If a quadratic function $[0,2] to [0,2]$ is bijective, find $f(2)$ [closed]

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If $f: [0,2] to [0,2]$ defined by $f(x)=ax^2+bx+c$ is a bijective function, find $f(2)$.



I don’t know how to start solving it. Please help







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closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    When you say you have no idea, I don't believe you. I believe you know more than you think.
    – Arthur
    Jul 22 at 12:18










  • Do you know what a bijective function is?
    – Joel Reyes Noche
    Jul 22 at 12:23










  • If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+min(f(a),f(c)))/2$, or the value $(f(b)+max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing.
    – user574889
    Jul 22 at 12:34














up vote
-2
down vote

favorite
1












If $f: [0,2] to [0,2]$ defined by $f(x)=ax^2+bx+c$ is a bijective function, find $f(2)$.



I don’t know how to start solving it. Please help







share|cite|improve this question













closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    When you say you have no idea, I don't believe you. I believe you know more than you think.
    – Arthur
    Jul 22 at 12:18










  • Do you know what a bijective function is?
    – Joel Reyes Noche
    Jul 22 at 12:23










  • If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+min(f(a),f(c)))/2$, or the value $(f(b)+max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing.
    – user574889
    Jul 22 at 12:34












up vote
-2
down vote

favorite
1









up vote
-2
down vote

favorite
1






1





If $f: [0,2] to [0,2]$ defined by $f(x)=ax^2+bx+c$ is a bijective function, find $f(2)$.



I don’t know how to start solving it. Please help







share|cite|improve this question













If $f: [0,2] to [0,2]$ defined by $f(x)=ax^2+bx+c$ is a bijective function, find $f(2)$.



I don’t know how to start solving it. Please help









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 12:24









mechanodroid

22.2k52041




22.2k52041









asked Jul 22 at 12:15









learner_avid

687313




687313




closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel Jul 23 at 1:46


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Arthur, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    When you say you have no idea, I don't believe you. I believe you know more than you think.
    – Arthur
    Jul 22 at 12:18










  • Do you know what a bijective function is?
    – Joel Reyes Noche
    Jul 22 at 12:23










  • If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+min(f(a),f(c)))/2$, or the value $(f(b)+max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing.
    – user574889
    Jul 22 at 12:34












  • 1




    When you say you have no idea, I don't believe you. I believe you know more than you think.
    – Arthur
    Jul 22 at 12:18










  • Do you know what a bijective function is?
    – Joel Reyes Noche
    Jul 22 at 12:23










  • If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+min(f(a),f(c)))/2$, or the value $(f(b)+max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing.
    – user574889
    Jul 22 at 12:34







1




1




When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18




When you say you have no idea, I don't believe you. I believe you know more than you think.
– Arthur
Jul 22 at 12:18












Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23




Do you know what a bijective function is?
– Joel Reyes Noche
Jul 22 at 12:23












If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+min(f(a),f(c)))/2$, or the value $(f(b)+max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing.
– user574889
Jul 22 at 12:34




If $f(x)$ were not monotonic on $[0,2]$ then there would be three points $a<b<c$ in $[0,2]$ such that either $f(a)>f(b)<f(c)$ or $f(a)<f(b)>f(c)$. In either case, by the intermediate value theorem, the value $(f(b)+min(f(a),f(c)))/2$, or the value $(f(b)+max(f(a),f(c)))/2$ is taken both in $(a,b)$ and in $b,c$. Therefore, the function wouldn't be injective. This means that $f$ must be monotonic. Therefore, $f(2)=2$ if it is increasing, or $f(2)=0$ if it is decreasing.
– user574889
Jul 22 at 12:34










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










Since you have the algebra-precalculus tag, you may appreciate a less "high-powered" answer.



The graph of a quadratic function is a parabola; its vertex is the extreme value of the function, the function is symmetric about the vertex, and the function is strictly increasing or decreasing on either side of the vertex.



If the vertex of the parabola were not at $x = 0$ or $x = 2$, then $f$ could not be injective, since moving slightly to either side of it would produce equal values. If the vertex were not at $y = 0$ or $y = 2$, then $f$ could not be surjective, since the values above (or below, depending on the direction of $f$) could not be reached.



It is not hard to picture the possibilities now. The parabola can start in the "top-left" corner and go down, the "bottom-left" corner and go up, and so on. This requires $f(2) = 0$ in some cases, and $f(2) = 2$ in other cases.



As a graphical reference, all of the (segments of) parabolas below satisfy the hypotheses of your question.



Drawing of valid parabolas






share|cite|improve this answer






























    up vote
    5
    down vote













    $f(2)$ is either $0$ or $2$.



    Indeed, $f|_[0,2)$ is a continuous bijection of $[0,2)$ and $[0,2] setminus f(2)$. The former is a connected set so the latter also must be connected, which isn't true unless $f(2) in 0, 2$.



    Both cases are possible:



    For $f(2) = 2$ consider $f(x) = frac12x^2$ and for $f(2) = 0$ consider $f(x) = 2-frac12 x^2$.






    share|cite|improve this answer




























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Since you have the algebra-precalculus tag, you may appreciate a less "high-powered" answer.



      The graph of a quadratic function is a parabola; its vertex is the extreme value of the function, the function is symmetric about the vertex, and the function is strictly increasing or decreasing on either side of the vertex.



      If the vertex of the parabola were not at $x = 0$ or $x = 2$, then $f$ could not be injective, since moving slightly to either side of it would produce equal values. If the vertex were not at $y = 0$ or $y = 2$, then $f$ could not be surjective, since the values above (or below, depending on the direction of $f$) could not be reached.



      It is not hard to picture the possibilities now. The parabola can start in the "top-left" corner and go down, the "bottom-left" corner and go up, and so on. This requires $f(2) = 0$ in some cases, and $f(2) = 2$ in other cases.



      As a graphical reference, all of the (segments of) parabolas below satisfy the hypotheses of your question.



      Drawing of valid parabolas






      share|cite|improve this answer



























        up vote
        1
        down vote



        accepted










        Since you have the algebra-precalculus tag, you may appreciate a less "high-powered" answer.



        The graph of a quadratic function is a parabola; its vertex is the extreme value of the function, the function is symmetric about the vertex, and the function is strictly increasing or decreasing on either side of the vertex.



        If the vertex of the parabola were not at $x = 0$ or $x = 2$, then $f$ could not be injective, since moving slightly to either side of it would produce equal values. If the vertex were not at $y = 0$ or $y = 2$, then $f$ could not be surjective, since the values above (or below, depending on the direction of $f$) could not be reached.



        It is not hard to picture the possibilities now. The parabola can start in the "top-left" corner and go down, the "bottom-left" corner and go up, and so on. This requires $f(2) = 0$ in some cases, and $f(2) = 2$ in other cases.



        As a graphical reference, all of the (segments of) parabolas below satisfy the hypotheses of your question.



        Drawing of valid parabolas






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Since you have the algebra-precalculus tag, you may appreciate a less "high-powered" answer.



          The graph of a quadratic function is a parabola; its vertex is the extreme value of the function, the function is symmetric about the vertex, and the function is strictly increasing or decreasing on either side of the vertex.



          If the vertex of the parabola were not at $x = 0$ or $x = 2$, then $f$ could not be injective, since moving slightly to either side of it would produce equal values. If the vertex were not at $y = 0$ or $y = 2$, then $f$ could not be surjective, since the values above (or below, depending on the direction of $f$) could not be reached.



          It is not hard to picture the possibilities now. The parabola can start in the "top-left" corner and go down, the "bottom-left" corner and go up, and so on. This requires $f(2) = 0$ in some cases, and $f(2) = 2$ in other cases.



          As a graphical reference, all of the (segments of) parabolas below satisfy the hypotheses of your question.



          Drawing of valid parabolas






          share|cite|improve this answer















          Since you have the algebra-precalculus tag, you may appreciate a less "high-powered" answer.



          The graph of a quadratic function is a parabola; its vertex is the extreme value of the function, the function is symmetric about the vertex, and the function is strictly increasing or decreasing on either side of the vertex.



          If the vertex of the parabola were not at $x = 0$ or $x = 2$, then $f$ could not be injective, since moving slightly to either side of it would produce equal values. If the vertex were not at $y = 0$ or $y = 2$, then $f$ could not be surjective, since the values above (or below, depending on the direction of $f$) could not be reached.



          It is not hard to picture the possibilities now. The parabola can start in the "top-left" corner and go down, the "bottom-left" corner and go up, and so on. This requires $f(2) = 0$ in some cases, and $f(2) = 2$ in other cases.



          As a graphical reference, all of the (segments of) parabolas below satisfy the hypotheses of your question.



          Drawing of valid parabolas







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 22 at 17:08


























          answered Jul 22 at 15:22









          rwbogl

          716415




          716415




















              up vote
              5
              down vote













              $f(2)$ is either $0$ or $2$.



              Indeed, $f|_[0,2)$ is a continuous bijection of $[0,2)$ and $[0,2] setminus f(2)$. The former is a connected set so the latter also must be connected, which isn't true unless $f(2) in 0, 2$.



              Both cases are possible:



              For $f(2) = 2$ consider $f(x) = frac12x^2$ and for $f(2) = 0$ consider $f(x) = 2-frac12 x^2$.






              share|cite|improve this answer

























                up vote
                5
                down vote













                $f(2)$ is either $0$ or $2$.



                Indeed, $f|_[0,2)$ is a continuous bijection of $[0,2)$ and $[0,2] setminus f(2)$. The former is a connected set so the latter also must be connected, which isn't true unless $f(2) in 0, 2$.



                Both cases are possible:



                For $f(2) = 2$ consider $f(x) = frac12x^2$ and for $f(2) = 0$ consider $f(x) = 2-frac12 x^2$.






                share|cite|improve this answer























                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote









                  $f(2)$ is either $0$ or $2$.



                  Indeed, $f|_[0,2)$ is a continuous bijection of $[0,2)$ and $[0,2] setminus f(2)$. The former is a connected set so the latter also must be connected, which isn't true unless $f(2) in 0, 2$.



                  Both cases are possible:



                  For $f(2) = 2$ consider $f(x) = frac12x^2$ and for $f(2) = 0$ consider $f(x) = 2-frac12 x^2$.






                  share|cite|improve this answer













                  $f(2)$ is either $0$ or $2$.



                  Indeed, $f|_[0,2)$ is a continuous bijection of $[0,2)$ and $[0,2] setminus f(2)$. The former is a connected set so the latter also must be connected, which isn't true unless $f(2) in 0, 2$.



                  Both cases are possible:



                  For $f(2) = 2$ consider $f(x) = frac12x^2$ and for $f(2) = 0$ consider $f(x) = 2-frac12 x^2$.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 22 at 12:22









                  mechanodroid

                  22.2k52041




                  22.2k52041












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