Is there something wrong with brackets? $f(2x+(f(y)+f(f(y))=4x+8y$ [on hold]












0















$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving







share|cite|improve this question















put on hold as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 5




    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    – Mauro ALLEGRANZA
    2 days ago












  • I think two more parentheses should be added
    – Mostafa Ayaz
    2 days ago










  • It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    – ancientmathematician
    2 days ago










  • @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    – Beginner
    2 days ago


















0















$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving







share|cite|improve this question















put on hold as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 5




    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    – Mauro ALLEGRANZA
    2 days ago












  • I think two more parentheses should be added
    – Mostafa Ayaz
    2 days ago










  • It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    – ancientmathematician
    2 days ago










  • @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    – Beginner
    2 days ago
















0












0








0








$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving







share|cite|improve this question
















$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving




algebra-precalculus contest-math problem-solving






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Beginner

1719




1719




put on hold as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    – Mauro ALLEGRANZA
    2 days ago












  • I think two more parentheses should be added
    – Mostafa Ayaz
    2 days ago










  • It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    – ancientmathematician
    2 days ago










  • @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    – Beginner
    2 days ago
















  • 5




    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    – Mauro ALLEGRANZA
    2 days ago












  • I think two more parentheses should be added
    – Mostafa Ayaz
    2 days ago










  • It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    – ancientmathematician
    2 days ago










  • @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    – Beginner
    2 days ago










5




5




Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
– Mauro ALLEGRANZA
2 days ago






Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
– Mauro ALLEGRANZA
2 days ago














I think two more parentheses should be added
– Mostafa Ayaz
2 days ago




I think two more parentheses should be added
– Mostafa Ayaz
2 days ago












It's not unamibiguous, though, is it @MauroALLEGRANZA ?
– ancientmathematician
2 days ago




It's not unamibiguous, though, is it @MauroALLEGRANZA ?
– ancientmathematician
2 days ago












@MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
– Beginner
2 days ago






@MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
– Beginner
2 days ago












1 Answer
1






active

oldest

votes


















4














Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer

















  • 1




    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    – Beginner
    2 days ago






  • 1




    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    – Ennar
    2 days ago










  • +1. Perfectly well explained.
    – Lucas Henrique
    2 days ago


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer

















  • 1




    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    – Beginner
    2 days ago






  • 1




    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    – Ennar
    2 days ago










  • +1. Perfectly well explained.
    – Lucas Henrique
    2 days ago
















4














Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer

















  • 1




    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    – Beginner
    2 days ago






  • 1




    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    – Ennar
    2 days ago










  • +1. Perfectly well explained.
    – Lucas Henrique
    2 days ago














4












4








4






Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer












Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









Ennar

14.4k32343




14.4k32343








  • 1




    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    – Beginner
    2 days ago






  • 1




    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    – Ennar
    2 days ago










  • +1. Perfectly well explained.
    – Lucas Henrique
    2 days ago














  • 1




    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    – Beginner
    2 days ago






  • 1




    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    – Ennar
    2 days ago










  • +1. Perfectly well explained.
    – Lucas Henrique
    2 days ago








1




1




(+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
– Beginner
2 days ago




(+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
– Beginner
2 days ago




1




1




@Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
– Ennar
2 days ago




@Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
– Ennar
2 days ago












+1. Perfectly well explained.
– Lucas Henrique
2 days ago




+1. Perfectly well explained.
– Lucas Henrique
2 days ago



-

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