Function composition - getting different outputs for the same input.












2












$begingroup$


Given:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



If you evaluate $g(f(1))$ by doing $f(1)$ first and inputting its value into $g(1)$, you get:



$$ f(1) = 3(1) - 1 = 2 $$



$$ g(f(1)) = 2^3+2 = 10 $$



But if you try to substitute $x$ with $f(x)$ in $g(x)$, i.e:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



$$ g(f(x)) = (3x-1)^3+2 $$



If you evaluate the last expression, you get:



$$ g(f(x)) = 27x^3-1+2 $$



i.e



$$ g(f(x)) = 27x^3+1 $$



$g(f(1))$ now gives a different value (28) instead of 10.



Did I make a mistake with the evaluation of the parenthesis or is there a rule here that the parenthesis shouldn't be evaluated in certain circumstances?



Thanks.






functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your error is in stating $(3x-1)^3 = 27x^3 - 1$. This is not true. Are you familiar with the binomial formula?
    $endgroup$
    – platty
    Dec 11 '18 at 23:23










  • $begingroup$
    $(3x-1)^3 neq 27x^3-1$. Indeed this is obvious if we let $x=1$, then we have $ (3x-1)^3=8 neq 27-1=26$.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:23


















2












$begingroup$


Given:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



If you evaluate $g(f(1))$ by doing $f(1)$ first and inputting its value into $g(1)$, you get:



$$ f(1) = 3(1) - 1 = 2 $$



$$ g(f(1)) = 2^3+2 = 10 $$



But if you try to substitute $x$ with $f(x)$ in $g(x)$, i.e:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



$$ g(f(x)) = (3x-1)^3+2 $$



If you evaluate the last expression, you get:



$$ g(f(x)) = 27x^3-1+2 $$



i.e



$$ g(f(x)) = 27x^3+1 $$



$g(f(1))$ now gives a different value (28) instead of 10.



Did I make a mistake with the evaluation of the parenthesis or is there a rule here that the parenthesis shouldn't be evaluated in certain circumstances?



Thanks.






functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your error is in stating $(3x-1)^3 = 27x^3 - 1$. This is not true. Are you familiar with the binomial formula?
    $endgroup$
    – platty
    Dec 11 '18 at 23:23










  • $begingroup$
    $(3x-1)^3 neq 27x^3-1$. Indeed this is obvious if we let $x=1$, then we have $ (3x-1)^3=8 neq 27-1=26$.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:23
















2












2








2





$begingroup$


Given:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



If you evaluate $g(f(1))$ by doing $f(1)$ first and inputting its value into $g(1)$, you get:



$$ f(1) = 3(1) - 1 = 2 $$



$$ g(f(1)) = 2^3+2 = 10 $$



But if you try to substitute $x$ with $f(x)$ in $g(x)$, i.e:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



$$ g(f(x)) = (3x-1)^3+2 $$



If you evaluate the last expression, you get:



$$ g(f(x)) = 27x^3-1+2 $$



i.e



$$ g(f(x)) = 27x^3+1 $$



$g(f(1))$ now gives a different value (28) instead of 10.



Did I make a mistake with the evaluation of the parenthesis or is there a rule here that the parenthesis shouldn't be evaluated in certain circumstances?



Thanks.






functions







share|cite|improve this question











$endgroup$




Given:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



If you evaluate $g(f(1))$ by doing $f(1)$ first and inputting its value into $g(1)$, you get:



$$ f(1) = 3(1) - 1 = 2 $$



$$ g(f(1)) = 2^3+2 = 10 $$



But if you try to substitute $x$ with $f(x)$ in $g(x)$, i.e:



$$ f(x) = 3x-1 $$



$$ g(x) = x^3+2 $$



$$ g(f(x)) = (3x-1)^3+2 $$



If you evaluate the last expression, you get:



$$ g(f(x)) = 27x^3-1+2 $$



i.e



$$ g(f(x)) = 27x^3+1 $$



$g(f(1))$ now gives a different value (28) instead of 10.



Did I make a mistake with the evaluation of the parenthesis or is there a rule here that the parenthesis shouldn't be evaluated in certain circumstances?



Thanks.






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 23:21









Dando18

4,67741235




4,67741235










asked Dec 11 '18 at 23:19









Click UpvoteClick Upvote

1257




1257








  • 1




    $begingroup$
    Your error is in stating $(3x-1)^3 = 27x^3 - 1$. This is not true. Are you familiar with the binomial formula?
    $endgroup$
    – platty
    Dec 11 '18 at 23:23










  • $begingroup$
    $(3x-1)^3 neq 27x^3-1$. Indeed this is obvious if we let $x=1$, then we have $ (3x-1)^3=8 neq 27-1=26$.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:23
















  • 1




    $begingroup$
    Your error is in stating $(3x-1)^3 = 27x^3 - 1$. This is not true. Are you familiar with the binomial formula?
    $endgroup$
    – platty
    Dec 11 '18 at 23:23










  • $begingroup$
    $(3x-1)^3 neq 27x^3-1$. Indeed this is obvious if we let $x=1$, then we have $ (3x-1)^3=8 neq 27-1=26$.
    $endgroup$
    – Mason
    Dec 11 '18 at 23:23










1




1




$begingroup$
Your error is in stating $(3x-1)^3 = 27x^3 - 1$. This is not true. Are you familiar with the binomial formula?
$endgroup$
– platty
Dec 11 '18 at 23:23




$begingroup$
Your error is in stating $(3x-1)^3 = 27x^3 - 1$. This is not true. Are you familiar with the binomial formula?
$endgroup$
– platty
Dec 11 '18 at 23:23












$begingroup$
$(3x-1)^3 neq 27x^3-1$. Indeed this is obvious if we let $x=1$, then we have $ (3x-1)^3=8 neq 27-1=26$.
$endgroup$
– Mason
Dec 11 '18 at 23:23






$begingroup$
$(3x-1)^3 neq 27x^3-1$. Indeed this is obvious if we let $x=1$, then we have $ (3x-1)^3=8 neq 27-1=26$.
$endgroup$
– Mason
Dec 11 '18 at 23:23












2 Answers
2






active

oldest

votes


















5












$begingroup$

Hint: You made a mistake. It is false to write $a^3-b^3$ for $(a-b)^3$. You have to perform polynomial multiplication to get $a^3-3a^2b+3ab^2-b^3$ instead.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. Ugh, I keep making this mistake.
    $endgroup$
    – Click Upvote
    Dec 11 '18 at 23:33










  • $begingroup$
    What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
    $endgroup$
    – Click Upvote
    Dec 11 '18 at 23:38






  • 1




    $begingroup$
    No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
    $endgroup$
    – MPW
    Dec 11 '18 at 23:41



















1












$begingroup$

Well:



$$(3(1)-1)^3+2=2^3+2=10$$



You just havent expanded out $(3x-1)^3$ correctly.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Hint: You made a mistake. It is false to write $a^3-b^3$ for $(a-b)^3$. You have to perform polynomial multiplication to get $a^3-3a^2b+3ab^2-b^3$ instead.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. Ugh, I keep making this mistake.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:33










    • $begingroup$
      What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:38






    • 1




      $begingroup$
      No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
      $endgroup$
      – MPW
      Dec 11 '18 at 23:41
















    5












    $begingroup$

    Hint: You made a mistake. It is false to write $a^3-b^3$ for $(a-b)^3$. You have to perform polynomial multiplication to get $a^3-3a^2b+3ab^2-b^3$ instead.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. Ugh, I keep making this mistake.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:33










    • $begingroup$
      What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:38






    • 1




      $begingroup$
      No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
      $endgroup$
      – MPW
      Dec 11 '18 at 23:41














    5












    5








    5





    $begingroup$

    Hint: You made a mistake. It is false to write $a^3-b^3$ for $(a-b)^3$. You have to perform polynomial multiplication to get $a^3-3a^2b+3ab^2-b^3$ instead.






    share|cite|improve this answer









    $endgroup$



    Hint: You made a mistake. It is false to write $a^3-b^3$ for $(a-b)^3$. You have to perform polynomial multiplication to get $a^3-3a^2b+3ab^2-b^3$ instead.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 11 '18 at 23:24









    MPWMPW

    30k12056




    30k12056












    • $begingroup$
      Thanks. Ugh, I keep making this mistake.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:33










    • $begingroup$
      What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:38






    • 1




      $begingroup$
      No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
      $endgroup$
      – MPW
      Dec 11 '18 at 23:41


















    • $begingroup$
      Thanks. Ugh, I keep making this mistake.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:33










    • $begingroup$
      What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
      $endgroup$
      – Click Upvote
      Dec 11 '18 at 23:38






    • 1




      $begingroup$
      No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
      $endgroup$
      – MPW
      Dec 11 '18 at 23:41
















    $begingroup$
    Thanks. Ugh, I keep making this mistake.
    $endgroup$
    – Click Upvote
    Dec 11 '18 at 23:33




    $begingroup$
    Thanks. Ugh, I keep making this mistake.
    $endgroup$
    – Click Upvote
    Dec 11 '18 at 23:33












    $begingroup$
    What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
    $endgroup$
    – Click Upvote
    Dec 11 '18 at 23:38




    $begingroup$
    What makes this confusing is if you have $(3-1)^3$ it would indeed evaluate to 27 - 1, but if you have $(3x-1)^3$ that needs to be expanded out with the polynomial multiplication.
    $endgroup$
    – Click Upvote
    Dec 11 '18 at 23:38




    1




    1




    $begingroup$
    No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
    $endgroup$
    – MPW
    Dec 11 '18 at 23:41




    $begingroup$
    No, that’s wrong for the same reason. $(3-1)^3=(2)^3=8$
    $endgroup$
    – MPW
    Dec 11 '18 at 23:41











    1












    $begingroup$

    Well:



    $$(3(1)-1)^3+2=2^3+2=10$$



    You just havent expanded out $(3x-1)^3$ correctly.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Well:



      $$(3(1)-1)^3+2=2^3+2=10$$



      You just havent expanded out $(3x-1)^3$ correctly.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Well:



        $$(3(1)-1)^3+2=2^3+2=10$$



        You just havent expanded out $(3x-1)^3$ correctly.






        share|cite|improve this answer









        $endgroup$



        Well:



        $$(3(1)-1)^3+2=2^3+2=10$$



        You just havent expanded out $(3x-1)^3$ correctly.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 23:30









        Rhys HughesRhys Hughes

        6,3311529




        6,3311529






























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