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Equation

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  • 18-03-2009 10:37am
    #1
    Granny Lover
    Closed Accounts Posts: 62 ✭✭


    (1/3*X^3)+(2X^2)-(5X)+7

    Can anyone find the Max and Min of this equation mathimatically
    Show all workings

    Prize for the first person who gets it right
    Tagged:


Comments

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    #2
    GhostInTheRuins
    Registered Users Posts: 5,963 ✭✭✭GhostInTheRuins


    I'm not doing your homework boyo


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    #3
    CreepingDeath
    Closed Accounts Posts: 8,015 ✭✭✭CreepingDeath


    Granny Lover wrote: »
    (1/3*X^3)+(2X^2)-(5X)+7

    Can anyone find the Max and Min of this equation mathimatically
    Show all workings

    Prize for the first person who gets it right

    The min is 1 and the max is 7.


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    #4
    Granny Lover
    Closed Accounts Posts: 62 ✭✭Granny Lover


    CreepingDeath wrote: »
    The min is 1 and the max is 7.
    Both are Min and you didnt show workings


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    #5
    Mr.Lizard
    Closed Accounts Posts: 1,476 ✭✭✭Mr.Lizard


    Granny Lover wrote: »
    (1/3*X^3)+(2X^2)-(5X)+7

    Can anyone find the Max and Min of this equation mathimatically
    Show all workings

    Prize for the first person who gets it right

    Surely you win the prize, that of someone doing your homework for you?


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    #6
    Granny Lover
    Closed Accounts Posts: 62 ✭✭Granny Lover


    Not Homework im just bored and enjoy Maths


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    #7
    Jesus Wept
    Registered Users Posts: 25,243 ✭✭✭✭Jesus Wept


    2+2 is 4


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    #8
    Bernard Black
    Closed Accounts Posts: 44 ✭Bernard Black


    GhostInTheRuins wrote: »
    I'm not doing your homework boyo
    Mr.Lizard wrote: »
    Surely you win the prize, that of someone doing your homework for you?

    Sounds like someone doesnt know how to work out the max and the min of a simple equation!


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    #9
    Zangetsu
    Registered Users Posts: 1,218 ✭✭✭Zangetsu


    Bernard Black wrote: »
    Sounds like someone doesnt know how to work out the max and the min of a simple equation!



    Ho ho ho ho ho BURNED!


    Seriously though, maths in AH? That's a new one!


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    #10
    [Deleted User]
    Posts: 17,378 ✭✭✭✭ [Deleted User]


    maths forum ftw


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    #11
    Princess Consuela Bananahammock
    Registered Users Posts: 33,400 ✭✭✭✭Princess Consuela Bananahammock


    Bernard Black wrote: »
    Sounds like someone doesnt know how to work out the max and the min of a simple equation!

    No idea.
    42?

    Everything I don't like is either woke or fascist - possibly both - pick one.



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    #12
    Gone Drinking
    Registered Users Posts: 2,809 ✭✭✭Gone Drinking


    Thought this was going to be a thread about the equator, which was going to lead me to commenting on the titties of local tribeswomen who live in or around the equator.

    Utterly disappointed


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    #13
    Kiera
    Closed Accounts Posts: 13,130 ✭✭✭✭Kiera


    Granny Lover wrote: »
    Not Homework im just bored and enjoy Maths
    Maths and grannies? You could just go count the wrinkles on your garnnies...... :P


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    #14
    GhostInTheRuins
    Registered Users Posts: 5,963 ✭✭✭GhostInTheRuins


    Bernard Black wrote: »
    Sounds like someone doesnt know how to work out the max and the min of a simple equation!

    I can't believe it, I've been rumbled :rolleyes:


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    #15
    zuutroy
    Registered Users Posts: 6,854 ✭✭✭zuutroy


    A timely reminder of the amount of pointless crap you learn in school.


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    #16
    stovelid
    Closed Accounts Posts: 27,252 ✭✭✭✭stovelid


    7X6


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    #17
    themadchef
    Closed Accounts Posts: 5,813 ✭✭✭themadchef


    Maths :eek: recurring Leaving cert nightmare alert.

    Hips to waist ratio, wish i paid more attention to that one.


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    #18
    token56
    Registered Users Posts: 2,616 ✭✭✭token56


    Granny Lover wrote: »
    (1/3*X^3)+(2X^2)-(5X)+7

    Can anyone find the Max and Min of this equation mathimatically
    Show all workings

    Prize for the first person who gets it right


    dy/dx = x^2 + 4x - 5 = 0

    (x + 5)(x - 1)

    x = -5, 1

    d^2y/dx^2 = 2x + 4

    for x = -5, d^2y/dx^2 = -2, > 0 therefore maximum

    for x = 1, d^2y/dx^2 = 6, > 0, therefore minimum

    max at -5, min at 1

    at least thats what i think it is if i remeber my leaving cert maths


    edit: mistake


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    #19
    Tigger
    Closed Accounts Posts: 16,705 ✭✭✭✭Tigger


    token56 wrote: »
    dy/dx = x^2 + 4x - 5 = 0

    (x + 5)(x - 1)

    x = -5, 1

    d^2y/dx^2 = 2x + 4

    for x = -5, d^2y/dx^2 = -2, > 0 therefore maximum

    for x = 1, d^2y/dx^2 = 6, > 0, therefore minimum

    max at -5, min at 1

    at least thats what i think it is if i remeber my leaving cert maths


    edit: mistake

    god i hated the leaving cert


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    #20
    javaboy
    Registered Users Posts: 15,094 ✭✭✭✭javaboy


    Maths+AH=Fail


This discussion has been closed.
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