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Math Help - absolute extrema

  1. #1
    Junior Member shepherdm1270's Avatar
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    absolute extrema

    again, if you could show me the steps, i would appreciate it... *sigh* :-/

    Find locations (ordered pairs) of all absolute extrema for the following functions with the specified domains:

    a) f(x) = 1-x/3+x [0,3]

    b) f(x) = x/x^2+2 [0,4]


    thanks again, i tried working all of these out on my own, but i'm getting nowhere close to a solution...
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shepherdm1270 View Post
    again, if you could show me the steps, i would appreciate it... *sigh* :-/

    Find locations (ordered pairs) of all absolute extrema for the following functions with the specified domains:

    a) f(x) = 1-x/3+x [0,3]

    b) f(x) = x/x^2+2 [0,4]


    thanks again, i tried working all of these out on my own, but i'm getting nowhere close to a solution...
    extrema occur where f(x) is undefined (we don't have to worry about that here, our functions are defined everywhere on the intervals given) and where f'(x) = 0

    with absolute extrema however, in addition to finding the value of the function at the points where the derivative is zero, we must also check the end-points. that is, for the first you must find f(0) and f(3) and for the second you must also find f(0) and f(4).

    the largest value you obtain is the absolute maximum, the smallest value you obtain is the absolute minimum.


    can you continue?
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  3. #3
    Junior Member shepherdm1270's Avatar
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    I understand the concept of absolute extrema, i think it's my simple algebra where i'm making mistakes...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shepherdm1270 View Post
    I understand the concept of absolute extrema, i think it's my simple algebra where i'm making mistakes...
    ok, tell me what you got for the derivatives and the values of the functions at the end points and we will take it from there
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  5. #5
    Junior Member shepherdm1270's Avatar
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    i only tried the first one, then realized i was getting nowhere


    i had 1-x/3+x

    (3+x)[-1]-(1-x)[1]/(3+x)^2 =

    -3-x-1+x/(3+x)^2 =

    -4/(3+x)^2

    Then I couldn't figure out where f(x) = 0 at... aka the critical points
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shepherdm1270 View Post
    i only tried the first one, then realized i was getting nowhere


    i had 1-x/3+x

    (3+x)[-1]-(1-x)[1]/(3+x)^2 =

    -3-x-1+x/(3+x)^2 =

    -4/(3+x)^2

    Then I couldn't figure out where f(x) = 0 at... aka the critical points
    that's just it. it has no critical points. the derivative is never zero.

    thus, all you need to do here, is check the endpoints. whichever gives you the highest number is the absolute max, and the other is the absolute min
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  7. #7
    Junior Member shepherdm1270's Avatar
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    for the second one:

    x / x^2 + 2 =

    (x^2 + 2)(1) - (x)(2x) / (x^2 +2)^2 =

    -x^2 + 2 / (x^2 +2) ^2

    ... can one of the (x^2 + 2)s on the bottom be canceled out? And then where would I go from there?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shepherdm1270 View Post
    for the second one:

    x / x^2 + 2 =

    (x^2 + 2)(1) - (x)(2x) / (x^2 +2)^2 =

    -x^2 + 2 / (x^2 +2) ^2

    ... can one of the (x^2 + 2)s on the bottom be canceled out? And then where would I go from there?
    no, you cannot cancel anything. from there, you set the derivative equal to zero and try to solve it to find the critical points. don't forget to check the endpoints since you're doing absolute max and min
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  9. #9
    Super Member angel.white's Avatar
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    Hi Jhevon ^_^

    Haven't seen you in a while. I saw your avatar. I'm using Ubuntu also, I bought a laptop so that I could really focus on my Data Structures and Algorithms course, and so I have Windows / Ubuntu, dual booting.

    We should totally start a coolguy club


    And for something on topic: shep, try to use parentheses, I read the first one as:
    f(x) = 1-\frac x3 + x

    maybe write it like this instead: f(x) = (1-x)/(3+x)
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by angel.white View Post
    Hi Jhevon ^_^

    Haven't seen you in a while.
    yes, i have been scarce these days. been busy
    I saw your avatar. I'm using Ubuntu also, I bought a laptop so that I could really focus on my Data Structures and Algorithms course, and so I have Windows / Ubuntu, dual booting.

    We should totally start a coolguy club
    hehe, yes we should. I am dual booting as well.

    And for something on topic: shep, try to use parentheses, I read the first one as:
    f(x) = 1-\frac x3 + x

    maybe write it like this instead: f(x) = (1-x)/(3+x)
    i've gotten tired of saying that. i usually just leave them alone if i can get what they are saying. (the way i knew what the fractions were suppose to be is from the users attempt at the quotient rule)
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