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Saddle point

  1. Apr 7, 2014 #1
    Is it necessary for some point ##x## of the function to be saddle that
    ##f'(x)=0##?
     
    Last edited: Apr 7, 2014
  2. jcsd
  3. Apr 7, 2014 #2

    Mark44

    Staff: Mentor

    What you have written isn't clear. For one thing, saddle points aren't applicable to functions of a single variable. For another, do you mean that f'(x) = 0 for some specific value of x? Or do you mean that f'(x) is identically equal to zero?

    If you meant f'(p) = 0, for some p, all this means is that at the point (p, f(p)), the tangent to the curve is horizontal. The following three functions all have horizontal tangents when x = 0.
    1. f(x) = x2 - there is a local (and global) minimum for x = 0.
    2. g(x) = x3 - there is an inflection point for x = 0. This function has no minimum and no maximum.
    3. h(x) = 1 - x2 - there is a local (and global) maximum for x = 0.

    If you meant f'(x) ##\equiv## 0 (i.e., identically equal to zero), it must be the case that f(x) = C, the graph of which is a horizontal line. This "curve" has no minimum and no maximum.
     
  4. Apr 7, 2014 #3
    Yes I mean for particular value of ##x##. For some ##x_0## is it possible situation that
    ##f'(x_0)\neq 0## and that in ##x_0## function has inflection point?
     
  5. Apr 7, 2014 #4

    Mark44

    Staff: Mentor

    Yes. Let f(x) = x1/3.
    f' is not defined at x = 0, nor is f'', but the curvature changes from concave up for x < 0 to concave down for x > 0. The fact that the curvature changes direction on either side of x = 0 is sufficient to be able to state that there is an inflection point for x = 0, which is in the domain of this function.
     
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