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
    Feb 17 at 21:40










  • $begingroup$
    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
    Feb 17 at 21:43










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    Feb 17 at 21:45








  • 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
    Feb 17 at 21:46
















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
    Feb 17 at 21:40










  • $begingroup$
    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
    Feb 17 at 21:43










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    Feb 17 at 21:45








  • 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
    Feb 17 at 21:46














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 Feb 17 at 22:05









Joseph Martin

690317




690317










asked Feb 17 at 21:39









ReddevilReddevil

11718




11718








  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    Feb 17 at 21:40










  • $begingroup$
    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
    Feb 17 at 21:43










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    Feb 17 at 21:45








  • 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
    Feb 17 at 21:46














  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    Feb 17 at 21:40










  • $begingroup$
    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
    Feb 17 at 21:43










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    Feb 17 at 21:45








  • 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
    Feb 17 at 21:46








2




2




$begingroup$
What is the domain?
$endgroup$
– Haris Gusic
Feb 17 at 21:40




$begingroup$
What is the domain?
$endgroup$
– Haris Gusic
Feb 17 at 21:40












$begingroup$
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
Feb 17 at 21:43




$begingroup$
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
Feb 17 at 21:43












$begingroup$
ive updated the question you guys, this time i have included all information needed/given.
$endgroup$
– Reddevil
Feb 17 at 21:45






$begingroup$
ive updated the question you guys, this time i have included all information needed/given.
$endgroup$
– Reddevil
Feb 17 at 21:45






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
Feb 17 at 21:46




$begingroup$
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
Feb 17 at 21:46










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











    $endgroup$





















      0












      $begingroup$

      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
        Feb 17 at 22:10










      • $begingroup$
        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
        $endgroup$
        – Joseph Martin
        Feb 17 at 22:18











      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 Feb 17 at 21:58

























          answered Feb 17 at 21:52









          Eleven-ElevenEleven-Eleven

          5,77372759




          5,77372759























              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 Feb 17 at 22:02

























                  answered Feb 17 at 21:57









                  zahbazzahbaz

                  8,43921938




                  8,43921938























                      0












                      $begingroup$

                      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
                        Feb 17 at 22:10










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Joseph Martin
                        Feb 17 at 22:18
















                      0












                      $begingroup$

                      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
                        Feb 17 at 22:10










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Joseph Martin
                        Feb 17 at 22:18














                      0












                      0








                      0





                      $begingroup$

                      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 Feb 17 at 22:16

























                      answered Feb 17 at 21:50









                      Joseph MartinJoseph Martin

                      690317




                      690317












                      • $begingroup$
                        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
                        Feb 17 at 22:10










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Joseph Martin
                        Feb 17 at 22:18


















                      • $begingroup$
                        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
                        Feb 17 at 22:10










                      • $begingroup$
                        Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                        $endgroup$
                        – Joseph Martin
                        Feb 17 at 22:18
















                      $begingroup$
                      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
                      Feb 17 at 22:10




                      $begingroup$
                      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
                      Feb 17 at 22:10












                      $begingroup$
                      Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                      $endgroup$
                      – Joseph Martin
                      Feb 17 at 22:18




                      $begingroup$
                      Yes, I was unclear about what I meant. @Reddevil talks about the highest and lowest values and so I have addressed that.
                      $endgroup$
                      – Joseph Martin
                      Feb 17 at 22:18


















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