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Infliction points

  1. Aug 9, 2009 #1
    Infliction point exist where concavity changes. Say y=a is a vertical asymptote. If f(x) approaches infinity from the left and negative infinity from the right. Since on the left is concave up and the right is concave down. Will "a" still be considered an infliction point? or does f'' have to equal zero and then change sign?
  2. jcsd
  3. Aug 10, 2009 #2


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    I think for an inflection point, it must lie on the curve.

    y=tan(x) is like the f(x) you describe. x=π/2 is vertical asymptote.
  4. Aug 10, 2009 #3
    In order for a value to be an inflection point (not infliction) it must be in the domain and the sign of the second derivative must change across that number. The second derivative does not have to equal zero (or even be defined) at the number.

    Some examples:

    y = x3. Inflection point at x = 0, y''(0) = 0.

    [tex]y=\root{3}{x}[/tex]. Inflection point at x = 0, y''(0) undefined due to vertical tangent.

    EDIT: [tex]y=\root{3}{x}[/tex] Sheesh. Tex not rendering correctly. This should be y = cuberoot of x.

    [tex]y=\left\left\left\{ \begin{array}{cc}
    x^{2}, & \text{{if }}x<0\\
    \sqrt{x}, & \text{{if }}x\geq0\end{array}\right[/tex]. Inflection point at x = 0. y''(0) undefined due to corner.

    EDIT: [tex]y=\left\left\left\{ \begin{array}{cc}
    x^{2}, & \text{{if }}x<0\\
    \sqrt{x}, & \text{{if }}x\geq0\end{array}\right[/tex]

    y = 1/x. No inflection point at x = 0 even though y'' changes sign across 0 since 0 is not in the domain of the function.

    Last edited: Aug 10, 2009
  5. Aug 10, 2009 #4
    Forgot about those darn cusps. Would make more sense if inflection point had to be in the domain. Since it is a point lol. Thanks mates.
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