Let $f(x) = ln x - 5x$, for $x > 0$.












1












$begingroup$


I’m in IB Math and we are working on some calculus problems but I wanted to get extra practice so this is a problem in my book. The number in parenthesis next to the parts are the “marks” we get for the question if we get it right. So, usually that’s about how much work we have to show or how many steps to take.



Let $f(x) = ln x - 5x$, for $x > 0$.



a) Find $f’(x)$. (2)



Would this be $1/x - 5$?



b) Find $f’’(x)$. (1)



Would this be simply $1/x$?



c) Solve for $f’(x) = f’’(x)$. (2)



We set them up equal to each other. So,



$1/x - 5$ = $1/x$



The $1/x$’s cancel out and we get $-5$. Is this correct?






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The derivative of $frac 1x$ is not $frac 1x$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:30
















1












$begingroup$


I’m in IB Math and we are working on some calculus problems but I wanted to get extra practice so this is a problem in my book. The number in parenthesis next to the parts are the “marks” we get for the question if we get it right. So, usually that’s about how much work we have to show or how many steps to take.



Let $f(x) = ln x - 5x$, for $x > 0$.



a) Find $f’(x)$. (2)



Would this be $1/x - 5$?



b) Find $f’’(x)$. (1)



Would this be simply $1/x$?



c) Solve for $f’(x) = f’’(x)$. (2)



We set them up equal to each other. So,



$1/x - 5$ = $1/x$



The $1/x$’s cancel out and we get $-5$. Is this correct?






calculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The derivative of $frac 1x$ is not $frac 1x$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:30














1












1








1





$begingroup$


I’m in IB Math and we are working on some calculus problems but I wanted to get extra practice so this is a problem in my book. The number in parenthesis next to the parts are the “marks” we get for the question if we get it right. So, usually that’s about how much work we have to show or how many steps to take.



Let $f(x) = ln x - 5x$, for $x > 0$.



a) Find $f’(x)$. (2)



Would this be $1/x - 5$?



b) Find $f’’(x)$. (1)



Would this be simply $1/x$?



c) Solve for $f’(x) = f’’(x)$. (2)



We set them up equal to each other. So,



$1/x - 5$ = $1/x$



The $1/x$’s cancel out and we get $-5$. Is this correct?






calculus







share|cite|improve this question











$endgroup$




I’m in IB Math and we are working on some calculus problems but I wanted to get extra practice so this is a problem in my book. The number in parenthesis next to the parts are the “marks” we get for the question if we get it right. So, usually that’s about how much work we have to show or how many steps to take.



Let $f(x) = ln x - 5x$, for $x > 0$.



a) Find $f’(x)$. (2)



Would this be $1/x - 5$?



b) Find $f’’(x)$. (1)



Would this be simply $1/x$?



c) Solve for $f’(x) = f’’(x)$. (2)



We set them up equal to each other. So,



$1/x - 5$ = $1/x$



The $1/x$’s cancel out and we get $-5$. Is this correct?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 15:37









gimusi

92.9k84494




92.9k84494










asked Dec 13 '18 at 15:29









EllaElla

33311




33311








  • 2




    $begingroup$
    The derivative of $frac 1x$ is not $frac 1x$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:30














  • 2




    $begingroup$
    The derivative of $frac 1x$ is not $frac 1x$.
    $endgroup$
    – lulu
    Dec 13 '18 at 15:30








2




2




$begingroup$
The derivative of $frac 1x$ is not $frac 1x$.
$endgroup$
– lulu
Dec 13 '18 at 15:30




$begingroup$
The derivative of $frac 1x$ is not $frac 1x$.
$endgroup$
– lulu
Dec 13 '18 at 15:30










3 Answers
3






active

oldest

votes


















3












$begingroup$

$a)$ Correct.



$b)$ Incorrect. The derivative of $frac{1}{x}$ is not $frac{1}{x}$. You can use the Power Rule.



$$f’(x) = frac{1}{x}-5= x^{-1}-5$$



$$f’’(x) = -1x^{-2} = -x^{-2} = -frac{1}{x^2}$$



$c)$ Let



$$f’(x) = f’’(x)$$



which results in



$$frac{1}{x}-5 = -frac{1}{x^2}$$



$$x^2bigg(frac{1}{x}-5 = -frac{1}{x^2}bigg)$$



$$x-5x^2 = -1$$



$$5x^2-x-1 = 0$$



Using the Quadratic Formula yields



$$x = frac{-(-1)pmsqrt{(-1)^2-4(5)(-1)}}{2(5)}$$



$$x = frac{1pmsqrt{21}}{10}$$






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $a)$ Yes



    $b)$ No. $$f'(x)=1/x-5$$



    $$implies f''(x)=-1/x^2$$



    because $$frac{d}{dx}(1/x)=-1/x^2$$



    $c)$ $$-1/x^2=1/x-5$$



    $$5x^2-x-1=0$$



    $$implies x=frac{1+sqrt{21}}{10}quad text{or}quad frac{1-sqrt{21}}{10}$$






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      Point "a)" is ok.



      For point "b)" recall that



      $$frac1x = x^{-1} implies frac{d}{dx}(x^{-1})=-x^{-2}=-frac1{x^2}$$



      Note also that for point "c)" in nay case



      $$frac1x - 5 = frac1x iff -5=0$$



      that would mean that the equation has not solutions.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        $a)$ Correct.



        $b)$ Incorrect. The derivative of $frac{1}{x}$ is not $frac{1}{x}$. You can use the Power Rule.



        $$f’(x) = frac{1}{x}-5= x^{-1}-5$$



        $$f’’(x) = -1x^{-2} = -x^{-2} = -frac{1}{x^2}$$



        $c)$ Let



        $$f’(x) = f’’(x)$$



        which results in



        $$frac{1}{x}-5 = -frac{1}{x^2}$$



        $$x^2bigg(frac{1}{x}-5 = -frac{1}{x^2}bigg)$$



        $$x-5x^2 = -1$$



        $$5x^2-x-1 = 0$$



        Using the Quadratic Formula yields



        $$x = frac{-(-1)pmsqrt{(-1)^2-4(5)(-1)}}{2(5)}$$



        $$x = frac{1pmsqrt{21}}{10}$$






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          $a)$ Correct.



          $b)$ Incorrect. The derivative of $frac{1}{x}$ is not $frac{1}{x}$. You can use the Power Rule.



          $$f’(x) = frac{1}{x}-5= x^{-1}-5$$



          $$f’’(x) = -1x^{-2} = -x^{-2} = -frac{1}{x^2}$$



          $c)$ Let



          $$f’(x) = f’’(x)$$



          which results in



          $$frac{1}{x}-5 = -frac{1}{x^2}$$



          $$x^2bigg(frac{1}{x}-5 = -frac{1}{x^2}bigg)$$



          $$x-5x^2 = -1$$



          $$5x^2-x-1 = 0$$



          Using the Quadratic Formula yields



          $$x = frac{-(-1)pmsqrt{(-1)^2-4(5)(-1)}}{2(5)}$$



          $$x = frac{1pmsqrt{21}}{10}$$






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            $a)$ Correct.



            $b)$ Incorrect. The derivative of $frac{1}{x}$ is not $frac{1}{x}$. You can use the Power Rule.



            $$f’(x) = frac{1}{x}-5= x^{-1}-5$$



            $$f’’(x) = -1x^{-2} = -x^{-2} = -frac{1}{x^2}$$



            $c)$ Let



            $$f’(x) = f’’(x)$$



            which results in



            $$frac{1}{x}-5 = -frac{1}{x^2}$$



            $$x^2bigg(frac{1}{x}-5 = -frac{1}{x^2}bigg)$$



            $$x-5x^2 = -1$$



            $$5x^2-x-1 = 0$$



            Using the Quadratic Formula yields



            $$x = frac{-(-1)pmsqrt{(-1)^2-4(5)(-1)}}{2(5)}$$



            $$x = frac{1pmsqrt{21}}{10}$$






            share|cite|improve this answer











            $endgroup$



            $a)$ Correct.



            $b)$ Incorrect. The derivative of $frac{1}{x}$ is not $frac{1}{x}$. You can use the Power Rule.



            $$f’(x) = frac{1}{x}-5= x^{-1}-5$$



            $$f’’(x) = -1x^{-2} = -x^{-2} = -frac{1}{x^2}$$



            $c)$ Let



            $$f’(x) = f’’(x)$$



            which results in



            $$frac{1}{x}-5 = -frac{1}{x^2}$$



            $$x^2bigg(frac{1}{x}-5 = -frac{1}{x^2}bigg)$$



            $$x-5x^2 = -1$$



            $$5x^2-x-1 = 0$$



            Using the Quadratic Formula yields



            $$x = frac{-(-1)pmsqrt{(-1)^2-4(5)(-1)}}{2(5)}$$



            $$x = frac{1pmsqrt{21}}{10}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 13 '18 at 16:28

























            answered Dec 13 '18 at 16:18









            KM101KM101

            6,0351525




            6,0351525























                3












                $begingroup$

                $a)$ Yes



                $b)$ No. $$f'(x)=1/x-5$$



                $$implies f''(x)=-1/x^2$$



                because $$frac{d}{dx}(1/x)=-1/x^2$$



                $c)$ $$-1/x^2=1/x-5$$



                $$5x^2-x-1=0$$



                $$implies x=frac{1+sqrt{21}}{10}quad text{or}quad frac{1-sqrt{21}}{10}$$






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  $a)$ Yes



                  $b)$ No. $$f'(x)=1/x-5$$



                  $$implies f''(x)=-1/x^2$$



                  because $$frac{d}{dx}(1/x)=-1/x^2$$



                  $c)$ $$-1/x^2=1/x-5$$



                  $$5x^2-x-1=0$$



                  $$implies x=frac{1+sqrt{21}}{10}quad text{or}quad frac{1-sqrt{21}}{10}$$






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $a)$ Yes



                    $b)$ No. $$f'(x)=1/x-5$$



                    $$implies f''(x)=-1/x^2$$



                    because $$frac{d}{dx}(1/x)=-1/x^2$$



                    $c)$ $$-1/x^2=1/x-5$$



                    $$5x^2-x-1=0$$



                    $$implies x=frac{1+sqrt{21}}{10}quad text{or}quad frac{1-sqrt{21}}{10}$$






                    share|cite|improve this answer











                    $endgroup$



                    $a)$ Yes



                    $b)$ No. $$f'(x)=1/x-5$$



                    $$implies f''(x)=-1/x^2$$



                    because $$frac{d}{dx}(1/x)=-1/x^2$$



                    $c)$ $$-1/x^2=1/x-5$$



                    $$5x^2-x-1=0$$



                    $$implies x=frac{1+sqrt{21}}{10}quad text{or}quad frac{1-sqrt{21}}{10}$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 13 '18 at 15:40









                    gimusi

                    92.9k84494




                    92.9k84494










                    answered Dec 13 '18 at 15:35









                    Ankit KumarAnkit Kumar

                    1,494221




                    1,494221























                        1












                        $begingroup$

                        Point "a)" is ok.



                        For point "b)" recall that



                        $$frac1x = x^{-1} implies frac{d}{dx}(x^{-1})=-x^{-2}=-frac1{x^2}$$



                        Note also that for point "c)" in nay case



                        $$frac1x - 5 = frac1x iff -5=0$$



                        that would mean that the equation has not solutions.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Point "a)" is ok.



                          For point "b)" recall that



                          $$frac1x = x^{-1} implies frac{d}{dx}(x^{-1})=-x^{-2}=-frac1{x^2}$$



                          Note also that for point "c)" in nay case



                          $$frac1x - 5 = frac1x iff -5=0$$



                          that would mean that the equation has not solutions.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Point "a)" is ok.



                            For point "b)" recall that



                            $$frac1x = x^{-1} implies frac{d}{dx}(x^{-1})=-x^{-2}=-frac1{x^2}$$



                            Note also that for point "c)" in nay case



                            $$frac1x - 5 = frac1x iff -5=0$$



                            that would mean that the equation has not solutions.






                            share|cite|improve this answer









                            $endgroup$



                            Point "a)" is ok.



                            For point "b)" recall that



                            $$frac1x = x^{-1} implies frac{d}{dx}(x^{-1})=-x^{-2}=-frac1{x^2}$$



                            Note also that for point "c)" in nay case



                            $$frac1x - 5 = frac1x iff -5=0$$



                            that would mean that the equation has not solutions.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 13 '18 at 15:36









                            gimusigimusi

                            92.9k84494




                            92.9k84494






























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