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
    1 hour 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
    1 hour ago










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










  • $begingroup$
    ive updated the question you guys, this time i have included all information needed/given.
    $endgroup$
    – Reddevil
    1 hour 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
    1 hour 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 1 hour ago









Lemniscate

402211




402211










asked 1 hour ago









ReddevilReddevil

11718




11718








  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    1 hour 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
    1 hour ago










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














  • 2




    $begingroup$
    What is the domain?
    $endgroup$
    – Haris Gusic
    1 hour 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
    1 hour ago










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








2




2




$begingroup$
What is the domain?
$endgroup$
– Haris Gusic
1 hour ago




$begingroup$
What is the domain?
$endgroup$
– Haris Gusic
1 hour 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
1 hour 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
1 hour ago












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






$begingroup$
ive updated the question you guys, this time i have included all information needed/given.
$endgroup$
– Reddevil
1 hour 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
1 hour 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
1 hour 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
        1 hour 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
        57 mins 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 1 hour ago

























          answered 1 hour ago









          Eleven-ElevenEleven-Eleven

          5,69572759




          5,69572759























              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 1 hour ago

























                  answered 1 hour 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
                        1 hour 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
                        57 mins 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
                        1 hour 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
                        57 mins 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 59 mins ago

























                      answered 1 hour 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
                        1 hour 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
                        57 mins 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
                        1 hour 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
                        57 mins 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
                      1 hour 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
                      1 hour 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
                      57 mins 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
                      57 mins ago


















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