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Math Help - Families of curves

  1. #1
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    Families of curves

    Find a function of the form y=ax^b lnx where a and b are nonzero constants and a local max at (e^2, 6e^-1)

    So I'm not sure where to start on this one. I'm not too sure what all the different families of curves are, I found this website:Exponential and Logarithmic Function Gallery
    but I can't tell which one of those my equation fits into.

    Can I get some advice of where to start?
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  2. #2
    Moo
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    Hello,

    First of all, you know that the point (e^2,6e^{-1}) belongs to the function.

    So :

    \underbrace{y}_{6e^{-1}}=a \ \underbrace{x^b}_{e^{2b}} \ \ln(\underbrace{x}_{e^2})

    So 6e^{-1}=ae^{2b} \ln(e^2)=2ae^{2b}

    Here is a first equation involving a & b.

    Then, as there is a maximum at this point, the derivative is null here. So find the derivative for y by x.

    \frac{dy}{dx}=abx^{b-1} \ln(x)+\frac{ax^b}{x}=abx^{b-1} \ln(x)+ax^{b-1} (used the product rule)

    \frac{dy}{dx}=ax^{b-1} [b+1]

    If x=e^2, \frac{dy}{dx}=0

    So do the same as in the first part.
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  3. #3
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    Here is what I got for an answer:

    y = 3e^(-7) x^3 lnx

    so a = 3e^-7 , b = 3

    I took the equation you had got,

    <br />
    " alt="\frac{dy}{dx}=ax^{b-1} [b+1]
    " />

    and plugged in e^2 for x, so

    [tex] 0 = ae^2 [b+1]

    so 2 = b-1, b=3
    I then plugged that into the original equation, and got 3e^-7=a

    but that equation I tried was wrong, so I'm not sure what I'm doing wrong. Any help?
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  4. #4
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    Sorry for double post.

    Can I get anymore help on this one? I'm still pretty lost.
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  5. #5
    Moo
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    Quote Originally Posted by billabong7329
    ...
    Ok, either I'm tired either what you wrote was not very clear... sorry about that, will try to explain..

    Local max at (e^2, 6e^{-1}).
    This means that this point belongs the curve.

    So f(e^2)=6e^{-1}

    f(e^2)=a (e^2)^b \ln(e^2)=ae^{2b} \ln(e^2)

    Let's get rid of \ln(e^2).
    We know that \ln(e^b)=b
    So \ln(e^2)=2

    Hence f(e^2)=6e^{-1}=2ae^{2b}
    Multiplying each side by e^1, and dividing by 2, we have :

    3*1=3=ae^{2b+1} (1)

    Now, we know that there is a maximum at this point. This means that f'(e^2)=0

    f'(x)=(ax^b \ln(x))'=a(x^b \ln(x))'
    Here, use the product rule...

    f'(x)=a (bx^{b-1} \ln(x)+\frac{x^b}{x} )=a (bx^{b-1} \ln(x)+x^{b-1} )=a \ x^{b-1} (b \ln(x)+1)

    f'(e^2)=a \ (e^2)^{b-1} (b \ln(e^2)+1)=a \ e^{2(b-1)} (2b+1)=0

    As the exponential function is never equal to 0 and a is a constant different from 0, 2b+1=0 \Longrightarrow b=\frac{-1}{2}

    Replacing in (1) :

    3=ae^{2b+1} \Longrightarrow 3=ae^0=a

    So a=3


    This is only the way to do, it seems that I've made a mistake (dunno where...), according to your answer... So check my demonstration, this will make a good exercise, trust me
    Last edited by Moo; April 18th 2008 at 12:12 PM.
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