The function $g(x)=3x+ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function.












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$begingroup$



The function $g(x)=3x+ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function.




So $g'(x)=3+frac{1}{x} $ for $x>0$ But I have no idea how I'm supposed to prove that this means $g(x)$ has an inverse function. Any help will be appreciated






functions derivatives inverse-function







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    1












    $begingroup$



    The function $g(x)=3x+ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function.




    So $g'(x)=3+frac{1}{x} $ for $x>0$ But I have no idea how I'm supposed to prove that this means $g(x)$ has an inverse function. Any help will be appreciated






    functions derivatives inverse-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$



      The function $g(x)=3x+ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function.




      So $g'(x)=3+frac{1}{x} $ for $x>0$ But I have no idea how I'm supposed to prove that this means $g(x)$ has an inverse function. Any help will be appreciated






      functions derivatives inverse-function







      share|cite|improve this question









      $endgroup$





      The function $g(x)=3x+ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function.




      So $g'(x)=3+frac{1}{x} $ for $x>0$ But I have no idea how I'm supposed to prove that this means $g(x)$ has an inverse function. Any help will be appreciated






      functions derivatives inverse-function




      functions derivatives inverse-function






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      asked Dec 8 '18 at 21:38









      H.LinkhornH.Linkhorn

      401113




      401113






















          4 Answers
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          active

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          0












          $begingroup$

          Hint: It follows from what you did that $x>0implies g'(x)>0$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            What is required for a function to be invertible? It must be one-to-one and onto. Your function is $g:mathbb{R}^+rightarrowmathbb{R}$. Establishing one-to-one-ness should be doable via the derivative you calculated, and establishing onto-ness should follow from continuity, the intermediate value theorem, and the appropriate limits to $0^+$ and $infty$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Since you have the condition $x>0$ and $g'(x) > 0$ for positive $x$, then you know $g$ is strictly increasing. This means that $g$ is injective on $(0,infty)$ and has an inverse on that interval.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                Generally speaking, there are two `standard' ways to show the existence of something. The first is to construct it, and the second is to show that if we suppose that such a thing does not exist, we get a contradiction. The intuitionist in me would always recommend the former before falling back on the latter. In this case you need to show that $g'(x)=3+frac{1}{x}$. To find the inverse of a function $f$, I was taught to substitute to let $y=f(x)$, swap the roles of $x$ and $y$, then rearrange for $y$. In this case, this amounts to setting $x=g'(y)=3+frac{1}{y}$ and rearranging for $y$. Good luck!



                EDIT:
                It's been pointed out that you've been asked to find the inverse of $g$ and not of $g'$. In that case, we recall the following lemma:




                Let $f:Ito mathbb{R}$ be a strictly monotonic, continuous function on an interval $I$. Then $f$ has an inverse.




                In light of this lemma, it suffices to show that $g$ is monotonic (since $mathbb{R}^+=(0,infty)$ is an interval). To do so we require that $g'(x)>0$ (or $g'(x)<0$) $forall xin mathbb{R}^+$. The result follows quickly from you calculation.






                share|cite|improve this answer











                $endgroup$









                • 1




                  $begingroup$
                  I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                  $endgroup$
                  – Dando18
                  Dec 8 '18 at 22:06










                • $begingroup$
                  Just re-read, thanks for setting me straight
                  $endgroup$
                  – Ben
                  Dec 8 '18 at 22:08










                • $begingroup$
                  Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                  $endgroup$
                  – Dando18
                  Dec 8 '18 at 22:10










                • $begingroup$
                  I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                  $endgroup$
                  – Ben
                  Dec 8 '18 at 22:20











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                4 Answers
                4






                active

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                4 Answers
                4






                active

                oldest

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                active

                oldest

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                active

                oldest

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                0












                $begingroup$

                Hint: It follows from what you did that $x>0implies g'(x)>0$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint: It follows from what you did that $x>0implies g'(x)>0$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint: It follows from what you did that $x>0implies g'(x)>0$.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: It follows from what you did that $x>0implies g'(x)>0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 8 '18 at 21:40









                    José Carlos SantosJosé Carlos Santos

                    160k22127232




                    160k22127232























                        0












                        $begingroup$

                        What is required for a function to be invertible? It must be one-to-one and onto. Your function is $g:mathbb{R}^+rightarrowmathbb{R}$. Establishing one-to-one-ness should be doable via the derivative you calculated, and establishing onto-ness should follow from continuity, the intermediate value theorem, and the appropriate limits to $0^+$ and $infty$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          What is required for a function to be invertible? It must be one-to-one and onto. Your function is $g:mathbb{R}^+rightarrowmathbb{R}$. Establishing one-to-one-ness should be doable via the derivative you calculated, and establishing onto-ness should follow from continuity, the intermediate value theorem, and the appropriate limits to $0^+$ and $infty$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            What is required for a function to be invertible? It must be one-to-one and onto. Your function is $g:mathbb{R}^+rightarrowmathbb{R}$. Establishing one-to-one-ness should be doable via the derivative you calculated, and establishing onto-ness should follow from continuity, the intermediate value theorem, and the appropriate limits to $0^+$ and $infty$.






                            share|cite|improve this answer









                            $endgroup$



                            What is required for a function to be invertible? It must be one-to-one and onto. Your function is $g:mathbb{R}^+rightarrowmathbb{R}$. Establishing one-to-one-ness should be doable via the derivative you calculated, and establishing onto-ness should follow from continuity, the intermediate value theorem, and the appropriate limits to $0^+$ and $infty$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 '18 at 21:41









                            RandomMathGuyRandomMathGuy

                            462




                            462























                                0












                                $begingroup$

                                Since you have the condition $x>0$ and $g'(x) > 0$ for positive $x$, then you know $g$ is strictly increasing. This means that $g$ is injective on $(0,infty)$ and has an inverse on that interval.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Since you have the condition $x>0$ and $g'(x) > 0$ for positive $x$, then you know $g$ is strictly increasing. This means that $g$ is injective on $(0,infty)$ and has an inverse on that interval.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Since you have the condition $x>0$ and $g'(x) > 0$ for positive $x$, then you know $g$ is strictly increasing. This means that $g$ is injective on $(0,infty)$ and has an inverse on that interval.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Since you have the condition $x>0$ and $g'(x) > 0$ for positive $x$, then you know $g$ is strictly increasing. This means that $g$ is injective on $(0,infty)$ and has an inverse on that interval.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 8 '18 at 21:48









                                    Dando18Dando18

                                    4,67741235




                                    4,67741235























                                        0












                                        $begingroup$

                                        Generally speaking, there are two `standard' ways to show the existence of something. The first is to construct it, and the second is to show that if we suppose that such a thing does not exist, we get a contradiction. The intuitionist in me would always recommend the former before falling back on the latter. In this case you need to show that $g'(x)=3+frac{1}{x}$. To find the inverse of a function $f$, I was taught to substitute to let $y=f(x)$, swap the roles of $x$ and $y$, then rearrange for $y$. In this case, this amounts to setting $x=g'(y)=3+frac{1}{y}$ and rearranging for $y$. Good luck!



                                        EDIT:
                                        It's been pointed out that you've been asked to find the inverse of $g$ and not of $g'$. In that case, we recall the following lemma:




                                        Let $f:Ito mathbb{R}$ be a strictly monotonic, continuous function on an interval $I$. Then $f$ has an inverse.




                                        In light of this lemma, it suffices to show that $g$ is monotonic (since $mathbb{R}^+=(0,infty)$ is an interval). To do so we require that $g'(x)>0$ (or $g'(x)<0$) $forall xin mathbb{R}^+$. The result follows quickly from you calculation.






                                        share|cite|improve this answer











                                        $endgroup$









                                        • 1




                                          $begingroup$
                                          I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:06










                                        • $begingroup$
                                          Just re-read, thanks for setting me straight
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:08










                                        • $begingroup$
                                          Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:10










                                        • $begingroup$
                                          I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:20
















                                        0












                                        $begingroup$

                                        Generally speaking, there are two `standard' ways to show the existence of something. The first is to construct it, and the second is to show that if we suppose that such a thing does not exist, we get a contradiction. The intuitionist in me would always recommend the former before falling back on the latter. In this case you need to show that $g'(x)=3+frac{1}{x}$. To find the inverse of a function $f$, I was taught to substitute to let $y=f(x)$, swap the roles of $x$ and $y$, then rearrange for $y$. In this case, this amounts to setting $x=g'(y)=3+frac{1}{y}$ and rearranging for $y$. Good luck!



                                        EDIT:
                                        It's been pointed out that you've been asked to find the inverse of $g$ and not of $g'$. In that case, we recall the following lemma:




                                        Let $f:Ito mathbb{R}$ be a strictly monotonic, continuous function on an interval $I$. Then $f$ has an inverse.




                                        In light of this lemma, it suffices to show that $g$ is monotonic (since $mathbb{R}^+=(0,infty)$ is an interval). To do so we require that $g'(x)>0$ (or $g'(x)<0$) $forall xin mathbb{R}^+$. The result follows quickly from you calculation.






                                        share|cite|improve this answer











                                        $endgroup$









                                        • 1




                                          $begingroup$
                                          I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:06










                                        • $begingroup$
                                          Just re-read, thanks for setting me straight
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:08










                                        • $begingroup$
                                          Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:10










                                        • $begingroup$
                                          I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:20














                                        0












                                        0








                                        0





                                        $begingroup$

                                        Generally speaking, there are two `standard' ways to show the existence of something. The first is to construct it, and the second is to show that if we suppose that such a thing does not exist, we get a contradiction. The intuitionist in me would always recommend the former before falling back on the latter. In this case you need to show that $g'(x)=3+frac{1}{x}$. To find the inverse of a function $f$, I was taught to substitute to let $y=f(x)$, swap the roles of $x$ and $y$, then rearrange for $y$. In this case, this amounts to setting $x=g'(y)=3+frac{1}{y}$ and rearranging for $y$. Good luck!



                                        EDIT:
                                        It's been pointed out that you've been asked to find the inverse of $g$ and not of $g'$. In that case, we recall the following lemma:




                                        Let $f:Ito mathbb{R}$ be a strictly monotonic, continuous function on an interval $I$. Then $f$ has an inverse.




                                        In light of this lemma, it suffices to show that $g$ is monotonic (since $mathbb{R}^+=(0,infty)$ is an interval). To do so we require that $g'(x)>0$ (or $g'(x)<0$) $forall xin mathbb{R}^+$. The result follows quickly from you calculation.






                                        share|cite|improve this answer











                                        $endgroup$



                                        Generally speaking, there are two `standard' ways to show the existence of something. The first is to construct it, and the second is to show that if we suppose that such a thing does not exist, we get a contradiction. The intuitionist in me would always recommend the former before falling back on the latter. In this case you need to show that $g'(x)=3+frac{1}{x}$. To find the inverse of a function $f$, I was taught to substitute to let $y=f(x)$, swap the roles of $x$ and $y$, then rearrange for $y$. In this case, this amounts to setting $x=g'(y)=3+frac{1}{y}$ and rearranging for $y$. Good luck!



                                        EDIT:
                                        It's been pointed out that you've been asked to find the inverse of $g$ and not of $g'$. In that case, we recall the following lemma:




                                        Let $f:Ito mathbb{R}$ be a strictly monotonic, continuous function on an interval $I$. Then $f$ has an inverse.




                                        In light of this lemma, it suffices to show that $g$ is monotonic (since $mathbb{R}^+=(0,infty)$ is an interval). To do so we require that $g'(x)>0$ (or $g'(x)<0$) $forall xin mathbb{R}^+$. The result follows quickly from you calculation.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Dec 8 '18 at 22:19

























                                        answered Dec 8 '18 at 22:01









                                        BenBen

                                        1499




                                        1499








                                        • 1




                                          $begingroup$
                                          I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:06










                                        • $begingroup$
                                          Just re-read, thanks for setting me straight
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:08










                                        • $begingroup$
                                          Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:10










                                        • $begingroup$
                                          I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:20














                                        • 1




                                          $begingroup$
                                          I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:06










                                        • $begingroup$
                                          Just re-read, thanks for setting me straight
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:08










                                        • $begingroup$
                                          Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                                          $endgroup$
                                          – Dando18
                                          Dec 8 '18 at 22:10










                                        • $begingroup$
                                          I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                                          $endgroup$
                                          – Ben
                                          Dec 8 '18 at 22:20








                                        1




                                        1




                                        $begingroup$
                                        I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                                        $endgroup$
                                        – Dando18
                                        Dec 8 '18 at 22:06




                                        $begingroup$
                                        I don't believe they're supposed to find $(g')^{-1}$, but rather $g^{-1}$.
                                        $endgroup$
                                        – Dando18
                                        Dec 8 '18 at 22:06












                                        $begingroup$
                                        Just re-read, thanks for setting me straight
                                        $endgroup$
                                        – Ben
                                        Dec 8 '18 at 22:08




                                        $begingroup$
                                        Just re-read, thanks for setting me straight
                                        $endgroup$
                                        – Ben
                                        Dec 8 '18 at 22:08












                                        $begingroup$
                                        Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                                        $endgroup$
                                        – Dando18
                                        Dec 8 '18 at 22:10




                                        $begingroup$
                                        Additionally, computing the inverse or contradicting that $g$ has no inverse seems like overkill here when we know that any strictly monotonic function is bijective and therefore has an inverse.
                                        $endgroup$
                                        – Dando18
                                        Dec 8 '18 at 22:10












                                        $begingroup$
                                        I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                                        $endgroup$
                                        – Ben
                                        Dec 8 '18 at 22:20




                                        $begingroup$
                                        I've added an edit, as I said those are standard ways of approaching any construction proof. Of course, as you say, we actually have a direct approach in this situation.
                                        $endgroup$
                                        – Ben
                                        Dec 8 '18 at 22:20


















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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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