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Math Help - More Anti-Derivatives

  1. #1
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    More Anti-Derivatives

    I need to find the most common anti-derivative of f '' (x) = x^-2 , where x > 0, f(1) = 0 and f(2) = 0

    As usual, this is how far I've gotten:

    f'(x) = - x^-1 + C
    f(x) = -Inx + Cx + D

    I think that's right. Now what am I supposed to do?

    I tried finding f(1) and f(2) of the function f(x) but it's not coming out as it should.

    The answer is supposed to be f(x) = -Inx + (In2)x - In2
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    I need to find the most common anti-derivative of f '' (x) = x^-2 , where x > 0, f(1) = 0 and f(2) = 0

    As usual, this is how far I've gotten:

    f'(x) = - x^-1 + C
    f(x) = -Inx + Cx + D

    I think that's right. Now what am I supposed to do?

    I tried finding f(1) and f(2) of the function f(x) but it's not coming out as it should.

    The answer is supposed to be f(x) = -Inx + (In2)x - In2
    You came up with:
    f(x) = -Inx + Cx + D
    => f(1) = -ln(1) + C + D = 0
    => C + D = 0 ...................(1) since, ln(1) = 0

    => f(2) = -ln(2) + 2C + D = 0
    => 2C + D = ln(2).............(2)

    So now to find C and D, we must solve the system:
    C + D = 0 ...................(1)
    2C + D = ln(2) .............(2)

    => C = ln(2) ...........(2) - (1)

    but C + D = 0
    => ln(2) + D = 0
    => D = -ln(2)

    so f(x) = -lnx + ln(2)x - ln(2)
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  3. #3
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    Quote Originally Posted by zachb View Post
    I need to find the most common anti-derivative of f '' (x) = x^-2 , where x > 0, f(1) = 0 and f(2) = 0

    As usual, this is how far I've gotten:

    f'(x) = - x^-1 + C
    f(x) = -Inx + Cx + D

    I think that's right. Now what am I supposed to do?

    I tried finding f(1) and f(2) of the function f(x) but it's not coming out as it should.

    The answer is supposed to be f(x) = -Inx + (In2)x - In2
    Hello,

    all your calculations are correct!

    You only have to plug in the values you know into the equation of f:

    f(1) = 0 = -ln(1) + C*1 + D ===> 0 = C + D (note that ln(1) = 0)
    f(2) = 0 = -ln(2) + C*2 * D ===> 0 = -ln(2) + 2C + D

    Subtract equ1 from equ2 and you'll get:

    0 = -ln(2) + C ===> C = ln(2). Plug in this value into the first equation D = -ln(2).

    The equation of f becomes:

    f(x) = -ln(x) +(ln(2))x - ln(2)
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