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Can we assume $f$ is linear?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Common method of calculating zero places of quadratic and linear function.Why can we assume WLOG that $x$ is zero?How can a unit step function be differentiable??Linear function definitionIf $f( cos^2(x) ) = cos^2(x)$ can I assume that $f(x) = x$?Can someone explain to a calculus student what “dual space is the space of linear functions” mean?Graphing systems of linear equations.Suppose f : R → R is continuous. Let λ be a positive real number, and assume that for everyIs $x^-1$ a linear function?If both $f(x)$ and $g(x)$ are one to one functions and $g(f(x))= f(g(x)) = x$, why does this prove they are inverse of each other?










0












$begingroup$


I read a hint of a question:



If $f(6)-f(3)=9$ then $f(12)-f(2)=...$



The hint: By assuming that $f(x)=ax+b$.



I'm confused why we can assume $f$ as linear function?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Without further assumptions on $f$, the question does not make sense, nor does the hint. Are you sure you haven't missed some part of the exercise?
    $endgroup$
    – MisterRiemann
    Mar 31 at 13:28











  • $begingroup$
    You sure it was a hint? Maybe it just said assume it's linear, and then find the value....
    $endgroup$
    – Displayname
    Mar 31 at 13:34















0












$begingroup$


I read a hint of a question:



If $f(6)-f(3)=9$ then $f(12)-f(2)=...$



The hint: By assuming that $f(x)=ax+b$.



I'm confused why we can assume $f$ as linear function?










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    Without further assumptions on $f$, the question does not make sense, nor does the hint. Are you sure you haven't missed some part of the exercise?
    $endgroup$
    – MisterRiemann
    Mar 31 at 13:28











  • $begingroup$
    You sure it was a hint? Maybe it just said assume it's linear, and then find the value....
    $endgroup$
    – Displayname
    Mar 31 at 13:34













0












0








0


1



$begingroup$


I read a hint of a question:



If $f(6)-f(3)=9$ then $f(12)-f(2)=...$



The hint: By assuming that $f(x)=ax+b$.



I'm confused why we can assume $f$ as linear function?










share|cite|improve this question











$endgroup$




I read a hint of a question:



If $f(6)-f(3)=9$ then $f(12)-f(2)=...$



The hint: By assuming that $f(x)=ax+b$.



I'm confused why we can assume $f$ as linear function?







calculus algebra-precalculus functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 31 at 13:38









Bernard

124k741117




124k741117










asked Mar 31 at 13:27









TinaTina

111




111







  • 2




    $begingroup$
    Without further assumptions on $f$, the question does not make sense, nor does the hint. Are you sure you haven't missed some part of the exercise?
    $endgroup$
    – MisterRiemann
    Mar 31 at 13:28











  • $begingroup$
    You sure it was a hint? Maybe it just said assume it's linear, and then find the value....
    $endgroup$
    – Displayname
    Mar 31 at 13:34












  • 2




    $begingroup$
    Without further assumptions on $f$, the question does not make sense, nor does the hint. Are you sure you haven't missed some part of the exercise?
    $endgroup$
    – MisterRiemann
    Mar 31 at 13:28











  • $begingroup$
    You sure it was a hint? Maybe it just said assume it's linear, and then find the value....
    $endgroup$
    – Displayname
    Mar 31 at 13:34







2




2




$begingroup$
Without further assumptions on $f$, the question does not make sense, nor does the hint. Are you sure you haven't missed some part of the exercise?
$endgroup$
– MisterRiemann
Mar 31 at 13:28





$begingroup$
Without further assumptions on $f$, the question does not make sense, nor does the hint. Are you sure you haven't missed some part of the exercise?
$endgroup$
– MisterRiemann
Mar 31 at 13:28













$begingroup$
You sure it was a hint? Maybe it just said assume it's linear, and then find the value....
$endgroup$
– Displayname
Mar 31 at 13:34




$begingroup$
You sure it was a hint? Maybe it just said assume it's linear, and then find the value....
$endgroup$
– Displayname
Mar 31 at 13:34










1 Answer
1






active

oldest

votes


















2












$begingroup$

No, you can not.



Say $f(x)= 1over 3x^2$, then $f(6)-f(3)= 9$



or $f(x) = sqrt33x-98$, then again $f(6)-f(3)= 9$






share|cite|improve this answer









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    1 Answer
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    2












    $begingroup$

    No, you can not.



    Say $f(x)= 1over 3x^2$, then $f(6)-f(3)= 9$



    or $f(x) = sqrt33x-98$, then again $f(6)-f(3)= 9$






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      No, you can not.



      Say $f(x)= 1over 3x^2$, then $f(6)-f(3)= 9$



      or $f(x) = sqrt33x-98$, then again $f(6)-f(3)= 9$






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        No, you can not.



        Say $f(x)= 1over 3x^2$, then $f(6)-f(3)= 9$



        or $f(x) = sqrt33x-98$, then again $f(6)-f(3)= 9$






        share|cite|improve this answer









        $endgroup$



        No, you can not.



        Say $f(x)= 1over 3x^2$, then $f(6)-f(3)= 9$



        or $f(x) = sqrt33x-98$, then again $f(6)-f(3)= 9$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 31 at 13:32









        Maria MazurMaria Mazur

        49.9k1361125




        49.9k1361125



























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