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Homework Help: Inflection points

  1. Aug 16, 2007 #1
    1. The problem statement, all variables and given/known data

    The inflection points are where the function changes its concavity, and can be found through the second derivative of the function... so, I have been given this equation:

    f(x) = x^4 - 2x^2 - 1

    and I have to find the inflection points.

    2. Relevant equations

    3. The attempt at a solution

    I derived it twice to get the second derivative:

    f`(x) = 4x^3 - 4x
    f``(x) = 12x^2 - 4

    which can be written as:
    4(3x^2 - 1)

    I ran a quadratic equation and got

    x = +- (sqrt(3) / 6)

    So my question is, are these the inflection points, since the function is continuous on those point...

    Is the difference between Inflection points and critical points that the critical points tell you where the slope reaches 0 if there is continuity at that point, and that they allow you to find whether the function is increasing or decreasing in those intervals, and the inflection points just tell you where exactly the function switches concavity? What more can inflection points do for you?
  2. jcsd
  3. Aug 16, 2007 #2


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    I get:
    3x^2 - 1 = 0
    x^2 = 1/3
    x = +- 1/sqrt(3) or x = +-sqrt(3)/3

    If the derivative at a point exists, then the function is continuous at that point... if it isn't continuous, then the derivative isn't defined (unless we talk about one-sided derivatives)

    Critical points are when the derivative of the function is 0 or undefined.

    Inflection points are when the second derivative is 0... they show you the point(s) when the slope of the function stops increasing and starts decreasing, or stops decreasing and starts increasing...

    Have a look at:

  4. Aug 16, 2007 #3
    I think there is a confusion in the vocabulary you use

    inflection points are as you said where concavity changes and obtained by equating second derivative zero
    4(3x^2 - 1)=0 implies x=-+ 1/sqrt(3)

    what you confuse is probably stationary points. stationary points are where the first derivative is zero
    so f'(x) = 4x^3 - 4x=0 implies x= 1 or x=-1
    Critical points are more general than stationary points . they include stationary points and points where derivative is not defined.
    for example for f(x)=|x| 0 is critical point because derivative do not exist there.But in your question there is no point of this type since you have a polynomial which are always wellbehaved
    Last edited: Aug 16, 2007
  5. Aug 16, 2007 #4
    Would the inflection points of this equation:

    f``(x) = 60x^3 - 120x


    60x = 0 and x^2 - 2 = 0

    so are the inflectoin points:

    x = 0, x= +- sqrt(2)

  6. Aug 16, 2007 #5
    yes, those are the inflection points
  7. Oct 31, 2007 #6
    But to say that it is an inflection point dont you have to provide some calculus reasoning. For me at least because I will be taking the Ap test i have to say something like-
    x=0 is a point of inflection because f"(x) changed signs.
    I think thats right....
  8. Oct 31, 2007 #7


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    antinerd, in the first post you calculated x = +- (sqrt(3) / 6) and asked "are these the inflection points". Later, for a different problem, you calculated 60x = 0 and x^2 - 2 = 0 and asked "so are the inflectoin points: x = 0, x= +- sqrt(2)?"

    It is important to understand that inflection points are points. Since you are giving only x values, the CANNOT be the inflection points. The points (0, y), (sqrt(2), y), (-sqrt(2),y), where each y is the function value corresponding to that x value might well be the inflection points.
  9. Oct 31, 2007 #8
    Hmmm....but I was thinking, if say we have a critical point where the function is not defined (ie its +-infinity). Would (critical point, +-infinity) be correct? (Eg. 1/(1-x) --> (1, ??) since we don't now what value the function will take at 1, limit doesn't exist)
  10. Nov 1, 2007 #9
    Is it really necessary to find the y value for the point? If it is a function it has to pass the vertical line test which basically makes sure that each x-value has only one y-value. So x=whatever is technically not a point but it is somewhat pointless/unnecessary to put the y-value
  11. Nov 1, 2007 #10


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    If a problem asks for a critical point then, yes, it is necessary to find the value in order to find the critical point. Now, sometimes a problem will ask for the critical number which is only the x-value. Yes, once you know the correct x value, you could calculate the corresponding y value- but being able to do it is not the same as doing it! If your boss asked you to look up the telephone number of a client and you handed him a telephone book do you think your boss would feel you had done what he asked? I don't consider it "pointless" to answer the question asked! Of course, I'm not the one grading your homework/tests. If I were and where I asked for a critical point, you gave me a critical number, YOU might be the one who was "pointless"!

    By the way, a point where the second derivative is 0 is not necessarily an inflection point. As antinerd said in his first post, "inflection points are where the function changes its concavity". Since "upward concavity" corresponds to negative second derivative and "downwardconcavity" corresponds to positive second derivative, that can only happen where the second derivative is 0, but, for example, y= x4 has second derivative 0 at x= 0 but does not change concavity- x= 0 is not an inflection point.
    Last edited by a moderator: Nov 1, 2007
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