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Thread: intervals using 1st derivative test??

  1. #1
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    intervals using 1st derivative test??

    Find the intervals of inc and dec & local max and min of h(x)=x^3/2-256x? using 1st derivative test.?


    thanks...this one is really getting me!!!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by monolith View Post
    Find the intervals of inc and dec & local max and min of h(x)=x^3/2-256x? using 1st derivative test.?


    thanks...this one is really getting me!!!
    Is it $\displaystyle h(x)=x^{\frac{3}{2}}-256x$ or $\displaystyle h(x)=\tfrac{1}{2}x^3-256x$??

    --Chris
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    sorry

    i guess i wasnt using proper scripting...it is the first one you have written

    $\displaystyle h(x)=x^{\frac{3}{2}}-256x$
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by monolith View Post
    i guess i wasnt using proper scripting...it is the first one you have written

    $\displaystyle h(x)=x^{\frac{3}{2}}-256x$
    First find $\displaystyle h'(x)$:

    $\displaystyle h'(x)=\tfrac{3}{2}x^{\frac{1}{2}}-256$

    Now set this equal to zero:

    $\displaystyle \tfrac{3}{2}x^{\frac{1}{2}}-256=0\implies x^{\frac{1}{2}}=\tfrac{512}{3}\implies x=\tfrac{262144}{9}\approx 29127.1$

    This is the only critical point. So we investigate the behavior of $\displaystyle h'(x)$ over the intervals $\displaystyle \left(0,\tfrac{262144}{9}\right)\text{ and }\left(\tfrac{262144}{9},\infty\right)$

    Pick a point in these intervals and plug them into $\displaystyle h'(x)$, for simplicity, I'll pick 1 and 360,000 (the last one seems large, but when we take the square root of it, we get a nice number to work with...no decimals involved).

    $\displaystyle h'(1)=\tfrac{3}{2}(1)^{\frac{1}{2}}-256=\color{red}\boxed{-254.5}$

    It is decreasing here.

    $\displaystyle h'(360,000)=\tfrac{3}{2}(360,000)^{\frac{1}{2}}-256=\color{red}\boxed{644}$

    It is increasing here.

    So to answer the first part of the question, the interval where its:

    increasing -- $\displaystyle \left(\tfrac{262144}{9},\infty\right)$
    decreasing -- $\displaystyle \left(0,\tfrac{262144}{9}\right)$

    where $\displaystyle \tfrac{262144}{9}\approx 29127.111$

    Now, take note that a maximum occurs when a function goes from increasing to decreasing at a critical point, and a minimum occurs when a function goes from decreasing to increasing at a critical point. Which one do you think it is? What is the coordinates of the maximum/minimum? I leave that for you to do.

    I hope this makes sense!

    --Chris
    Last edited by Chris L T521; Aug 13th 2008 at 09:52 PM. Reason: clarification
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