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Critical Numbers

  1. Oct 31, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    If f'(x) = x^2(x-1)^2(x-3)^2, how many local extrema does f(x) have?

    2. Relevant equations

    Extrema occur at critical points.

    Critical points are values of x such that f'(x) = 0 or = ±∞

    3. The attempt at a solution

    Not all values that zero the derivative are critical points. For example, if the problem had been f'(x) = -1/x^2, 1/x would have been the original function. 0 would seem like a critical point since it causes the derivative to approach negative infinity. However, it cannot be a critical point since 0 is not on the domain of the original function.

    Is there any case to worry with this particular problem, however? I don't think so because the derivative appears to be a polynomial and the original function appears to also be a polynomial and polynomials are continuous through all real numbers.

    Am I correct? Should I just say the critical points are x = 0, x = 1, and x = 3 and rest assured they exist on the domain of f(x) since f(x) appears to be a polynomial given its polynomial derivative?

    ----

    Also, regardless of the critical numbers, the derivative appears to be all positive throughout its domain, making the discussion of relative extrema moot since there are no changes in the sign of the first derivative. All the terms are squared and we know that any term squared is a positive number.
     
  2. jcsd
  3. Oct 31, 2013 #2

    mfb

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    Yes.

    The domain is not an issue, right.

    Good, as this settles the question about saddle points.
     
  4. Oct 31, 2013 #3

    Qube

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    What's a saddle point? I'm getting something about a stationary point ... what does stationary mean in this context?
     
  5. Nov 1, 2013 #4

    Ray Vickson

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    A saddle point is a stationary point that is neither a maximum nor a minimum; for example, the point x = 0 is a saddle point of f(x) = x^3.

    Anyway, you have missed two other critical points.
     
  6. Nov 1, 2013 #5

    mfb

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    Be careful, the function given in the first post is f', not f itself.
     
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