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Finding points of inflection

  1. Apr 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose the graph f''(x) of a function is given by:
    (see attachment)

    (a) Find all points of inflection of f(x)

    3. The attempt at a solution
    First I figured, by looking at the graph and seeing the intercept points [(-2,0),(0,0),(2,0)] that f''(x) = [tex]x^{4}-4x^{2}[/tex]
    solving for f''(x)=0 gives us x = -2 or 0 or 2.
    Between -2 < x < 2 there is no change of sign, which indicates no point of inflection.
    Thus the points of inflection for f(x) are at x = -2 and x = 2.
    But how can I find the y intercepts with just what I have? I only know how to find them by putting x back into f(x). As we don't have f(x) here, I'm stuck. Integrating back doesn't help because of the unknown variables.
    Or am I trying too hard, and what I've done is the answer?

    Attached Files:

  2. jcsd
  3. Apr 22, 2008 #2

    Gib Z

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    Homework Helper

    Well, remember that an inflection point of f(x) is where the first derivative is at either a maximum or minimum. And find first the maxima and minima of the first derivative, we must solve the zeros of the third derivative =D
  4. Apr 22, 2008 #3


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    Staff Emeritus
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    Okay, that looks good.

    ?? If f(x)= x4 - 4x2 then f '(x)= 4x3- 8x and f "(x)= 12x2- 8. That is not 0 at x= 0, 2, and -2!

  5. Apr 22, 2008 #4
    third derivative is [tex]4x^{3}-8x[/tex]
    solving for f'''(x) = 0 gives us
    x = +/-[tex]\sqrt{2}[/tex] and x = 0.

    minima points of f''(x) are at x = (-[tex]\sqrt{2}[/tex], -4) & ([tex]\sqrt{2}[/tex], -4)
    maxima point of f''(x) is (0, 0).

    I still confused here! How does this help me find the y co-ordinates for inflection points of f(x)?
  6. Apr 22, 2008 #5
    the graph x4 - 4x2 is the second derivative, not f(x).
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