1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

First Derivative is 0

  1. Jan 11, 2015 #1
    I would like to know how to correctly define and classify turning points using elementary calculus. The points I wish to clarify are maxima, minima, inflection points and saddle points.

    So I am aware of the basic info available everywhere, such as that a point is a maximum if and only if the first derivative is 0 and second derivative is negative there, and a minimum if and only if the first derivative is 0 and second derivative is positive there. Sites are much less clear on how to determine (or even define) inflection points and saddle points for certain. Furthermore, I believe the first two pieces of information are insufficient - if the second derivative is 0, it is also possible the point is a minimum or maximum (e.g. y=x4).

    Can I get an explanation of what analysis to do in order to classify the turning points correctly?
     
  2. jcsd
  3. Jan 11, 2015 #2
    You can try visualising points of inflection as parts of a graph in which the curve changes it's bend pattern (such as when it starts bending downwards when it was previously bending upwards). You can look at the graphs of basic trigonometric functions and notice where this happens(such as at π radians for the sine curve), calculate the 2nd order derivative at this point and find that it is 0.(This means that the gradient itself is not showing fluctuations in it's rate of change)

    For turning points, consider a quadratic equation with a positive leading coefficient - the graph always curves upwards, and the gradient is 0 at the turning point/minima (where the first derivative is 0 and the 2nd derivative is positive). If the leading coefficient is negative, the graph always curves downwards, and the gradient at the turning point/maxima is 0 (where the 2nd order derivate is negative, since the gradient decreases at a greater rate).
    In general, for a continuous function in a defined interval, the maxima is the greatest value of f(x) in that interval, and the minima is the lowest value. (I know this sounds very obvious and stupid but this is the textbook explanation, so x^4 (for example) does not have any inflection points as it always bends upwards, but it does have a minima)
     
    Last edited: Jan 11, 2015
  4. Jan 11, 2015 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That is not true. If all derivatives exist, then the first non-zero derivative at that point has to be an even derivative (2., 4., ...) and has to be smaller/larger than zero for a maximum/minimum.

    You can find a similar rule for inflection points, but in addition to the second derivative, you'll need the odd derivatives (usually the third is sufficient).
     
  5. Jan 11, 2015 #4
    The second derivative test doesn't always contain useful information. When it yields an answer of 0, then you will have to try something else. The best way would be checking the sign of the derivative slightly before and after the zero of the derivative.
     
  6. Jan 11, 2015 #5

    Mark44

    Staff: Mentor

    You also need to consider the domain of the function in cases where the domain is restricted. For example, ##y = \sqrt{x - x^2}##. For this function, the maximum is at (1/2, 1/2), which is easy to find by setting the derivative to zero. There are two minimum points -- (0, 0) and (1, 0). At neither of these is the derivative equal to zero.

    Another example is the function y = |x|. The derivative is never zero, yet there is a global minimum at (0, 0), a point at which the derivative is undefined.

    My point is that it is not enough to merely look for points at which the derivative is zero. It's possible for an extremum to occur at endpoints of the domain or at points in the domain at which the derivative is undefined.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook