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Thread: relative extrema

  1. #1
    Junior Member
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    relative extrema

    Can someone help me with the first derivative of this one please.

    $\displaystyle f(x)=7x+ln(-4x)$
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  2. #2
    Junior Member Ziaris's Avatar
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    Take the derivative of $\displaystyle f(x)=7x+\ln(-4x)$ using the chain rule for ln(-4x),

    $\displaystyle \frac{d}{dx}\left(7x+\ln(-4x)\right)=7+\frac{1}{x}$.

    Set the derivative equal to zero and solve for x,

    $\displaystyle 7+\frac{1}{x}=0\rightarrow x=-\frac{1}{7}$.

    This is the x-coordinate for your sole relative maximum. Plug this back into the original function,

    $\displaystyle \ln\left(\frac{4}{7}\right)-1=f_{max}(x)$

    So your relative maximum occurs at $\displaystyle (x,f(x))=\left(-\frac{1}{7},\ln\left(\frac{4}{7}\right)-1\right)$.

    Pretty sure that's what you were asking for.
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