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Maximum and minimum functions

  1. Aug 11, 2005 #1
    I want to let f and g be functions to be defined on a common domain D. Then the maximum and minimum functions are defined as follows for any x belonging to D....

    max(f,g)(x) = f(x), if f(x) >= g(x)
    max(f,g)(x) = g(x), if g(x) > f(x)

    and

    min(f, g)(x) = f(x), if f(x) <= g(x)
    min(f, g)(x) = g(x), if g(x) < f(x)

    How do I define my max(sinx, cosx) and min(x, x^2)?
     
  2. jcsd
  3. Aug 11, 2005 #2
    think about it: when is cos greater than sin over the same domain? clearly the min and max functionals will depend upon the value of theta.

    same for x and x^2 - clearly x is greater than x^2 for values less than 1, and less than x^2 for values greater than 1.
     
  4. Aug 11, 2005 #3

    lurflurf

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    |f-g|=max(f,g)-min(f,g)
    so
    max(f,g)=(f+g+|f-g|)/2
    min(f,g)=(f+g-|f-g|)/2
    also the functions you are considering are continuous so if you find when they are equal, you can find which is greater on an interval where they are always not equal by considering any point in that interval.
    for example
    f=3x+2
    g=2x-3
    f=g->x=-5
    so we consider
    x<-5 then g>f
    x>-5 then f>g
    or we could try
    max(f,g)=(3x+2+2x-3+|3x+2-(2x-3)|)/2
    =(5x-1+|x+5|)/2
    =(6x+4)/2=3x+2 x>-5
    =(4x-6)/2=2x-3 x<-5
     
  5. Aug 11, 2005 #4
    I want to know how you got |f-g|=max(f,g)-min(f,g)....

    Also, will that example you posted be my guide to solve this? So, I have to say the interval in which the two functions are as designated....
     
  6. Aug 11, 2005 #5
    quetzalcoatl9,

    You mentioned that "think about it: when is cos greater than sin over the same domain? clearly the min and max functionals will depend upon the value of theta"

    The graph seems to look like "lots of McDonald's M"

    You also mentioned that "same for x and x^2 - clearly x is greater than x^2 for values less than 1, and less than x^2 for values greater than 1. "

    x > x^2 when 0 < x < 1... not less that 1.. :D
    x < x^2 when x > 1 or x < 0.

    x = x^2 when x = 0 or 1.
     
  7. Aug 11, 2005 #6
    Sine and Cosine are uniquely defined by their behavior on [0, 2pi ). So if you define the behavior of max(Sin(x), Cos(x)) on that interval, by extension you know what it looks like everywhere.
     
  8. Aug 12, 2005 #7

    lurflurf

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    max(f,g)=min(f,g)->f=g
    max(f,g)>=min(f,g)
    so max(f,g)-min(f,g)>=0
    and either max(f,g)-min(f,g)=f-g or max(f,g)-min(f,g)=g-f
    hence
    max(f,g)-min(f,g)=|f-g|
    There are two easy ways to do these.
    They are easy because the functions are continuous and because they are equal only on discrete sets. That is for every x such that f(x)=g(x) there exist a positive number h so that x is the only point of equality in (x-h,x+h).
    Method 1: piecewise definition
    here we aim to partition the real numbers into intervals and specify which function to use on the interval. ie
    x<a
    a<x<b
    b<x<c
    c<x
    with f(x)=g(x) if x=a,b,c
    so on any interval where f and g are continuous and never equal |f-g| never changes sign thus if f>g for any point in such an inteval f>g for all points.
    method 2: absolute value
    max(f,g)=(f+g+|f-g|)/2
    min(f,g)=(f+g-|f-g|)/2
    allow us to right out the min and max right away
    sometimes further simplification is possible
    sometimes we use this to give the min and max as piecewise defined functions
     
  9. Aug 12, 2005 #8
    Hmm, thank you for the help...

    If I were to define my function, can I just "plugged in" the f(x) and g(x) with their corresponding functions.. then just analyze the behavior of their graphs, of course, I will have to find those intervals?

    max(sinx, cosx)

    max(f,g)(x) = sinx, if sinx >= cosx
    max(f,g)(x) = cosx, if cosx > sinx
     
  10. Aug 12, 2005 #9

    lurflurf

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    find out when sin(x)=cos(x)
    then show sin(x)-cos(x) changes sigh at all of its zeros
    then which function (sin(x) or cos(x)) is used in each interval can be found by finding which is used for any point in the interval
    as has been said you can do any interval of length 2pi and then extend to all numbers.
    max(f,g)(x) = sinx, if sinx >= cosx
    max(f,g)(x) = cosx, if cosx > sinx
    is a fine definition
    as is
    max(f,g)(x)=(sin(x)+cos(x)+|sin(x)-cos(x)|)/2
    but you probably what to express the conditions
    sin(x) > cos(x)
    cos(x) > sin(x)
    in a more useful form
    what properties of max(sin(x),cos(x)) do you want to highlight?
     
  11. Aug 14, 2005 #10
    Ok... I got it... thank you for the help
     
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