Inflection point for function with fractional exponents

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$f(x) = 4x^left(1/3right)-x^left(4/3right)$.



I correctly get $f'(x) = 4(1-x)/(3x^left(2/3right)) $ and correctly calculate the second derivative as $f''(x) =- 4 (x+2)/(9x^left(5/3right))$.



It is clear to me that there is an inflection point at $x=-2$ since this value of $x$ makes the second derivative zero. The text shows that there is also an inflection point at $x=1$. I see that this value makes the first derivative $=0$, but I don't understand why this causes an inflection point. Can anyone help clarify this point?






calculus







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    up vote
    4
    down vote

    favorite












    $f(x) = 4x^left(1/3right)-x^left(4/3right)$.



    I correctly get $f'(x) = 4(1-x)/(3x^left(2/3right)) $ and correctly calculate the second derivative as $f''(x) =- 4 (x+2)/(9x^left(5/3right))$.



    It is clear to me that there is an inflection point at $x=-2$ since this value of $x$ makes the second derivative zero. The text shows that there is also an inflection point at $x=1$. I see that this value makes the first derivative $=0$, but I don't understand why this causes an inflection point. Can anyone help clarify this point?






    calculus







    share|cite|improve this question























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      $f(x) = 4x^left(1/3right)-x^left(4/3right)$.



      I correctly get $f'(x) = 4(1-x)/(3x^left(2/3right)) $ and correctly calculate the second derivative as $f''(x) =- 4 (x+2)/(9x^left(5/3right))$.



      It is clear to me that there is an inflection point at $x=-2$ since this value of $x$ makes the second derivative zero. The text shows that there is also an inflection point at $x=1$. I see that this value makes the first derivative $=0$, but I don't understand why this causes an inflection point. Can anyone help clarify this point?






      calculus







      share|cite|improve this question













      $f(x) = 4x^left(1/3right)-x^left(4/3right)$.



      I correctly get $f'(x) = 4(1-x)/(3x^left(2/3right)) $ and correctly calculate the second derivative as $f''(x) =- 4 (x+2)/(9x^left(5/3right))$.



      It is clear to me that there is an inflection point at $x=-2$ since this value of $x$ makes the second derivative zero. The text shows that there is also an inflection point at $x=1$. I see that this value makes the first derivative $=0$, but I don't understand why this causes an inflection point. Can anyone help clarify this point?






      calculus




      calculus






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      asked 2 hours ago









      user163862

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      7441914




















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          Because $f''(1)<0$, this point is a local maximum, not an inflection point, and the book is wrong.






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            $x=1$ is not an inflection point (as you can see from the graph below).



            enter image description here



            On the other hand you missed another inflection point $x=0$! Perhaps your book mistakenly wrote $x=1$ instead of $x=0$. The reason you missed $0$ is because the function is not differentiable at $0$ (and for such points, the condition of $f''(x)=0$ for inflection points obviously fails).






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              2 Answers
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              active

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              2 Answers
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              active

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              active

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              up vote
              2
              down vote



              accepted










              Because $f''(1)<0$, this point is a local maximum, not an inflection point, and the book is wrong.






              share|cite|improve this answer
























                up vote
                2
                down vote



                accepted










                Because $f''(1)<0$, this point is a local maximum, not an inflection point, and the book is wrong.






                share|cite|improve this answer






















                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  Because $f''(1)<0$, this point is a local maximum, not an inflection point, and the book is wrong.






                  share|cite|improve this answer












                  Because $f''(1)<0$, this point is a local maximum, not an inflection point, and the book is wrong.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Parcly Taxel

                  37.9k137096




                  37.9k137096




















                      up vote
                      0
                      down vote













                      $x=1$ is not an inflection point (as you can see from the graph below).



                      enter image description here



                      On the other hand you missed another inflection point $x=0$! Perhaps your book mistakenly wrote $x=1$ instead of $x=0$. The reason you missed $0$ is because the function is not differentiable at $0$ (and for such points, the condition of $f''(x)=0$ for inflection points obviously fails).






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        $x=1$ is not an inflection point (as you can see from the graph below).



                        enter image description here



                        On the other hand you missed another inflection point $x=0$! Perhaps your book mistakenly wrote $x=1$ instead of $x=0$. The reason you missed $0$ is because the function is not differentiable at $0$ (and for such points, the condition of $f''(x)=0$ for inflection points obviously fails).






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          $x=1$ is not an inflection point (as you can see from the graph below).



                          enter image description here



                          On the other hand you missed another inflection point $x=0$! Perhaps your book mistakenly wrote $x=1$ instead of $x=0$. The reason you missed $0$ is because the function is not differentiable at $0$ (and for such points, the condition of $f''(x)=0$ for inflection points obviously fails).






                          share|cite|improve this answer












                          $x=1$ is not an inflection point (as you can see from the graph below).



                          enter image description here



                          On the other hand you missed another inflection point $x=0$! Perhaps your book mistakenly wrote $x=1$ instead of $x=0$. The reason you missed $0$ is because the function is not differentiable at $0$ (and for such points, the condition of $f''(x)=0$ for inflection points obviously fails).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          yathish

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