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Math Help - [SOLVED] Find a function of the form...?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Find a function of the form...?

    Find a function of the f(x) = \lambda e^{2x} + \mu e^{x} - \nu x where \lambda, \mu, \nu are independent of x and f(0) = -1, f'(\log_{e} 2) = 30 and \int_{0}^{\log_{e} 4} \{f(x) + \nu x\} dx = 28.5
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  2. #2
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    Just need to make a system of equations and substitute.

    lamda=L mew=M v=V

    f(0)=Le^(2*0)+M*e^(0)-v*0=-1
    L+M=-1

    f'(x)= 2*L*e^(2x)+M*e^(x) -V
    f'(ln(2))=8*L+2*M-V=30

    integral(L*e^(2*x)+M*e^(x),x,0,ln(4))

    solve(L/2*e^(2*X)+M*e^x),x,0,ln(4))
    8L+4M-(L/2+M)=28.5

    System of equations:
    1*L+1*M+0*V=-1
    8*L+2*M-1*V=30
    15/2*L+3*M=28.5
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  3. #3
    Super Member fardeen_gen's Avatar
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    Answer:
    Spoiler:
    7e^{2x} - 8e^{x} - 10x
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