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Thread: Families of curves

  1. #1
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    Families of curves

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

    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 $\displaystyle (e^2,6e^{-1})$ belongs to the function.

    So :

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

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

    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.

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

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

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

    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,

    $\displaystyle
    $
    $\displaystyle \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 $\displaystyle (e^2, 6e^{-1})$.
    This means that this point belongs the curve.

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

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

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

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

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

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

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

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

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

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

    Replacing in (1) :

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

    So $\displaystyle a=3$


    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; Apr 18th 2008 at 12:12 PM.
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