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Critical Numbers Confusion

  1. Mar 3, 2013 #1
    I am a bit confused over something that should be relatively easy to research , however, I am having a hard time finding a direct answer to my question.

    When finding the extrema of a function , we find at what points the first derivative is 0 or undefined .. with the stipulation , if Im not mistaken , that the function itself IS defined at those values(continuous but not necessarily differentiable at those points) .. if they are not , then they will not be critical numbers per the definition . .

    when finding concavity / points of inflection .. i am assuming the continuity requirement for the original function with respect to the critical numbers of the second derivative is dropped . am i right ? ..also , can I automatically assume if there are discontinuties in the original function , those will serve as critical numbers for sake of determining concavity ?

    i am getting conflicting information but my intuition tells me this has to be so . just by looking at a couple graphs.concavity changes between vertical asymptotes. but i just wanted to make sure .. my book sucks. =D
  2. jcsd
  3. Mar 4, 2013 #2
    Finding the critical points of a function means where the function's derivative is either zero or undefined. So when you get a set of critical points, you are solving for both where the derivative is 0 and/or undefined.

    Take the tangent function for example, which has a vertical asymptote every pi/2 + pi*k where k is an integer. Solving for critical numbers, we would get sec(x)^2 = 0. Since the secant never equals 0, the tangent never has a horizontal tangent line. But the secant is undefined at pi/2 + pi*k, which gives us where the derivative is undefined, because of the vertical asymptotes. To be clear, these asymptotes are included in these critical numbers, but are not extrema.

    To solve for inflection points, we take the second derivative. Thus, we have 2tanx(secx)^2 = 0. 2(secx)^2 will never be zero, so we cancel it out. tanx = 0 at pi/4 + pi*k. However, because of the secant, the second derivative is also undefined at pi/2 + pi*k, and the function changes concavity at both of these sets. So, to clarify: the second derivative's zeros and undefined values gives us where the concavity changes. But only the zeros are considered inflection points, because the function must be defined in order for an inflection point to exist.

    Hope this helps.
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