        109k7103186




                        109k7103186






















                            up vote
                            5
                            down vote













                            Note, that when a function is defined, it's not only about a formula but also about its domain.



                            For the function given



                            $$f(x) = frac {x^2 + x - 6}{x - 2}$$



                            if it is to be expressed in that way, it must clearly be $xneq 2$ and thus its domain is $D_f = mathbb R setminus {2}$. This means, that even after the simplification, the value $x=2$ does not belong to the domain of the given function and thus cannot be substituted in.






                            share|cite|improve this answer

























                              up vote
                              5
                              down vote













                              Note, that when a function is defined, it's not only about a formula but also about its domain.



                              For the function given



                              $$f(x) = frac {x^2 + x - 6}{x - 2}$$



                              if it is to be expressed in that way, it must clearly be $xneq 2$ and thus its domain is $D_f = mathbb R setminus {2}$. This means, that even after the simplification, the value $x=2$ does not belong to the domain of the given function and thus cannot be substituted in.






                              share|cite|improve this answer























                                up vote
                                5
                                down vote










                                up vote
                                5
                                down vote









                                Note, that when a function is defined, it's not only about a formula but also about its domain.



                                For the function given



                                $$f(x) = frac {x^2 + x - 6}{x - 2}$$



                                if it is to be expressed in that way, it must clearly be $xneq 2$ and thus its domain is $D_f = mathbb R setminus {2}$. This means, that even after the simplification, the value $x=2$ does not belong to the domain of the given function and thus cannot be substituted in.






                                share|cite|improve this answer












                                Note, that when a function is defined, it's not only about a formula but also about its domain.



                                For the function given



                                $$f(x) = frac {x^2 + x - 6}{x - 2}$$



                                if it is to be expressed in that way, it must clearly be $xneq 2$ and thus its domain is $D_f = mathbb R setminus {2}$. This means, that even after the simplification, the value $x=2$ does not belong to the domain of the given function and thus cannot be substituted in.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 29 at 12:30









                                Rebellos

                                12.9k21042




                                12.9k21042






















                                    up vote
                                    4
                                    down vote













                                    This is known as a removable discontinuity. If you check the graph, you would see that the two functions produce the same output at every point except $x = 2$ where the original function is undefined. By simplifying, you removed the discontinuity at that point.



                                    Take this as another example:



                                    $$f_1(x) = frac{x(x+1)}{x+1} quad f_2(x) = x$$



                                    At first, it might seem the two functions are equivalent. Yet, you have to notice $f_2(x)$ is defined for all $x in mathbb{R}$ whereas $f_1(x)$ is defined for $x neq -1$, so the two functions aren't the "same" as their domains differ. They are equivalent at every other point except $x = -1$, where the first function is undefined. (You can think of it as a gap in $f_1(x)$. Otherwise, the graphs look exactly the same.) The exact same idea applies to your example.






                                    share|cite|improve this answer



























                                      up vote
                                      4
                                      down vote













                                      This is known as a removable discontinuity. If you check the graph, you would see that the two functions produce the same output at every point except $x = 2$ where the original function is undefined. By simplifying, you removed the discontinuity at that point.



                                      Take this as another example:



                                      $$f_1(x) = frac{x(x+1)}{x+1} quad f_2(x) = x$$



                                      At first, it might seem the two functions are equivalent. Yet, you have to notice $f_2(x)$ is defined for all $x in mathbb{R}$ whereas $f_1(x)$ is defined for $x neq -1$, so the two functions aren't the "same" as their domains differ. They are equivalent at every other point except $x = -1$, where the first function is undefined. (You can think of it as a gap in $f_1(x)$. Otherwise, the graphs look exactly the same.) The exact same idea applies to your example.






                                      share|cite|improve this answer

























                                        up vote
                                        4
                                        down vote










                                        up vote
                                        4
                                        down vote









                                        This is known as a removable discontinuity. If you check the graph, you would see that the two functions produce the same output at every point except $x = 2$ where the original function is undefined. By simplifying, you removed the discontinuity at that point.



                                        Take this as another example:



                                        $$f_1(x) = frac{x(x+1)}{x+1} quad f_2(x) = x$$



                                        At first, it might seem the two functions are equivalent. Yet, you have to notice $f_2(x)$ is defined for all $x in mathbb{R}$ whereas $f_1(x)$ is defined for $x neq -1$, so the two functions aren't the "same" as their domains differ. They are equivalent at every other point except $x = -1$, where the first function is undefined. (You can think of it as a gap in $f_1(x)$. Otherwise, the graphs look exactly the same.) The exact same idea applies to your example.






                                        share|cite|improve this answer














                                        This is known as a removable discontinuity. If you check the graph, you would see that the two functions produce the same output at every point except $x = 2$ where the original function is undefined. By simplifying, you removed the discontinuity at that point.



                                        Take this as another example:



                                        $$f_1(x) = frac{x(x+1)}{x+1} quad f_2(x) = x$$



                                        At first, it might seem the two functions are equivalent. Yet, you have to notice $f_2(x)$ is defined for all $x in mathbb{R}$ whereas $f_1(x)$ is defined for $x neq -1$, so the two functions aren't the "same" as their domains differ. They are equivalent at every other point except $x = -1$, where the first function is undefined. (You can think of it as a gap in $f_1(x)$. Otherwise, the graphs look exactly the same.) The exact same idea applies to your example.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Nov 29 at 12:38

























                                        answered Nov 29 at 12:33









                                        KM101

                                        3,288417




                                        3,288417






















                                            up vote
                                            2
                                            down vote













                                            Surely not because you are manipulating the definition of continuity. For example$$f(x)=begin{cases}1quad,quad xne 0end{cases}\g(x)=1$$ are clearly different. One is continuous on $Bbb R$ and one is not but we have $$f(x)={xover x}ne 1=g(x)$$You can see the difference in the illustration of the both function. $f(x)$ contains a gap in $x=0$ while $g(x)$ doesn't.






                                            share|cite|improve this answer

























                                              up vote
                                              2
                                              down vote













                                              Surely not because you are manipulating the definition of continuity. For example$$f(x)=begin{cases}1quad,quad xne 0end{cases}\g(x)=1$$ are clearly different. One is continuous on $Bbb R$ and one is not but we have $$f(x)={xover x}ne 1=g(x)$$You can see the difference in the illustration of the both function. $f(x)$ contains a gap in $x=0$ while $g(x)$ doesn't.






                                              share|cite|improve this answer























                                                up vote
                                                2
                                                down vote










                                                up vote
                                                2
                                                down vote









                                                Surely not because you are manipulating the definition of continuity. For example$$f(x)=begin{cases}1quad,quad xne 0end{cases}\g(x)=1$$ are clearly different. One is continuous on $Bbb R$ and one is not but we have $$f(x)={xover x}ne 1=g(x)$$You can see the difference in the illustration of the both function. $f(x)$ contains a gap in $x=0$ while $g(x)$ doesn't.






                                                share|cite|improve this answer












                                                Surely not because you are manipulating the definition of continuity. For example$$f(x)=begin{cases}1quad,quad xne 0end{cases}\g(x)=1$$ are clearly different. One is continuous on $Bbb R$ and one is not but we have $$f(x)={xover x}ne 1=g(x)$$You can see the difference in the illustration of the both function. $f(x)$ contains a gap in $x=0$ while $g(x)$ doesn't.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Nov 29 at 12:32









                                                Mostafa Ayaz

                                                13.3k3836




                                                13.3k3836















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