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Homework Help: Max, min and average of the sum of 3 functions

  1. Aug 22, 2011 #1
    hi, i have these 3 functions

    [PLAIN]http://img804.imageshack.us/img804/2230/provakd.jpg [Broken]

    i must calculate the resultant function, and to find the max, min and the average.. i have tried to calculate it, i have found these values (max=53; min=38; average=43) but the solution in my book are different (max=54; min=19; average=34).. can you help me?? thank you!!
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 22, 2011 #2


    Staff: Mentor

    Your f2 is somewhat confusing. Is your formula saying that f2(t) = 19 for 10 < t <= 18, and f2(t) = 4, for 0 < t <= 10, and for 18 < t <= 24?
    Last edited by a moderator: May 5, 2017
  4. Aug 22, 2011 #3
    yes exactly !! , all functions are 0< t<=24
  5. Aug 22, 2011 #4


    Staff: Mentor

    I would sketch all three functions on the same coordinate system. The one that's a little tricky is the first one. Its period is 8, so it will have three complete cycles in your interval of 0 through 24.
  6. Aug 22, 2011 #5
    yes i know, but i would compare my solution with your solution, i mean i would like to know what max, min absolute you have found.. because i have found max=53 and min=38.. but my book say me different solution (max=54; min=19) so i must understand what i was wrong!! (or if the book is wrong)
  7. Aug 22, 2011 #6


    Staff: Mentor

    If you look at f1(t) + f3(t), you get 34 + 10 cos(π/4 * t). The maximum value of this function is 44 and the minimum value is 24.

    If we add f2(t) to the above, we are adding 4 on the interval (0, 10], adding 19 on the interval (10, 18], and adding 4 on the interval (18, 24].

    So on (0, 10], the max value is 48 and the min value is 28.
    On (10, 18], the max value is 63 and the min value is 43.
    On (18, 24], the max value is 48 and the min value is 28.

    I didn't calculate average values. To do this requires an integral.

    Are you sure you copied the numbers in your problem correctly?
  8. Aug 22, 2011 #7


    User Avatar
    Homework Helper

    The nice thing about finding the average value of the sum of the functions is that it is the sum of the average values of the functions. Two of the functions ( f2 and f3 ) only involve rectangles, so their averages over the entire interval [0, 24] are easy to calculate; so is the average for the first term of f1 . So you will only really need an "average value integration" for the cosine term in f1 ...
  9. Aug 22, 2011 #8
    i have tried this..

    in (0, 10] => f_x=(25+4+9)+10cos(pi/4*t)=38+10cos(pi/4*t)
    in (10, 18] =>f_x=(25+19+9)+10cos(pi/4*t)=53+10cos(pi/4*t)
    in (18, 24] =>f_x=(25+4+9)+10cos(pi/4*t)=38+10cos(pi/4*t)

    for t=0 => f1=48
    for t=10 => f2=38
    for t=18 => f3=53
    for t=24 => f4=48

    so for me max=53 and min=38. why my argument is wrong?

    for average i have thought:

  10. Aug 22, 2011 #9


    Staff: Mentor

    What you have above looks fine.
    You are evaluating your function only at the endpoints of the domains of f2. In the interval (10, 18], the function is 53 + 10 cos(pi/4 * t). The max and min values don't occur at 10 and 18. Actually, the max value on this interval is 63 and the min value is 43.

    If you haven't sketched a graph of the combined function, you really should do so.
    This is not how it works. The average value of a function f(x) on an interval [a, b] is defined as an integral:
    [tex]\frac{1}{b - a}\int_a^b f(x)dx[/tex]
  11. Aug 22, 2011 #10
    you are right .. for t=4 => f_x=28 and 4 is inside the interval (0,10]; in t=12 => f_x=43 and in t=20 => f_x=28, the same for the max.. you are right !! thank you very much!!!
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