How to find the range of a composite function?












2












$begingroup$


I have been stuck at this question: I have $$f(x)=cos(pi cdot x)$$$$g(x)=frac{7cdot x}{6}$$ and $$h(x)=f(g(x))$$



and i am asked to compute the range for $h(x)$.



My solution:
$$h(x)=cos(pi cdot frac{7x}{6})$$
so the highest value the function can output is $1$ and the lowest is $-1$. My instructor says it's not the right answer. How do i go about finding the range of this function? Thank you in advance.



EDIT



$f:mathbb{Q}tomathbb R$



$g:mathbb{Z}tomathbb{Q}$






functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    7 hours ago










  • $begingroup$
    You're only affecting the domain when you affect the input of a function. So your range should remain the same. I would imagine that you have a domain restriction as others have suggested...
    $endgroup$
    – Eleven-Eleven
    7 hours ago










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    7 hours ago








  • 2




    $begingroup$
    SInce you are only inputting integer values, you will not get a continuous range... You are only going to get specific outputs for integer inputs. This is because your angles are multiples of $7pi /6$
    $endgroup$
    – Eleven-Eleven
    7 hours ago
















2












$begingroup$


I have been stuck at this question: I have $$f(x)=cos(pi cdot x)$$$$g(x)=frac{7cdot x}{6}$$ and $$h(x)=f(g(x))$$



and i am asked to compute the range for $h(x)$.



My solution:
$$h(x)=cos(pi cdot frac{7x}{6})$$
so the highest value the function can output is $1$ and the lowest is $-1$. My instructor says it's not the right answer. How do i go about finding the range of this function? Thank you in advance.



EDIT



$f:mathbb{Q}tomathbb R$



$g:mathbb{Z}tomathbb{Q}$






functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    7 hours ago










  • $begingroup$
    You're only affecting the domain when you affect the input of a function. So your range should remain the same. I would imagine that you have a domain restriction as others have suggested...
    $endgroup$
    – Eleven-Eleven
    7 hours ago










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    7 hours ago








  • 2




    $begingroup$
    SInce you are only inputting integer values, you will not get a continuous range... You are only going to get specific outputs for integer inputs. This is because your angles are multiples of $7pi /6$
    $endgroup$
    – Eleven-Eleven
    7 hours ago














2












2








2





$begingroup$


I have been stuck at this question: I have $$f(x)=cos(pi cdot x)$$$$g(x)=frac{7cdot x}{6}$$ and $$h(x)=f(g(x))$$



and i am asked to compute the range for $h(x)$.



My solution:
$$h(x)=cos(pi cdot frac{7x}{6})$$
so the highest value the function can output is $1$ and the lowest is $-1$. My instructor says it's not the right answer. How do i go about finding the range of this function? Thank you in advance.



EDIT



$f:mathbb{Q}tomathbb R$



$g:mathbb{Z}tomathbb{Q}$






functions







share|cite|improve this question











$endgroup$




I have been stuck at this question: I have $$f(x)=cos(pi cdot x)$$$$g(x)=frac{7cdot x}{6}$$ and $$h(x)=f(g(x))$$



and i am asked to compute the range for $h(x)$.



My solution:
$$h(x)=cos(pi cdot frac{7x}{6})$$
so the highest value the function can output is $1$ and the lowest is $-1$. My instructor says it's not the right answer. How do i go about finding the range of this function? Thank you in advance.



EDIT



$f:mathbb{Q}tomathbb R$



$g:mathbb{Z}tomathbb{Q}$






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









Lemniscate

402211




402211










asked 7 hours ago









ReddevilReddevil

11718




11718








  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    7 hours ago










  • $begingroup$
    You're only affecting the domain when you affect the input of a function. So your range should remain the same. I would imagine that you have a domain restriction as others have suggested...
    $endgroup$
    – Eleven-Eleven
    7 hours ago










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    7 hours ago








  • 2




    $begingroup$
    SInce you are only inputting integer values, you will not get a continuous range... You are only going to get specific outputs for integer inputs. This is because your angles are multiples of $7pi /6$
    $endgroup$
    – Eleven-Eleven
    7 hours ago














  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    7 hours ago










  • $begingroup$
    You're only affecting the domain when you affect the input of a function. So your range should remain the same. I would imagine that you have a domain restriction as others have suggested...
    $endgroup$
    – Eleven-Eleven
    7 hours ago










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    7 hours ago








  • 2




    $begingroup$
    SInce you are only inputting integer values, you will not get a continuous range... You are only going to get specific outputs for integer inputs. This is because your angles are multiples of $7pi /6$
    $endgroup$
    – Eleven-Eleven
    7 hours ago








2




2




$begingroup$
What is the domain?
$endgroup$
– Haris Gusic
7 hours ago




$begingroup$
What is the domain?
$endgroup$
– Haris Gusic
7 hours ago












$begingroup$
You're only affecting the domain when you affect the input of a function. So your range should remain the same. I would imagine that you have a domain restriction as others have suggested...
$endgroup$
– Eleven-Eleven
7 hours ago




$begingroup$
You're only affecting the domain when you affect the input of a function. So your range should remain the same. I would imagine that you have a domain restriction as others have suggested...
$endgroup$
– Eleven-Eleven
7 hours ago












$begingroup$
ive updated the question you guys, this time i have included all information needed/given.
$endgroup$
– Reddevil
7 hours ago






$begingroup$
ive updated the question you guys, this time i have included all information needed/given.
$endgroup$
– Reddevil
7 hours ago






2




2




$begingroup$
SInce you are only inputting integer values, you will not get a continuous range... You are only going to get specific outputs for integer inputs. This is because your angles are multiples of $7pi /6$
$endgroup$
– Eleven-Eleven
7 hours ago




$begingroup$
SInce you are only inputting integer values, you will not get a continuous range... You are only going to get specific outputs for integer inputs. This is because your angles are multiples of $7pi /6$
$endgroup$
– Eleven-Eleven
7 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

Your composite function then is



$$fcirc g:mathbb{Z}rightarrow mathbb{R}$$



But now, by inputting only multiples of $frac{7pi}{6}$ into the cosine function, and because cosine is a periodic function, you are only going to get specific values for your range. These are your "special" angle values because you will never have an angle as an input to cosine that is NOT a multiple of $frac{pi}{6}$. So your range will be $$left{-1,-frac{sqrt{3}}{2},-frac{1}{2},0,frac{1}{2},frac{sqrt{3}}{2},1right}$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Since $f(x) = cos(pi x)$ has periodicity $2$, we have $h(x) = f(g(x)) = f(frac{7x}{6}+2m)$ with $m$ integer. Now note that $h(x)=h(x+12)$.



    Thus your codomain will be determined by $xin{0,1,dots,11}$. The range is then determined by the unique subset of values from ${h(0),h(1),dots,h(11)}$.






    share|cite|improve this answer











    $endgroup$





















      -1












      $begingroup$

      Normally range is taken to mean the image (set of possible outputs). Perhaps I'm wrong but you seem to want to know the global minimum and global maximum of $h$. The global minimum of $h$ is $-1$ and the global maximum is $1$.



      $h(12) = mathrm{cos}(14 pi) = 1$, and $h(6) = mathrm{cos}(7 pi) = -1$.



      Since $forall x in mathbb{R}$ $mathrm{cos}(x) in [-1,1]$, we have $forall x in mathbb{Q}$ $f(x) in [-1,1]$. This means that $h$ does not take values above $1$ or below $-1$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
        $endgroup$
        – Milan Stojanovic
        7 hours ago










      • $begingroup$
        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
        $endgroup$
        – Lemniscate
        7 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Your composite function then is



      $$fcirc g:mathbb{Z}rightarrow mathbb{R}$$



      But now, by inputting only multiples of $frac{7pi}{6}$ into the cosine function, and because cosine is a periodic function, you are only going to get specific values for your range. These are your "special" angle values because you will never have an angle as an input to cosine that is NOT a multiple of $frac{pi}{6}$. So your range will be $$left{-1,-frac{sqrt{3}}{2},-frac{1}{2},0,frac{1}{2},frac{sqrt{3}}{2},1right}$$






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Your composite function then is



        $$fcirc g:mathbb{Z}rightarrow mathbb{R}$$



        But now, by inputting only multiples of $frac{7pi}{6}$ into the cosine function, and because cosine is a periodic function, you are only going to get specific values for your range. These are your "special" angle values because you will never have an angle as an input to cosine that is NOT a multiple of $frac{pi}{6}$. So your range will be $$left{-1,-frac{sqrt{3}}{2},-frac{1}{2},0,frac{1}{2},frac{sqrt{3}}{2},1right}$$






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Your composite function then is



          $$fcirc g:mathbb{Z}rightarrow mathbb{R}$$



          But now, by inputting only multiples of $frac{7pi}{6}$ into the cosine function, and because cosine is a periodic function, you are only going to get specific values for your range. These are your "special" angle values because you will never have an angle as an input to cosine that is NOT a multiple of $frac{pi}{6}$. So your range will be $$left{-1,-frac{sqrt{3}}{2},-frac{1}{2},0,frac{1}{2},frac{sqrt{3}}{2},1right}$$






          share|cite|improve this answer











          $endgroup$



          Your composite function then is



          $$fcirc g:mathbb{Z}rightarrow mathbb{R}$$



          But now, by inputting only multiples of $frac{7pi}{6}$ into the cosine function, and because cosine is a periodic function, you are only going to get specific values for your range. These are your "special" angle values because you will never have an angle as an input to cosine that is NOT a multiple of $frac{pi}{6}$. So your range will be $$left{-1,-frac{sqrt{3}}{2},-frac{1}{2},0,frac{1}{2},frac{sqrt{3}}{2},1right}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          Eleven-ElevenEleven-Eleven

          5,71072759




          5,71072759























              1












              $begingroup$

              Since $f(x) = cos(pi x)$ has periodicity $2$, we have $h(x) = f(g(x)) = f(frac{7x}{6}+2m)$ with $m$ integer. Now note that $h(x)=h(x+12)$.



              Thus your codomain will be determined by $xin{0,1,dots,11}$. The range is then determined by the unique subset of values from ${h(0),h(1),dots,h(11)}$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Since $f(x) = cos(pi x)$ has periodicity $2$, we have $h(x) = f(g(x)) = f(frac{7x}{6}+2m)$ with $m$ integer. Now note that $h(x)=h(x+12)$.



                Thus your codomain will be determined by $xin{0,1,dots,11}$. The range is then determined by the unique subset of values from ${h(0),h(1),dots,h(11)}$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Since $f(x) = cos(pi x)$ has periodicity $2$, we have $h(x) = f(g(x)) = f(frac{7x}{6}+2m)$ with $m$ integer. Now note that $h(x)=h(x+12)$.



                  Thus your codomain will be determined by $xin{0,1,dots,11}$. The range is then determined by the unique subset of values from ${h(0),h(1),dots,h(11)}$.






                  share|cite|improve this answer











                  $endgroup$



                  Since $f(x) = cos(pi x)$ has periodicity $2$, we have $h(x) = f(g(x)) = f(frac{7x}{6}+2m)$ with $m$ integer. Now note that $h(x)=h(x+12)$.



                  Thus your codomain will be determined by $xin{0,1,dots,11}$. The range is then determined by the unique subset of values from ${h(0),h(1),dots,h(11)}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 7 hours ago

























                  answered 7 hours ago









                  zahbazzahbaz

                  8,38921937




                  8,38921937























                      -1












                      $begingroup$

                      Normally range is taken to mean the image (set of possible outputs). Perhaps I'm wrong but you seem to want to know the global minimum and global maximum of $h$. The global minimum of $h$ is $-1$ and the global maximum is $1$.



                      $h(12) = mathrm{cos}(14 pi) = 1$, and $h(6) = mathrm{cos}(7 pi) = -1$.



                      Since $forall x in mathbb{R}$ $mathrm{cos}(x) in [-1,1]$, we have $forall x in mathbb{Q}$ $f(x) in [-1,1]$. This means that $h$ does not take values above $1$ or below $-1$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
                        $endgroup$
                        – Milan Stojanovic
                        7 hours ago










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Lemniscate
                        7 hours ago
















                      -1












                      $begingroup$

                      Normally range is taken to mean the image (set of possible outputs). Perhaps I'm wrong but you seem to want to know the global minimum and global maximum of $h$. The global minimum of $h$ is $-1$ and the global maximum is $1$.



                      $h(12) = mathrm{cos}(14 pi) = 1$, and $h(6) = mathrm{cos}(7 pi) = -1$.



                      Since $forall x in mathbb{R}$ $mathrm{cos}(x) in [-1,1]$, we have $forall x in mathbb{Q}$ $f(x) in [-1,1]$. This means that $h$ does not take values above $1$ or below $-1$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
                        $endgroup$
                        – Milan Stojanovic
                        7 hours ago










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Lemniscate
                        7 hours ago














                      -1












                      -1








                      -1





                      $begingroup$

                      Normally range is taken to mean the image (set of possible outputs). Perhaps I'm wrong but you seem to want to know the global minimum and global maximum of $h$. The global minimum of $h$ is $-1$ and the global maximum is $1$.



                      $h(12) = mathrm{cos}(14 pi) = 1$, and $h(6) = mathrm{cos}(7 pi) = -1$.



                      Since $forall x in mathbb{R}$ $mathrm{cos}(x) in [-1,1]$, we have $forall x in mathbb{Q}$ $f(x) in [-1,1]$. This means that $h$ does not take values above $1$ or below $-1$.






                      share|cite|improve this answer











                      $endgroup$



                      Normally range is taken to mean the image (set of possible outputs). Perhaps I'm wrong but you seem to want to know the global minimum and global maximum of $h$. The global minimum of $h$ is $-1$ and the global maximum is $1$.



                      $h(12) = mathrm{cos}(14 pi) = 1$, and $h(6) = mathrm{cos}(7 pi) = -1$.



                      Since $forall x in mathbb{R}$ $mathrm{cos}(x) in [-1,1]$, we have $forall x in mathbb{Q}$ $f(x) in [-1,1]$. This means that $h$ does not take values above $1$ or below $-1$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 7 hours ago

























                      answered 7 hours ago









                      LemniscateLemniscate

                      402211




                      402211












                      • $begingroup$
                        You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
                        $endgroup$
                        – Milan Stojanovic
                        7 hours ago










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Lemniscate
                        7 hours ago


















                      • $begingroup$
                        You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
                        $endgroup$
                        – Milan Stojanovic
                        7 hours ago










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Lemniscate
                        7 hours ago
















                      $begingroup$
                      You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
                      $endgroup$
                      – Milan Stojanovic
                      7 hours ago




                      $begingroup$
                      You are wrong. It doesnt extend above or below 1 or -1 but it doesnt map to everything between 1 and -1 . Maybe you got confused by range en.m.wikipedia.org/wiki/Range_(mathematics)
                      $endgroup$
                      – Milan Stojanovic
                      7 hours ago












                      $begingroup$
                      Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                      $endgroup$
                      – Lemniscate
                      7 hours ago




                      $begingroup$
                      Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                      $endgroup$
                      – Lemniscate
                      7 hours ago


















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                      Península de Txukotka

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                      Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
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