Nick 發問於 科學及數學數學 · 9 年前

Point of inflection

For any quadratic function, I'm sure there has no point of inflection and there has only one stationary point either max or min.

For cubic function, there must have one point of inflection, and can have 2 stationary point or none.

But for quartic function, how many point of inflection will it have?!

and how about polynomial of degree n?

Is it true that degree of n will have (n-2) point of inflection?!

because the when we find point of inflection, we need to differentiate twice, so the power will goes down twice. therefore with n degree to start with, there will have (n-2) left.

更新:

you didnt really answer my question.

my question was "Is it true that degree of n will have (n-2) point of inflection?!"

and the way you prove the quartic function isnt really proving how many point of inflection it has got.

2 個解答

評分
  • hung
    Lv 5
    9 年前
    最愛解答

    If we consider a degree 4 polynomial f(x) ,

    f ' ' (x) = 0 may not have real root

    say f(x) = x^4 + 6x^2

    f ' ' (x) = 12x^2 + 12 = 12(x^2 + 1 )

    But f ' '(x) = 0 => x^2 + 1 = 0 , which has no real root

    so f has no pt of inflection

    consider f( x ) = ax^2 + bx + c , a =/=0

    f ' ' ( x ) = 2a cannot be 0 .

    For any quadratic function, I'm sure there has no point of inflection

    consider f(x) = ax^3 +bx^2+cx+d

    f ' '(x) = 6ax + 2b

    as a =/= 0 , f ' ' ( -2b/6a) = 0 .

    For cubic function, there must have one point of inflection

    2012-01-04 09:45:02 補充:

    my pt is that your statement "Is it true that degree of n will have (n-2) point of inflection?!"

    is wrong as I provided a counter example .

    2012-01-04 09:48:57 補充:

    For quartic function , there are three case ,

    0 , 1 or 2 points of inflection

    2012-01-04 09:49:03 補充:

    Let f(x) = ax^4 + bx^3 + cx^2 + dx + e

    f ' ' ( x ) = 12ax^2 + 6b x + 2 c

    denote 12ax^2 +6bx +2c =0 as (*)

    (*) has no real root => f has no point of inflection

    (*) has repeated root => f has 1 point of inflection

    (*) has distinct roots => f has 2 points of inflection

    2012-01-04 09:52:02 補充:

    you may also use the quadratic properties to analyze (*)

    (*) has no real root => (6b)^2 - 4(12a)(2c) < 0

    (*) has repeated root => (6b)^2 - 4(12a)(2c) = 0

    (*) has distinct roots => (6b)^2 - 4(12a)(2c) > 0

    2012-01-05 09:48:30 補充:

    Yes . You are right . In that case , we shall consider the graph locally

    near the point x = 0 . You can see that f ' ' > 0 , whenever x < 0 or x > 0 .

    that is , the second derivative does NOT change sign .

    2012-01-05 09:48:40 補充:

    Generally , when we

    determine what point(s) are pt of inflection , we can use the table to help us .

    Just like the first derivative table to find the critical points .

    It is of the form

    x <0 =0 >0

    f ' ' + 0 + In case f = x^4

    2012-01-05 09:49:35 補充:

    you can take a look for more info. about pt of inflection on this web

    http://en.wikipedia.org/wiki/Inflection_point

  • 9 年前

    I totally agree with your example given.

    But if I use f(x) = x^4 as an example.

    f'(x) = 4x^3

    f"(x) = 12x^2

    for f"(x) = 12x^2 = 0

    x = 0, therefore repeated root. but it clearly isnt a inflection point, it is only a turning point of minimum.

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