Determining $f(2)$











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



$$f(-1)= dfrac{2}{9}$$



$$f(2) = ? $$




Substituting $x = -2$ to get $f(-1)$ since it is known



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



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



$$dfrac{2}{9}=3f(-2) implies f(-2) = dfrac{2}{27}$$



I think I found $f(-2)$. However, how can I find $f(2)$ instead?






functions







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  • @lulu I'm still unable to get it.
    – Mark
    Nov 15 at 15:39










  • Hint: $frac29,frac23,2,6$.
    – Yves Daoust
    Nov 15 at 15:58










  • Why on Earth do you decrease $x$ when you need to go from $-1$ to $2$ ?
    – Yves Daoust
    Nov 15 at 15:59










  • @YvesDaoust lol
    – Mark
    Nov 15 at 16:00










  • Stop just regurgitating your homework onto this site at such a rapid rate please
    – Chase Ryan Taylor
    Nov 15 at 20:25















up vote
1
down vote

favorite













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



$$f(-1)= dfrac{2}{9}$$



$$f(2) = ? $$




Substituting $x = -2$ to get $f(-1)$ since it is known



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



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



$$dfrac{2}{9}=3f(-2) implies f(-2) = dfrac{2}{27}$$



I think I found $f(-2)$. However, how can I find $f(2)$ instead?






functions







share|cite|improve this question






















  • @lulu I'm still unable to get it.
    – Mark
    Nov 15 at 15:39










  • Hint: $frac29,frac23,2,6$.
    – Yves Daoust
    Nov 15 at 15:58










  • Why on Earth do you decrease $x$ when you need to go from $-1$ to $2$ ?
    – Yves Daoust
    Nov 15 at 15:59










  • @YvesDaoust lol
    – Mark
    Nov 15 at 16:00










  • Stop just regurgitating your homework onto this site at such a rapid rate please
    – Chase Ryan Taylor
    Nov 15 at 20:25













up vote
1
down vote

favorite









up vote
1
down vote

favorite












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



$$f(-1)= dfrac{2}{9}$$



$$f(2) = ? $$




Substituting $x = -2$ to get $f(-1)$ since it is known



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



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



$$dfrac{2}{9}=3f(-2) implies f(-2) = dfrac{2}{27}$$



I think I found $f(-2)$. However, how can I find $f(2)$ instead?






functions







share|cite|improve this question














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



$$f(-1)= dfrac{2}{9}$$



$$f(2) = ? $$




Substituting $x = -2$ to get $f(-1)$ since it is known



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



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



$$dfrac{2}{9}=3f(-2) implies f(-2) = dfrac{2}{27}$$



I think I found $f(-2)$. However, how can I find $f(2)$ instead?






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 15 at 15:32









Mark

515




515












  • @lulu I'm still unable to get it.
    – Mark
    Nov 15 at 15:39










  • Hint: $frac29,frac23,2,6$.
    – Yves Daoust
    Nov 15 at 15:58










  • Why on Earth do you decrease $x$ when you need to go from $-1$ to $2$ ?
    – Yves Daoust
    Nov 15 at 15:59










  • @YvesDaoust lol
    – Mark
    Nov 15 at 16:00










  • Stop just regurgitating your homework onto this site at such a rapid rate please
    – Chase Ryan Taylor
    Nov 15 at 20:25


















  • @lulu I'm still unable to get it.
    – Mark
    Nov 15 at 15:39










  • Hint: $frac29,frac23,2,6$.
    – Yves Daoust
    Nov 15 at 15:58










  • Why on Earth do you decrease $x$ when you need to go from $-1$ to $2$ ?
    – Yves Daoust
    Nov 15 at 15:59










  • @YvesDaoust lol
    – Mark
    Nov 15 at 16:00










  • Stop just regurgitating your homework onto this site at such a rapid rate please
    – Chase Ryan Taylor
    Nov 15 at 20:25
















@lulu I'm still unable to get it.
– Mark
Nov 15 at 15:39




@lulu I'm still unable to get it.
– Mark
Nov 15 at 15:39












Hint: $frac29,frac23,2,6$.
– Yves Daoust
Nov 15 at 15:58




Hint: $frac29,frac23,2,6$.
– Yves Daoust
Nov 15 at 15:58












Why on Earth do you decrease $x$ when you need to go from $-1$ to $2$ ?
– Yves Daoust
Nov 15 at 15:59




Why on Earth do you decrease $x$ when you need to go from $-1$ to $2$ ?
– Yves Daoust
Nov 15 at 15:59












@YvesDaoust lol
– Mark
Nov 15 at 16:00




@YvesDaoust lol
– Mark
Nov 15 at 16:00












Stop just regurgitating your homework onto this site at such a rapid rate please
– Chase Ryan Taylor
Nov 15 at 20:25




Stop just regurgitating your homework onto this site at such a rapid rate please
– Chase Ryan Taylor
Nov 15 at 20:25










3 Answers
3






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up vote
1
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$$f(2) = 3f(1)= 9f(0)= 27f(-1) = ...$$






share|cite|improve this answer




























    up vote
    1
    down vote













    From $x=-1$, you can extend the values both ways,



    $$cdots,frac2{243},frac2{81},frac2{27},frac29,frac23,2,6,18,cdots$$



    Choose the right one.






    share|cite|improve this answer

















    • 1




      Yves.Nice answer:)+
      – Peter Szilas
      Nov 15 at 16:27










    • @PeterSzilas ;-)
      – Yves Daoust
      Nov 15 at 16:32


















    up vote
    0
    down vote













    Exactly what you just did, but the other way around:



    $f(0) = f(-1 + 1) = 3f(-1) = 3left(frac{2}{9}right) = frac{2}{3}$. Repeat.






    share|cite|improve this answer





















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      3 Answers
      3






      active

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      3 Answers
      3






      active

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      active

      oldest

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      up vote
      1
      down vote













      $$f(2) = 3f(1)= 9f(0)= 27f(-1) = ...$$






      share|cite|improve this answer

























        up vote
        1
        down vote













        $$f(2) = 3f(1)= 9f(0)= 27f(-1) = ...$$






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          $$f(2) = 3f(1)= 9f(0)= 27f(-1) = ...$$






          share|cite|improve this answer












          $$f(2) = 3f(1)= 9f(0)= 27f(-1) = ...$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 15 at 15:35









          greedoid

          34.3k114488




          34.3k114488






















              up vote
              1
              down vote













              From $x=-1$, you can extend the values both ways,



              $$cdots,frac2{243},frac2{81},frac2{27},frac29,frac23,2,6,18,cdots$$



              Choose the right one.






              share|cite|improve this answer

















              • 1




                Yves.Nice answer:)+
                – Peter Szilas
                Nov 15 at 16:27










              • @PeterSzilas ;-)
                – Yves Daoust
                Nov 15 at 16:32















              up vote
              1
              down vote













              From $x=-1$, you can extend the values both ways,



              $$cdots,frac2{243},frac2{81},frac2{27},frac29,frac23,2,6,18,cdots$$



              Choose the right one.






              share|cite|improve this answer

















              • 1




                Yves.Nice answer:)+
                – Peter Szilas
                Nov 15 at 16:27










              • @PeterSzilas ;-)
                – Yves Daoust
                Nov 15 at 16:32













              up vote
              1
              down vote










              up vote
              1
              down vote









              From $x=-1$, you can extend the values both ways,



              $$cdots,frac2{243},frac2{81},frac2{27},frac29,frac23,2,6,18,cdots$$



              Choose the right one.






              share|cite|improve this answer












              From $x=-1$, you can extend the values both ways,



              $$cdots,frac2{243},frac2{81},frac2{27},frac29,frac23,2,6,18,cdots$$



              Choose the right one.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 15 at 16:05









              Yves Daoust

              121k668218




              121k668218








              • 1




                Yves.Nice answer:)+
                – Peter Szilas
                Nov 15 at 16:27










              • @PeterSzilas ;-)
                – Yves Daoust
                Nov 15 at 16:32














              • 1




                Yves.Nice answer:)+
                – Peter Szilas
                Nov 15 at 16:27










              • @PeterSzilas ;-)
                – Yves Daoust
                Nov 15 at 16:32








              1




              1




              Yves.Nice answer:)+
              – Peter Szilas
              Nov 15 at 16:27




              Yves.Nice answer:)+
              – Peter Szilas
              Nov 15 at 16:27












              @PeterSzilas ;-)
              – Yves Daoust
              Nov 15 at 16:32




              @PeterSzilas ;-)
              – Yves Daoust
              Nov 15 at 16:32










              up vote
              0
              down vote













              Exactly what you just did, but the other way around:



              $f(0) = f(-1 + 1) = 3f(-1) = 3left(frac{2}{9}right) = frac{2}{3}$. Repeat.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Exactly what you just did, but the other way around:



                $f(0) = f(-1 + 1) = 3f(-1) = 3left(frac{2}{9}right) = frac{2}{3}$. Repeat.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Exactly what you just did, but the other way around:



                  $f(0) = f(-1 + 1) = 3f(-1) = 3left(frac{2}{9}right) = frac{2}{3}$. Repeat.






                  share|cite|improve this answer












                  Exactly what you just did, but the other way around:



                  $f(0) = f(-1 + 1) = 3f(-1) = 3left(frac{2}{9}right) = frac{2}{3}$. Repeat.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 at 15:35









                  user3482749

                  981411




                  981411






























                       

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