What's the difference between $f(x)=sqrt{x^2+9}$ and $k(x^2+9)=sqrt{x^2+9}$?












5












$begingroup$


Let's say we 've got a function $f(x)=sqrt{x^2+9}$, which is a composite function. $f(x)=sqrt{g(x)}$ and $g(x)=x^2+9$.



When we have a function like $h(x)=x$, we are allowed to set $x$ to $x+9$ and have $h(x+9)=x+9$.



So why do we need $g(x)$ and can't just set $x=x^2+9$, which with a function like $k(x)=sqrt{x}$ leads to $k(x^2+9)=sqrt{x^2+9}$ (same as f(x) above)?



Where's the difference between these two ($f(x),k(x)$)? Is $x^2+9$ even a valid argument for $k(x)$?






functions







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$endgroup$












  • $begingroup$
    Function composition is not unique. Another could be $f(x)=sqrt{x+9}, g(x)=x^2$.
    $endgroup$
    – Eleven-Eleven
    Nov 30 '18 at 19:09
















5












$begingroup$


Let's say we 've got a function $f(x)=sqrt{x^2+9}$, which is a composite function. $f(x)=sqrt{g(x)}$ and $g(x)=x^2+9$.



When we have a function like $h(x)=x$, we are allowed to set $x$ to $x+9$ and have $h(x+9)=x+9$.



So why do we need $g(x)$ and can't just set $x=x^2+9$, which with a function like $k(x)=sqrt{x}$ leads to $k(x^2+9)=sqrt{x^2+9}$ (same as f(x) above)?



Where's the difference between these two ($f(x),k(x)$)? Is $x^2+9$ even a valid argument for $k(x)$?






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Function composition is not unique. Another could be $f(x)=sqrt{x+9}, g(x)=x^2$.
    $endgroup$
    – Eleven-Eleven
    Nov 30 '18 at 19:09














5












5








5





$begingroup$


Let's say we 've got a function $f(x)=sqrt{x^2+9}$, which is a composite function. $f(x)=sqrt{g(x)}$ and $g(x)=x^2+9$.



When we have a function like $h(x)=x$, we are allowed to set $x$ to $x+9$ and have $h(x+9)=x+9$.



So why do we need $g(x)$ and can't just set $x=x^2+9$, which with a function like $k(x)=sqrt{x}$ leads to $k(x^2+9)=sqrt{x^2+9}$ (same as f(x) above)?



Where's the difference between these two ($f(x),k(x)$)? Is $x^2+9$ even a valid argument for $k(x)$?






functions







share|cite|improve this question











$endgroup$




Let's say we 've got a function $f(x)=sqrt{x^2+9}$, which is a composite function. $f(x)=sqrt{g(x)}$ and $g(x)=x^2+9$.



When we have a function like $h(x)=x$, we are allowed to set $x$ to $x+9$ and have $h(x+9)=x+9$.



So why do we need $g(x)$ and can't just set $x=x^2+9$, which with a function like $k(x)=sqrt{x}$ leads to $k(x^2+9)=sqrt{x^2+9}$ (same as f(x) above)?



Where's the difference between these two ($f(x),k(x)$)? Is $x^2+9$ even a valid argument for $k(x)$?






functions




functions






share|cite|improve this question















share|cite|improve this question













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edited Dec 1 '18 at 8:19







yot

















asked Nov 30 '18 at 18:57









yotyot

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285












  • $begingroup$
    Function composition is not unique. Another could be $f(x)=sqrt{x+9}, g(x)=x^2$.
    $endgroup$
    – Eleven-Eleven
    Nov 30 '18 at 19:09


















  • $begingroup$
    Function composition is not unique. Another could be $f(x)=sqrt{x+9}, g(x)=x^2$.
    $endgroup$
    – Eleven-Eleven
    Nov 30 '18 at 19:09
















$begingroup$
Function composition is not unique. Another could be $f(x)=sqrt{x+9}, g(x)=x^2$.
$endgroup$
– Eleven-Eleven
Nov 30 '18 at 19:09




$begingroup$
Function composition is not unique. Another could be $f(x)=sqrt{x+9}, g(x)=x^2$.
$endgroup$
– Eleven-Eleven
Nov 30 '18 at 19:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

A function is a mapping from elements of a domain set to elements of a range set (also called the "codomain"). Let's consider your example of $k(x) = sqrt{x}$. This actually is an incomplete definition of a function; you also need to specify the domain and range. So suppose $k$ takes nonnegative real numbers to nonnegative real numbers.



Then when we take the function $g(x) = x^2+9$, we again have to specify the domain and range. So let's say $g$ takes real numbers to real numbers greater than or equal to 9.



Now consider the composition $f(x) = k(g(x))$. The equation is $sqrt{x^2+9}$. But now the domain and range have changed a bit from the original $k$ or $g$. In particular, now the domain is all real numbers, and the range is real numbers greater than or equal to 3. So there is a subtle difference between the functions $f(x)$ and $k(x)$. It's important to keep the domain and range/codomain in mind whenever you do function composition.



Now to address your confusion regarding how we are "allowed" to set $x = x^2+9$ and write $k(x^2+9) = sqrt{x^2+9}$. Again, think about a function as taking inputs to outputs. So when you write $k(x^2+9) = sqrt{x^2+9}$, what you're saying is that given a number $x$, $k$ maps the number $x^2+9$ to $sqrt{x^2+9}$. This is really just a variable substitution. There is nothing wrong with writing down $k(x^2+9)$, or $k(e^x)$; just like with $k(2)$ or $k(pi)$, it represents passing some value into the function $k$.



Hopefully that addresses your questions, and let me know in the comments if I can clarify further!






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $f(y)$ is a function which maps $y$ to $sqrt{y^2+9}$, and you've fed $y=x$ into it to get your first function.



    $k(y)$ is a function which maps $y$ to $sqrt y$, and you've fed $y=x^2+9$ into it to get your second function.






    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









      1












      $begingroup$

      A function is a mapping from elements of a domain set to elements of a range set (also called the "codomain"). Let's consider your example of $k(x) = sqrt{x}$. This actually is an incomplete definition of a function; you also need to specify the domain and range. So suppose $k$ takes nonnegative real numbers to nonnegative real numbers.



      Then when we take the function $g(x) = x^2+9$, we again have to specify the domain and range. So let's say $g$ takes real numbers to real numbers greater than or equal to 9.



      Now consider the composition $f(x) = k(g(x))$. The equation is $sqrt{x^2+9}$. But now the domain and range have changed a bit from the original $k$ or $g$. In particular, now the domain is all real numbers, and the range is real numbers greater than or equal to 3. So there is a subtle difference between the functions $f(x)$ and $k(x)$. It's important to keep the domain and range/codomain in mind whenever you do function composition.



      Now to address your confusion regarding how we are "allowed" to set $x = x^2+9$ and write $k(x^2+9) = sqrt{x^2+9}$. Again, think about a function as taking inputs to outputs. So when you write $k(x^2+9) = sqrt{x^2+9}$, what you're saying is that given a number $x$, $k$ maps the number $x^2+9$ to $sqrt{x^2+9}$. This is really just a variable substitution. There is nothing wrong with writing down $k(x^2+9)$, or $k(e^x)$; just like with $k(2)$ or $k(pi)$, it represents passing some value into the function $k$.



      Hopefully that addresses your questions, and let me know in the comments if I can clarify further!






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        A function is a mapping from elements of a domain set to elements of a range set (also called the "codomain"). Let's consider your example of $k(x) = sqrt{x}$. This actually is an incomplete definition of a function; you also need to specify the domain and range. So suppose $k$ takes nonnegative real numbers to nonnegative real numbers.



        Then when we take the function $g(x) = x^2+9$, we again have to specify the domain and range. So let's say $g$ takes real numbers to real numbers greater than or equal to 9.



        Now consider the composition $f(x) = k(g(x))$. The equation is $sqrt{x^2+9}$. But now the domain and range have changed a bit from the original $k$ or $g$. In particular, now the domain is all real numbers, and the range is real numbers greater than or equal to 3. So there is a subtle difference between the functions $f(x)$ and $k(x)$. It's important to keep the domain and range/codomain in mind whenever you do function composition.



        Now to address your confusion regarding how we are "allowed" to set $x = x^2+9$ and write $k(x^2+9) = sqrt{x^2+9}$. Again, think about a function as taking inputs to outputs. So when you write $k(x^2+9) = sqrt{x^2+9}$, what you're saying is that given a number $x$, $k$ maps the number $x^2+9$ to $sqrt{x^2+9}$. This is really just a variable substitution. There is nothing wrong with writing down $k(x^2+9)$, or $k(e^x)$; just like with $k(2)$ or $k(pi)$, it represents passing some value into the function $k$.



        Hopefully that addresses your questions, and let me know in the comments if I can clarify further!






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          A function is a mapping from elements of a domain set to elements of a range set (also called the "codomain"). Let's consider your example of $k(x) = sqrt{x}$. This actually is an incomplete definition of a function; you also need to specify the domain and range. So suppose $k$ takes nonnegative real numbers to nonnegative real numbers.



          Then when we take the function $g(x) = x^2+9$, we again have to specify the domain and range. So let's say $g$ takes real numbers to real numbers greater than or equal to 9.



          Now consider the composition $f(x) = k(g(x))$. The equation is $sqrt{x^2+9}$. But now the domain and range have changed a bit from the original $k$ or $g$. In particular, now the domain is all real numbers, and the range is real numbers greater than or equal to 3. So there is a subtle difference between the functions $f(x)$ and $k(x)$. It's important to keep the domain and range/codomain in mind whenever you do function composition.



          Now to address your confusion regarding how we are "allowed" to set $x = x^2+9$ and write $k(x^2+9) = sqrt{x^2+9}$. Again, think about a function as taking inputs to outputs. So when you write $k(x^2+9) = sqrt{x^2+9}$, what you're saying is that given a number $x$, $k$ maps the number $x^2+9$ to $sqrt{x^2+9}$. This is really just a variable substitution. There is nothing wrong with writing down $k(x^2+9)$, or $k(e^x)$; just like with $k(2)$ or $k(pi)$, it represents passing some value into the function $k$.



          Hopefully that addresses your questions, and let me know in the comments if I can clarify further!






          share|cite|improve this answer











          $endgroup$



          A function is a mapping from elements of a domain set to elements of a range set (also called the "codomain"). Let's consider your example of $k(x) = sqrt{x}$. This actually is an incomplete definition of a function; you also need to specify the domain and range. So suppose $k$ takes nonnegative real numbers to nonnegative real numbers.



          Then when we take the function $g(x) = x^2+9$, we again have to specify the domain and range. So let's say $g$ takes real numbers to real numbers greater than or equal to 9.



          Now consider the composition $f(x) = k(g(x))$. The equation is $sqrt{x^2+9}$. But now the domain and range have changed a bit from the original $k$ or $g$. In particular, now the domain is all real numbers, and the range is real numbers greater than or equal to 3. So there is a subtle difference between the functions $f(x)$ and $k(x)$. It's important to keep the domain and range/codomain in mind whenever you do function composition.



          Now to address your confusion regarding how we are "allowed" to set $x = x^2+9$ and write $k(x^2+9) = sqrt{x^2+9}$. Again, think about a function as taking inputs to outputs. So when you write $k(x^2+9) = sqrt{x^2+9}$, what you're saying is that given a number $x$, $k$ maps the number $x^2+9$ to $sqrt{x^2+9}$. This is really just a variable substitution. There is nothing wrong with writing down $k(x^2+9)$, or $k(e^x)$; just like with $k(2)$ or $k(pi)$, it represents passing some value into the function $k$.



          Hopefully that addresses your questions, and let me know in the comments if I can clarify further!







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 30 '18 at 19:26

























          answered Nov 30 '18 at 19:20









          fractal1729fractal1729

          1,387213




          1,387213























              1












              $begingroup$

              $f(y)$ is a function which maps $y$ to $sqrt{y^2+9}$, and you've fed $y=x$ into it to get your first function.



              $k(y)$ is a function which maps $y$ to $sqrt y$, and you've fed $y=x^2+9$ into it to get your second function.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $f(y)$ is a function which maps $y$ to $sqrt{y^2+9}$, and you've fed $y=x$ into it to get your first function.



                $k(y)$ is a function which maps $y$ to $sqrt y$, and you've fed $y=x^2+9$ into it to get your second function.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $f(y)$ is a function which maps $y$ to $sqrt{y^2+9}$, and you've fed $y=x$ into it to get your first function.



                  $k(y)$ is a function which maps $y$ to $sqrt y$, and you've fed $y=x^2+9$ into it to get your second function.






                  share|cite|improve this answer









                  $endgroup$



                  $f(y)$ is a function which maps $y$ to $sqrt{y^2+9}$, and you've fed $y=x$ into it to get your first function.



                  $k(y)$ is a function which maps $y$ to $sqrt y$, and you've fed $y=x^2+9$ into it to get your second function.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 21:23









                  timtfjtimtfj

                  1,183318




                  1,183318






























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