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Thread: MTH101 Assignment 3 Deadline 7 July 2010

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    Senior Member viki's Avatar
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    Last edited by viki; 07-03-2010 at 07:24 AM.
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    IS this is MGT101 (Financial Accounting)????? I dont think so..!!

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    its MTH 101 assignment # 3

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    Senior Member viki's Avatar
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    Quote Originally Posted by Taim View Post
    IS this is MGT101 (Financial Accounting)????? I dont think so..!!
    i have made the correction
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    where is solution of mth101 assignment no 3 please Help

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    Brother is ka solution hay tu post karo please

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    Senior Member viki's Avatar
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    Quote Originally Posted by gulammmmm View Post
    Brother is ka solution hay tu post karo please
    INSHALLAH i will try
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    assignment 03 mth 101

    Best Answer – Q1

    y = x^2+k\x = x^2 + kx^(-1)

    Max/min points are found when the gradient = 0
    Find the gradient:

    dy/dx = 2x - kx^(-2)
    2x - kx^(-2) = 0

    Taking LCM

    2x^3 - k =0
    2x^3 = k

    If x=3:

    k = 2(3)^3 = 2(27) = 54

    k = 54



    Q2
    f'(x) = 1 - 1/x^2 > or = 0 in [1, +oo]
    So, f(x) is increasing.

    fmin = f(1) = 1 + 1/1 = 2
    x + 1/x > fmin = 2


    f'(x) = 1 - 1/x^2 = (x^2 - 1)/x^2.

    Then, if f(x) is increasing, then f'(x) > 0 and so:

    (x^2 - 1)/x^2 > 0.

    Then, since x^2 is positive for all x, we have:

    x^2 - 1 > 0
    ==> x < -1 and x > 1.

    Thus, f(x) is increasing for (1, infinity).

    For the second part, since f(x) is increasing for x > 1:
    x > 1 <==> f(x) > f(1).

    Since f(1) = 1 + 1/1 = 2, we see that for x > 1:
    f(x) > 1 + 1/1 = 2. ∴

    I hope this helps!
    Last edited by viki; 07-07-2010 at 01:42 PM.

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