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Homework Help: 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

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    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

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    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.
     
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