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












1















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  • 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?










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marked as duplicate by Martin R, Hans Lundmark, Community Nov 29 '18 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 '18 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 '18 at 12:48
















1















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 '18 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 '18 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 '18 at 12:48














1












1








1








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 '18 at 13:48









user21820

38.8k543153




38.8k543153










asked Nov 29 '18 at 12:26









WorldGovWorldGov

2629




2629




marked as duplicate by Martin R, Hans Lundmark, Community Nov 29 '18 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 '18 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 '18 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 '18 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 '18 at 12:47












  • See also: Why can we resolve indeterminate forms?
    – StackTD
    Nov 29 '18 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 '18 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 '18 at 12:47














See also: Why can we resolve indeterminate forms?
– StackTD
Nov 29 '18 at 12:48




See also: Why can we resolve indeterminate forms?
– StackTD
Nov 29 '18 at 12:48










5 Answers
5






active

oldest

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8














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





























    5














    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





























      5














      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





























        4














        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































          2














          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









            8














            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


























              8














              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
























                8












                8








                8






                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 '18 at 12:29









                MPWMPW

                29.8k12056




                29.8k12056























                    5














                    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


























                      5














                      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
























                        5












                        5








                        5






                        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 '18 at 12:30









                        ArthurArthur

                        111k7107189




                        111k7107189























                            5














                            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


























                              5














                              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
























                                5












                                5








                                5






                                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 '18 at 12:30









                                RebellosRebellos

                                14.5k31246




                                14.5k31246























                                    4














                                    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




























                                      4














                                      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


























                                        4












                                        4








                                        4






                                        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 '18 at 12:38

























                                        answered Nov 29 '18 at 12:33









                                        KM101KM101

                                        5,7411423




                                        5,7411423























                                            2














                                            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


























                                              2














                                              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
























                                                2












                                                2








                                                2






                                                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 '18 at 12:32









                                                Mostafa AyazMostafa Ayaz

                                                14.5k3937




                                                14.5k3937















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