Relationship with derivative












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$begingroup$


Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.




  1. $F(x)-F(y) = (x-y)F’(x)$

  2. $F(x)-F(y) geq (x-y)F’(x)$

  3. $F(x)-F(y) leq (x-y)F’(x)$


4.$F(x) -F(y) = F’(x) -F’(y)$



If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.






calculus limits







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    0












    $begingroup$


    Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.




    1. $F(x)-F(y) = (x-y)F’(x)$

    2. $F(x)-F(y) geq (x-y)F’(x)$

    3. $F(x)-F(y) leq (x-y)F’(x)$


    4.$F(x) -F(y) = F’(x) -F’(y)$



    If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.






    calculus limits







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    $endgroup$















      0












      0








      0





      $begingroup$


      Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.




      1. $F(x)-F(y) = (x-y)F’(x)$

      2. $F(x)-F(y) geq (x-y)F’(x)$

      3. $F(x)-F(y) leq (x-y)F’(x)$


      4.$F(x) -F(y) = F’(x) -F’(y)$



      If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.






      calculus limits







      share|cite|improve this question









      $endgroup$




      Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.




      1. $F(x)-F(y) = (x-y)F’(x)$

      2. $F(x)-F(y) geq (x-y)F’(x)$

      3. $F(x)-F(y) leq (x-y)F’(x)$


      4.$F(x) -F(y) = F’(x) -F’(y)$



      If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.






      calculus limits




      calculus limits






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      share|cite|improve this question











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      asked Jan 21 at 7:41









      user601297user601297

      37119




      37119






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Credit to Kavi Rama Murthy.



          MVT:



          For $x>y:$



          $F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.



          Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$



          Hence 1)-3)?



          4) $f(t)=t^2$;



          $f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.



          $=f'(x)-f'(y)=2(x-y)$.



          Now choose $x,y in (0,1)$(Why?) to rule out 4).



          Appended:



          Proof of option 3:



          We have MVT:



          $F(x)-F(y)= F'(a)(x-y) le$



          $F'(x)(x-y)$,



          since $F'(a)le F'(x)$, recall $y <a<x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
            $endgroup$
            – user601297
            Jan 21 at 9:54










          • $begingroup$
            user601297.Does this appended part answer your question?
            $endgroup$
            – Peter Szilas
            Jan 21 at 10:30










          • $begingroup$
            The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
            $endgroup$
            – user601297
            Jan 21 at 11:50












          • $begingroup$
            Ok after thinking about it for a while, I got the proof, thanks
            $endgroup$
            – user601297
            Jan 21 at 12:45



















          2












          $begingroup$

          3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you tell me the process to come to that conclusion?
            $endgroup$
            – user601297
            Jan 21 at 7:44











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Credit to Kavi Rama Murthy.



          MVT:



          For $x>y:$



          $F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.



          Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$



          Hence 1)-3)?



          4) $f(t)=t^2$;



          $f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.



          $=f'(x)-f'(y)=2(x-y)$.



          Now choose $x,y in (0,1)$(Why?) to rule out 4).



          Appended:



          Proof of option 3:



          We have MVT:



          $F(x)-F(y)= F'(a)(x-y) le$



          $F'(x)(x-y)$,



          since $F'(a)le F'(x)$, recall $y <a<x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
            $endgroup$
            – user601297
            Jan 21 at 9:54










          • $begingroup$
            user601297.Does this appended part answer your question?
            $endgroup$
            – Peter Szilas
            Jan 21 at 10:30










          • $begingroup$
            The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
            $endgroup$
            – user601297
            Jan 21 at 11:50












          • $begingroup$
            Ok after thinking about it for a while, I got the proof, thanks
            $endgroup$
            – user601297
            Jan 21 at 12:45
















          2












          $begingroup$

          Credit to Kavi Rama Murthy.



          MVT:



          For $x>y:$



          $F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.



          Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$



          Hence 1)-3)?



          4) $f(t)=t^2$;



          $f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.



          $=f'(x)-f'(y)=2(x-y)$.



          Now choose $x,y in (0,1)$(Why?) to rule out 4).



          Appended:



          Proof of option 3:



          We have MVT:



          $F(x)-F(y)= F'(a)(x-y) le$



          $F'(x)(x-y)$,



          since $F'(a)le F'(x)$, recall $y <a<x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
            $endgroup$
            – user601297
            Jan 21 at 9:54










          • $begingroup$
            user601297.Does this appended part answer your question?
            $endgroup$
            – Peter Szilas
            Jan 21 at 10:30










          • $begingroup$
            The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
            $endgroup$
            – user601297
            Jan 21 at 11:50












          • $begingroup$
            Ok after thinking about it for a while, I got the proof, thanks
            $endgroup$
            – user601297
            Jan 21 at 12:45














          2












          2








          2





          $begingroup$

          Credit to Kavi Rama Murthy.



          MVT:



          For $x>y:$



          $F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.



          Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$



          Hence 1)-3)?



          4) $f(t)=t^2$;



          $f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.



          $=f'(x)-f'(y)=2(x-y)$.



          Now choose $x,y in (0,1)$(Why?) to rule out 4).



          Appended:



          Proof of option 3:



          We have MVT:



          $F(x)-F(y)= F'(a)(x-y) le$



          $F'(x)(x-y)$,



          since $F'(a)le F'(x)$, recall $y <a<x$.






          share|cite|improve this answer











          $endgroup$



          Credit to Kavi Rama Murthy.



          MVT:



          For $x>y:$



          $F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.



          Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$



          Hence 1)-3)?



          4) $f(t)=t^2$;



          $f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.



          $=f'(x)-f'(y)=2(x-y)$.



          Now choose $x,y in (0,1)$(Why?) to rule out 4).



          Appended:



          Proof of option 3:



          We have MVT:



          $F(x)-F(y)= F'(a)(x-y) le$



          $F'(x)(x-y)$,



          since $F'(a)le F'(x)$, recall $y <a<x$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 21 at 15:49

























          answered Jan 21 at 8:48









          Peter SzilasPeter Szilas

          11.4k2822




          11.4k2822












          • $begingroup$
            Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
            $endgroup$
            – user601297
            Jan 21 at 9:54










          • $begingroup$
            user601297.Does this appended part answer your question?
            $endgroup$
            – Peter Szilas
            Jan 21 at 10:30










          • $begingroup$
            The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
            $endgroup$
            – user601297
            Jan 21 at 11:50












          • $begingroup$
            Ok after thinking about it for a while, I got the proof, thanks
            $endgroup$
            – user601297
            Jan 21 at 12:45


















          • $begingroup$
            Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
            $endgroup$
            – user601297
            Jan 21 at 9:54










          • $begingroup$
            user601297.Does this appended part answer your question?
            $endgroup$
            – Peter Szilas
            Jan 21 at 10:30










          • $begingroup$
            The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
            $endgroup$
            – user601297
            Jan 21 at 11:50












          • $begingroup$
            Ok after thinking about it for a while, I got the proof, thanks
            $endgroup$
            – user601297
            Jan 21 at 12:45
















          $begingroup$
          Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
          $endgroup$
          – user601297
          Jan 21 at 9:54




          $begingroup$
          Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
          $endgroup$
          – user601297
          Jan 21 at 9:54












          $begingroup$
          user601297.Does this appended part answer your question?
          $endgroup$
          – Peter Szilas
          Jan 21 at 10:30




          $begingroup$
          user601297.Does this appended part answer your question?
          $endgroup$
          – Peter Szilas
          Jan 21 at 10:30












          $begingroup$
          The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
          $endgroup$
          – user601297
          Jan 21 at 11:50






          $begingroup$
          The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
          $endgroup$
          – user601297
          Jan 21 at 11:50














          $begingroup$
          Ok after thinking about it for a while, I got the proof, thanks
          $endgroup$
          – user601297
          Jan 21 at 12:45




          $begingroup$
          Ok after thinking about it for a while, I got the proof, thanks
          $endgroup$
          – user601297
          Jan 21 at 12:45











          2












          $begingroup$

          3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you tell me the process to come to that conclusion?
            $endgroup$
            – user601297
            Jan 21 at 7:44
















          2












          $begingroup$

          3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you tell me the process to come to that conclusion?
            $endgroup$
            – user601297
            Jan 21 at 7:44














          2












          2








          2





          $begingroup$

          3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.






          share|cite|improve this answer











          $endgroup$



          3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 21 at 7:46

























          answered Jan 21 at 7:43









          Kavi Rama MurthyKavi Rama Murthy

          62.2k42262




          62.2k42262












          • $begingroup$
            Can you tell me the process to come to that conclusion?
            $endgroup$
            – user601297
            Jan 21 at 7:44


















          • $begingroup$
            Can you tell me the process to come to that conclusion?
            $endgroup$
            – user601297
            Jan 21 at 7:44
















          $begingroup$
          Can you tell me the process to come to that conclusion?
          $endgroup$
          – user601297
          Jan 21 at 7:44




          $begingroup$
          Can you tell me the process to come to that conclusion?
          $endgroup$
          – user601297
          Jan 21 at 7:44


















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