1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stationary Points of Inflection

  1. May 31, 2004 #1
    Now, given y=x^3 -9x^2+23x-16 on the interval [-3,7]

    the maximum and minimum values would be the turning points right?

    also, a stationary point of inflextion is where the grandient is zero, with a positive or negative gradient on both sides right?

    i am asked to find the EXACT values of the x-coordinates of the points of inflection on the graph of -x^4 + 3x^3 + 5x^2 + 2x + 11

    but, there are two maximums and one minimum, not a stationary point of inflection.

    So, are minimum and maximum values points of inflections?

    Also, how do i obtain an EXACT value for the x coordinates.
     
  2. jcsd
  3. May 31, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You are mixing together the concepts "point of inflection" and "saddle point"

    A "saddle point" is a point where the derivative is zero, but where the function does not achieve a local extremal value.

    A "point of inflection" is a point where the curvature changes sign; in particular, the 2.derivative is zero at a "point of inflection".
     
  4. May 31, 2004 #3
    Since you're given an interval, then you're dealing with local extrema here. I'm not sure what you mean by 'turning point'. I suggest you take a look at the definition of a local max. and min. again.

    As arildno said, you're probably getting some of these concepts mixed up. There is a so-called second derivative test which you can use to find whether a given point on the function is a a local maximum or minimum. It basically says:

    If for some p, f'(p) = 0 then
    - If f''(p) > 0, there is a local minimum at (p, f(p))
    - If f''(p) < 0, there is a local maximum at (p, f(p))
    - If f''(p) = 0, then you're out of luck.

    Hope that helps,
    e(ho0n3
     
  5. May 31, 2004 #4
    Be careful with that. A point of inflection only occurs if the 2nd derivative changes signs, not simply if it is zero. For example, if [tex]f(x)=3x[/tex], then [tex]f''(x)=0[/tex] for all values of x, but there are clearly no inflection points.

    But if you look at [tex]f(x)=x^3[/tex], [tex]f''(x)=6x[/tex] and we see that [tex]f''(0)=0, f''(-.1)=-.6, f''(.1)=.6[/tex] and the 2nd derivative is clearly changing signs at x=0, so there is an inflection point there.
     
  6. May 31, 2004 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Hmm..that's what I meant with ("a point where the curvature changes sign"), but I see now that the last part of the sentence made the meaning ambiguous.
    Thx.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Stationary Points of Inflection
  1. Stationary waves (Replies: 4)

  2. Stationary waves (Replies: 9)

Loading...