Computing $f(5)$












0












$begingroup$



f is linear and one-to-one function such that



$$(f+f)(x) = (f circ f)(x)$$
Compute $f(5)$.




We're told that f is linear, which implies



$$f(x) = ax+b$$



Then plugging this into the given equation



$$biggr [2(2ax+b)biggr ] (x) = f(ax+b)$$



Let $ax+b = 5$



$$(10)(x) = f(5)$$



Here I think I got the correct answer, 10. However, are my steps mathematically correct?



Regards






functions







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$endgroup$












  • $begingroup$
    You have a typo there in the middle: it should be $;left[2(ax+b)right];$ ...
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 17:34
















0












$begingroup$



f is linear and one-to-one function such that



$$(f+f)(x) = (f circ f)(x)$$
Compute $f(5)$.




We're told that f is linear, which implies



$$f(x) = ax+b$$



Then plugging this into the given equation



$$biggr [2(2ax+b)biggr ] (x) = f(ax+b)$$



Let $ax+b = 5$



$$(10)(x) = f(5)$$



Here I think I got the correct answer, 10. However, are my steps mathematically correct?



Regards






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You have a typo there in the middle: it should be $;left[2(ax+b)right];$ ...
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 17:34














0












0








0





$begingroup$



f is linear and one-to-one function such that



$$(f+f)(x) = (f circ f)(x)$$
Compute $f(5)$.




We're told that f is linear, which implies



$$f(x) = ax+b$$



Then plugging this into the given equation



$$biggr [2(2ax+b)biggr ] (x) = f(ax+b)$$



Let $ax+b = 5$



$$(10)(x) = f(5)$$



Here I think I got the correct answer, 10. However, are my steps mathematically correct?



Regards






functions







share|cite|improve this question









$endgroup$





f is linear and one-to-one function such that



$$(f+f)(x) = (f circ f)(x)$$
Compute $f(5)$.




We're told that f is linear, which implies



$$f(x) = ax+b$$



Then plugging this into the given equation



$$biggr [2(2ax+b)biggr ] (x) = f(ax+b)$$



Let $ax+b = 5$



$$(10)(x) = f(5)$$



Here I think I got the correct answer, 10. However, are my steps mathematically correct?



Regards






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 17:29









EnzoEnzo

19917




19917












  • $begingroup$
    You have a typo there in the middle: it should be $;left[2(ax+b)right];$ ...
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 17:34


















  • $begingroup$
    You have a typo there in the middle: it should be $;left[2(ax+b)right];$ ...
    $endgroup$
    – DonAntonio
    Dec 21 '18 at 17:34
















$begingroup$
You have a typo there in the middle: it should be $;left[2(ax+b)right];$ ...
$endgroup$
– DonAntonio
Dec 21 '18 at 17:34




$begingroup$
You have a typo there in the middle: it should be $;left[2(ax+b)right];$ ...
$endgroup$
– DonAntonio
Dec 21 '18 at 17:34










2 Answers
2






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oldest

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2












$begingroup$

I don't understand very well what you have done.



Here's my try:



$$(f+f)(x)=2ax+2b$$
$$fcirc f(x)=a(ax+b)+b=a^2x+ab+b$$



Since $f+f=fcirc f$,
$$left{begin{align}a^2&=2a\2b&=ab+bend{align}right.$$



This system has two solutions: $a=b=0$ and $a=2$, $b=0$. In the first case, $f$ would not be injective, and in the other $f(5)=10$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But, in the first case, $f$ is not injective.
    $endgroup$
    – Andrés E. Caicedo
    Dec 21 '18 at 17:38










  • $begingroup$
    Oh, yes, I missed that condition.
    $endgroup$
    – ajotatxe
    Dec 21 '18 at 17:39



















1












$begingroup$

I'm afraid you haven't, but you've come close.



By definition, $$(f+f)(x)=f(x)+f(x),$$ so since $f$ is linear, then $$(f+f)(x)=ax+b+ax+b=2ax+2b.$$



You've used the definition of $(fcirc f)(x)$ correctly, but didn't follow through. $$f(ax+b)=a(ax+b)+b=a^2x+ab+b.$$ So, we know that $$2ax+2b=a^2x+ab+b.$$ Can you take it from there to figure out what $a$ and $b$ are? Once you do that, finding $f(5)$ should be straightforward.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    I don't understand very well what you have done.



    Here's my try:



    $$(f+f)(x)=2ax+2b$$
    $$fcirc f(x)=a(ax+b)+b=a^2x+ab+b$$



    Since $f+f=fcirc f$,
    $$left{begin{align}a^2&=2a\2b&=ab+bend{align}right.$$



    This system has two solutions: $a=b=0$ and $a=2$, $b=0$. In the first case, $f$ would not be injective, and in the other $f(5)=10$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But, in the first case, $f$ is not injective.
      $endgroup$
      – Andrés E. Caicedo
      Dec 21 '18 at 17:38










    • $begingroup$
      Oh, yes, I missed that condition.
      $endgroup$
      – ajotatxe
      Dec 21 '18 at 17:39
















    2












    $begingroup$

    I don't understand very well what you have done.



    Here's my try:



    $$(f+f)(x)=2ax+2b$$
    $$fcirc f(x)=a(ax+b)+b=a^2x+ab+b$$



    Since $f+f=fcirc f$,
    $$left{begin{align}a^2&=2a\2b&=ab+bend{align}right.$$



    This system has two solutions: $a=b=0$ and $a=2$, $b=0$. In the first case, $f$ would not be injective, and in the other $f(5)=10$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But, in the first case, $f$ is not injective.
      $endgroup$
      – Andrés E. Caicedo
      Dec 21 '18 at 17:38










    • $begingroup$
      Oh, yes, I missed that condition.
      $endgroup$
      – ajotatxe
      Dec 21 '18 at 17:39














    2












    2








    2





    $begingroup$

    I don't understand very well what you have done.



    Here's my try:



    $$(f+f)(x)=2ax+2b$$
    $$fcirc f(x)=a(ax+b)+b=a^2x+ab+b$$



    Since $f+f=fcirc f$,
    $$left{begin{align}a^2&=2a\2b&=ab+bend{align}right.$$



    This system has two solutions: $a=b=0$ and $a=2$, $b=0$. In the first case, $f$ would not be injective, and in the other $f(5)=10$.






    share|cite|improve this answer











    $endgroup$



    I don't understand very well what you have done.



    Here's my try:



    $$(f+f)(x)=2ax+2b$$
    $$fcirc f(x)=a(ax+b)+b=a^2x+ab+b$$



    Since $f+f=fcirc f$,
    $$left{begin{align}a^2&=2a\2b&=ab+bend{align}right.$$



    This system has two solutions: $a=b=0$ and $a=2$, $b=0$. In the first case, $f$ would not be injective, and in the other $f(5)=10$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 17:39

























    answered Dec 21 '18 at 17:37









    ajotatxeajotatxe

    54.1k24190




    54.1k24190












    • $begingroup$
      But, in the first case, $f$ is not injective.
      $endgroup$
      – Andrés E. Caicedo
      Dec 21 '18 at 17:38










    • $begingroup$
      Oh, yes, I missed that condition.
      $endgroup$
      – ajotatxe
      Dec 21 '18 at 17:39


















    • $begingroup$
      But, in the first case, $f$ is not injective.
      $endgroup$
      – Andrés E. Caicedo
      Dec 21 '18 at 17:38










    • $begingroup$
      Oh, yes, I missed that condition.
      $endgroup$
      – ajotatxe
      Dec 21 '18 at 17:39
















    $begingroup$
    But, in the first case, $f$ is not injective.
    $endgroup$
    – Andrés E. Caicedo
    Dec 21 '18 at 17:38




    $begingroup$
    But, in the first case, $f$ is not injective.
    $endgroup$
    – Andrés E. Caicedo
    Dec 21 '18 at 17:38












    $begingroup$
    Oh, yes, I missed that condition.
    $endgroup$
    – ajotatxe
    Dec 21 '18 at 17:39




    $begingroup$
    Oh, yes, I missed that condition.
    $endgroup$
    – ajotatxe
    Dec 21 '18 at 17:39











    1












    $begingroup$

    I'm afraid you haven't, but you've come close.



    By definition, $$(f+f)(x)=f(x)+f(x),$$ so since $f$ is linear, then $$(f+f)(x)=ax+b+ax+b=2ax+2b.$$



    You've used the definition of $(fcirc f)(x)$ correctly, but didn't follow through. $$f(ax+b)=a(ax+b)+b=a^2x+ab+b.$$ So, we know that $$2ax+2b=a^2x+ab+b.$$ Can you take it from there to figure out what $a$ and $b$ are? Once you do that, finding $f(5)$ should be straightforward.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I'm afraid you haven't, but you've come close.



      By definition, $$(f+f)(x)=f(x)+f(x),$$ so since $f$ is linear, then $$(f+f)(x)=ax+b+ax+b=2ax+2b.$$



      You've used the definition of $(fcirc f)(x)$ correctly, but didn't follow through. $$f(ax+b)=a(ax+b)+b=a^2x+ab+b.$$ So, we know that $$2ax+2b=a^2x+ab+b.$$ Can you take it from there to figure out what $a$ and $b$ are? Once you do that, finding $f(5)$ should be straightforward.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I'm afraid you haven't, but you've come close.



        By definition, $$(f+f)(x)=f(x)+f(x),$$ so since $f$ is linear, then $$(f+f)(x)=ax+b+ax+b=2ax+2b.$$



        You've used the definition of $(fcirc f)(x)$ correctly, but didn't follow through. $$f(ax+b)=a(ax+b)+b=a^2x+ab+b.$$ So, we know that $$2ax+2b=a^2x+ab+b.$$ Can you take it from there to figure out what $a$ and $b$ are? Once you do that, finding $f(5)$ should be straightforward.






        share|cite|improve this answer









        $endgroup$



        I'm afraid you haven't, but you've come close.



        By definition, $$(f+f)(x)=f(x)+f(x),$$ so since $f$ is linear, then $$(f+f)(x)=ax+b+ax+b=2ax+2b.$$



        You've used the definition of $(fcirc f)(x)$ correctly, but didn't follow through. $$f(ax+b)=a(ax+b)+b=a^2x+ab+b.$$ So, we know that $$2ax+2b=a^2x+ab+b.$$ Can you take it from there to figure out what $a$ and $b$ are? Once you do that, finding $f(5)$ should be straightforward.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 17:36









        Cameron BuieCameron Buie

        86.2k772161




        86.2k772161






























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