$f(2x)=f(x+y)f(x-y)$ for all $x,y in mathbb{R}$












-2












$begingroup$


$f(2x)=f(x+y)f(x-y)$ for all $x,y in mathbb{R}$, and we also know $f(10)=4, f(0)neq 0$, and $f'(0)=2$.



Then what is $f'(10)$ ...?






functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 7:22
















-2












$begingroup$


$f(2x)=f(x+y)f(x-y)$ for all $x,y in mathbb{R}$, and we also know $f(10)=4, f(0)neq 0$, and $f'(0)=2$.



Then what is $f'(10)$ ...?






functions







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 7:22














-2












-2








-2





$begingroup$


$f(2x)=f(x+y)f(x-y)$ for all $x,y in mathbb{R}$, and we also know $f(10)=4, f(0)neq 0$, and $f'(0)=2$.



Then what is $f'(10)$ ...?






functions







share|cite|improve this question











$endgroup$




$f(2x)=f(x+y)f(x-y)$ for all $x,y in mathbb{R}$, and we also know $f(10)=4, f(0)neq 0$, and $f'(0)=2$.



Then what is $f'(10)$ ...?






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 7:33









InsideOut

4,96631033




4,96631033










asked Dec 7 '18 at 7:16









Apple PieApple Pie

32




32








  • 3




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 7:22














  • 3




    $begingroup$
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    $endgroup$
    – José Carlos Santos
    Dec 7 '18 at 7:22








3




3




$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 7:22




$begingroup$
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 7:22










3 Answers
3






active

oldest

votes


















0












$begingroup$

Differntiate w.r.t. $y$ to get $f(x+y)f'(x-y)=f'(x+y)f(x-y)$. Put $x=y$ to get $2f(2x)=f'(2x)f(0)$. Hence $2f(x)=f'(x)f(0)$ for all $x$. Solving this we get $f(x) =ae^{cx}$. Can you take it from here?.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Let $x=y=0$, then
    $$f(0)=f(0)f(0)$$
    which implies $f(0)=1$ when $f(0)neq 0$.



    If $f(x)$ is differentiable then we have
    $$2f'(2x)=f'(x+y)f(x-y)+f(x+y)f'(x-y)$$
    let $x=y=5$, then
    $$2f'(10)=f'(10)f(0)+f(10)f'(0)=f'(10)+8$$
    hence we have $f'(10)=8$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can u tell me what are u differentiating the equation with respect to ?
      $endgroup$
      – Apple Pie
      Dec 8 '18 at 14:09










    • $begingroup$
      with respect to $x$
      $endgroup$
      – Lau
      Dec 9 '18 at 2:05










    • $begingroup$
      Sry but im just confused why isn’t there a y’. Is it considered constant
      $endgroup$
      – Apple Pie
      Dec 9 '18 at 2:58












    • $begingroup$
      I think $x,y$ are independent.
      $endgroup$
      – Lau
      Dec 9 '18 at 6:35



















    0












    $begingroup$

    We need not assume that $f$ is differentiable or even continuous except at $0$.



    With $x=frac{u+v}2$, $y=frac{u-v}2$, we obtain
    $$tag1 f(u+v)=f(u)f(v)qquad text{for all }u,vinBbb R .$$
    Then also
    $$tag2operatorname{sgn}(f(u+v))= operatorname{sgn}(f(u))operatorname{sgn}(f(v))qquad text{for all }u,vinBbb R .$$
    As $f$ is continuous at $0$ and $f(0)ne 0$, $operatorname{sgn}circ f$ is constant in a neighbourhood of $0$. Then from $(2)$ with $vapprox 0$, we see that $operatorname{sgn}circ f$ is locally constant, hence constant throughout, and finally must be $=1$, i.e.,
    $$tag3f(x)>0qquad text{for all }xinBbb R.$$
    Hence we can define $g(x)=ln f(x)$ for all $xin Bbb R$ and conclude from $(1)$ that $g$ is additive.
    It is well-known that continuity at a single point (as here implied by the existence of $f'(0)$) implies that all solutions are of the form
    $$ g(x)=cx.$$
    Therefore
    $$ f(x)=e^{cx}$$
    for a suitable constant $c$.
    It follows that $f$ is smooth and that
    $$ f'(x)=cf(x)$$
    so that
    $$f'(10)=frac{f'(0)}{f(0)}f(10)=8. $$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Differntiate w.r.t. $y$ to get $f(x+y)f'(x-y)=f'(x+y)f(x-y)$. Put $x=y$ to get $2f(2x)=f'(2x)f(0)$. Hence $2f(x)=f'(x)f(0)$ for all $x$. Solving this we get $f(x) =ae^{cx}$. Can you take it from here?.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Differntiate w.r.t. $y$ to get $f(x+y)f'(x-y)=f'(x+y)f(x-y)$. Put $x=y$ to get $2f(2x)=f'(2x)f(0)$. Hence $2f(x)=f'(x)f(0)$ for all $x$. Solving this we get $f(x) =ae^{cx}$. Can you take it from here?.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Differntiate w.r.t. $y$ to get $f(x+y)f'(x-y)=f'(x+y)f(x-y)$. Put $x=y$ to get $2f(2x)=f'(2x)f(0)$. Hence $2f(x)=f'(x)f(0)$ for all $x$. Solving this we get $f(x) =ae^{cx}$. Can you take it from here?.






          share|cite|improve this answer









          $endgroup$



          Differntiate w.r.t. $y$ to get $f(x+y)f'(x-y)=f'(x+y)f(x-y)$. Put $x=y$ to get $2f(2x)=f'(2x)f(0)$. Hence $2f(x)=f'(x)f(0)$ for all $x$. Solving this we get $f(x) =ae^{cx}$. Can you take it from here?.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 7:26









          Kavi Rama MurthyKavi Rama Murthy

          58.2k42160




          58.2k42160























              2












              $begingroup$

              Let $x=y=0$, then
              $$f(0)=f(0)f(0)$$
              which implies $f(0)=1$ when $f(0)neq 0$.



              If $f(x)$ is differentiable then we have
              $$2f'(2x)=f'(x+y)f(x-y)+f(x+y)f'(x-y)$$
              let $x=y=5$, then
              $$2f'(10)=f'(10)f(0)+f(10)f'(0)=f'(10)+8$$
              hence we have $f'(10)=8$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Can u tell me what are u differentiating the equation with respect to ?
                $endgroup$
                – Apple Pie
                Dec 8 '18 at 14:09










              • $begingroup$
                with respect to $x$
                $endgroup$
                – Lau
                Dec 9 '18 at 2:05










              • $begingroup$
                Sry but im just confused why isn’t there a y’. Is it considered constant
                $endgroup$
                – Apple Pie
                Dec 9 '18 at 2:58












              • $begingroup$
                I think $x,y$ are independent.
                $endgroup$
                – Lau
                Dec 9 '18 at 6:35
















              2












              $begingroup$

              Let $x=y=0$, then
              $$f(0)=f(0)f(0)$$
              which implies $f(0)=1$ when $f(0)neq 0$.



              If $f(x)$ is differentiable then we have
              $$2f'(2x)=f'(x+y)f(x-y)+f(x+y)f'(x-y)$$
              let $x=y=5$, then
              $$2f'(10)=f'(10)f(0)+f(10)f'(0)=f'(10)+8$$
              hence we have $f'(10)=8$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Can u tell me what are u differentiating the equation with respect to ?
                $endgroup$
                – Apple Pie
                Dec 8 '18 at 14:09










              • $begingroup$
                with respect to $x$
                $endgroup$
                – Lau
                Dec 9 '18 at 2:05










              • $begingroup$
                Sry but im just confused why isn’t there a y’. Is it considered constant
                $endgroup$
                – Apple Pie
                Dec 9 '18 at 2:58












              • $begingroup$
                I think $x,y$ are independent.
                $endgroup$
                – Lau
                Dec 9 '18 at 6:35














              2












              2








              2





              $begingroup$

              Let $x=y=0$, then
              $$f(0)=f(0)f(0)$$
              which implies $f(0)=1$ when $f(0)neq 0$.



              If $f(x)$ is differentiable then we have
              $$2f'(2x)=f'(x+y)f(x-y)+f(x+y)f'(x-y)$$
              let $x=y=5$, then
              $$2f'(10)=f'(10)f(0)+f(10)f'(0)=f'(10)+8$$
              hence we have $f'(10)=8$.






              share|cite|improve this answer









              $endgroup$



              Let $x=y=0$, then
              $$f(0)=f(0)f(0)$$
              which implies $f(0)=1$ when $f(0)neq 0$.



              If $f(x)$ is differentiable then we have
              $$2f'(2x)=f'(x+y)f(x-y)+f(x+y)f'(x-y)$$
              let $x=y=5$, then
              $$2f'(10)=f'(10)f(0)+f(10)f'(0)=f'(10)+8$$
              hence we have $f'(10)=8$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 7 '18 at 7:26









              LauLau

              527315




              527315












              • $begingroup$
                Can u tell me what are u differentiating the equation with respect to ?
                $endgroup$
                – Apple Pie
                Dec 8 '18 at 14:09










              • $begingroup$
                with respect to $x$
                $endgroup$
                – Lau
                Dec 9 '18 at 2:05










              • $begingroup$
                Sry but im just confused why isn’t there a y’. Is it considered constant
                $endgroup$
                – Apple Pie
                Dec 9 '18 at 2:58












              • $begingroup$
                I think $x,y$ are independent.
                $endgroup$
                – Lau
                Dec 9 '18 at 6:35


















              • $begingroup$
                Can u tell me what are u differentiating the equation with respect to ?
                $endgroup$
                – Apple Pie
                Dec 8 '18 at 14:09










              • $begingroup$
                with respect to $x$
                $endgroup$
                – Lau
                Dec 9 '18 at 2:05










              • $begingroup$
                Sry but im just confused why isn’t there a y’. Is it considered constant
                $endgroup$
                – Apple Pie
                Dec 9 '18 at 2:58












              • $begingroup$
                I think $x,y$ are independent.
                $endgroup$
                – Lau
                Dec 9 '18 at 6:35
















              $begingroup$
              Can u tell me what are u differentiating the equation with respect to ?
              $endgroup$
              – Apple Pie
              Dec 8 '18 at 14:09




              $begingroup$
              Can u tell me what are u differentiating the equation with respect to ?
              $endgroup$
              – Apple Pie
              Dec 8 '18 at 14:09












              $begingroup$
              with respect to $x$
              $endgroup$
              – Lau
              Dec 9 '18 at 2:05




              $begingroup$
              with respect to $x$
              $endgroup$
              – Lau
              Dec 9 '18 at 2:05












              $begingroup$
              Sry but im just confused why isn’t there a y’. Is it considered constant
              $endgroup$
              – Apple Pie
              Dec 9 '18 at 2:58






              $begingroup$
              Sry but im just confused why isn’t there a y’. Is it considered constant
              $endgroup$
              – Apple Pie
              Dec 9 '18 at 2:58














              $begingroup$
              I think $x,y$ are independent.
              $endgroup$
              – Lau
              Dec 9 '18 at 6:35




              $begingroup$
              I think $x,y$ are independent.
              $endgroup$
              – Lau
              Dec 9 '18 at 6:35











              0












              $begingroup$

              We need not assume that $f$ is differentiable or even continuous except at $0$.



              With $x=frac{u+v}2$, $y=frac{u-v}2$, we obtain
              $$tag1 f(u+v)=f(u)f(v)qquad text{for all }u,vinBbb R .$$
              Then also
              $$tag2operatorname{sgn}(f(u+v))= operatorname{sgn}(f(u))operatorname{sgn}(f(v))qquad text{for all }u,vinBbb R .$$
              As $f$ is continuous at $0$ and $f(0)ne 0$, $operatorname{sgn}circ f$ is constant in a neighbourhood of $0$. Then from $(2)$ with $vapprox 0$, we see that $operatorname{sgn}circ f$ is locally constant, hence constant throughout, and finally must be $=1$, i.e.,
              $$tag3f(x)>0qquad text{for all }xinBbb R.$$
              Hence we can define $g(x)=ln f(x)$ for all $xin Bbb R$ and conclude from $(1)$ that $g$ is additive.
              It is well-known that continuity at a single point (as here implied by the existence of $f'(0)$) implies that all solutions are of the form
              $$ g(x)=cx.$$
              Therefore
              $$ f(x)=e^{cx}$$
              for a suitable constant $c$.
              It follows that $f$ is smooth and that
              $$ f'(x)=cf(x)$$
              so that
              $$f'(10)=frac{f'(0)}{f(0)}f(10)=8. $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                We need not assume that $f$ is differentiable or even continuous except at $0$.



                With $x=frac{u+v}2$, $y=frac{u-v}2$, we obtain
                $$tag1 f(u+v)=f(u)f(v)qquad text{for all }u,vinBbb R .$$
                Then also
                $$tag2operatorname{sgn}(f(u+v))= operatorname{sgn}(f(u))operatorname{sgn}(f(v))qquad text{for all }u,vinBbb R .$$
                As $f$ is continuous at $0$ and $f(0)ne 0$, $operatorname{sgn}circ f$ is constant in a neighbourhood of $0$. Then from $(2)$ with $vapprox 0$, we see that $operatorname{sgn}circ f$ is locally constant, hence constant throughout, and finally must be $=1$, i.e.,
                $$tag3f(x)>0qquad text{for all }xinBbb R.$$
                Hence we can define $g(x)=ln f(x)$ for all $xin Bbb R$ and conclude from $(1)$ that $g$ is additive.
                It is well-known that continuity at a single point (as here implied by the existence of $f'(0)$) implies that all solutions are of the form
                $$ g(x)=cx.$$
                Therefore
                $$ f(x)=e^{cx}$$
                for a suitable constant $c$.
                It follows that $f$ is smooth and that
                $$ f'(x)=cf(x)$$
                so that
                $$f'(10)=frac{f'(0)}{f(0)}f(10)=8. $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We need not assume that $f$ is differentiable or even continuous except at $0$.



                  With $x=frac{u+v}2$, $y=frac{u-v}2$, we obtain
                  $$tag1 f(u+v)=f(u)f(v)qquad text{for all }u,vinBbb R .$$
                  Then also
                  $$tag2operatorname{sgn}(f(u+v))= operatorname{sgn}(f(u))operatorname{sgn}(f(v))qquad text{for all }u,vinBbb R .$$
                  As $f$ is continuous at $0$ and $f(0)ne 0$, $operatorname{sgn}circ f$ is constant in a neighbourhood of $0$. Then from $(2)$ with $vapprox 0$, we see that $operatorname{sgn}circ f$ is locally constant, hence constant throughout, and finally must be $=1$, i.e.,
                  $$tag3f(x)>0qquad text{for all }xinBbb R.$$
                  Hence we can define $g(x)=ln f(x)$ for all $xin Bbb R$ and conclude from $(1)$ that $g$ is additive.
                  It is well-known that continuity at a single point (as here implied by the existence of $f'(0)$) implies that all solutions are of the form
                  $$ g(x)=cx.$$
                  Therefore
                  $$ f(x)=e^{cx}$$
                  for a suitable constant $c$.
                  It follows that $f$ is smooth and that
                  $$ f'(x)=cf(x)$$
                  so that
                  $$f'(10)=frac{f'(0)}{f(0)}f(10)=8. $$






                  share|cite|improve this answer









                  $endgroup$



                  We need not assume that $f$ is differentiable or even continuous except at $0$.



                  With $x=frac{u+v}2$, $y=frac{u-v}2$, we obtain
                  $$tag1 f(u+v)=f(u)f(v)qquad text{for all }u,vinBbb R .$$
                  Then also
                  $$tag2operatorname{sgn}(f(u+v))= operatorname{sgn}(f(u))operatorname{sgn}(f(v))qquad text{for all }u,vinBbb R .$$
                  As $f$ is continuous at $0$ and $f(0)ne 0$, $operatorname{sgn}circ f$ is constant in a neighbourhood of $0$. Then from $(2)$ with $vapprox 0$, we see that $operatorname{sgn}circ f$ is locally constant, hence constant throughout, and finally must be $=1$, i.e.,
                  $$tag3f(x)>0qquad text{for all }xinBbb R.$$
                  Hence we can define $g(x)=ln f(x)$ for all $xin Bbb R$ and conclude from $(1)$ that $g$ is additive.
                  It is well-known that continuity at a single point (as here implied by the existence of $f'(0)$) implies that all solutions are of the form
                  $$ g(x)=cx.$$
                  Therefore
                  $$ f(x)=e^{cx}$$
                  for a suitable constant $c$.
                  It follows that $f$ is smooth and that
                  $$ f'(x)=cf(x)$$
                  so that
                  $$f'(10)=frac{f'(0)}{f(0)}f(10)=8. $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 7 '18 at 20:27









                  Hagen von EitzenHagen von Eitzen

                  279k23271501




                  279k23271501






























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                      Willebadessen

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                      Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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                      Residenzschloss Arolsen

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                      Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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