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Math Help - Find the absolute max and absolute min

  1. #1
    s3a
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    Find the absolute max and absolute min

    Question:
    (1 pt) Let on the interval . Find the absolute maximum and absolute minimum of on this interval.

    The absolute max occurs at ?
    The absolute min occurs at ?

    The absolute max is supposedly x = -1 and I do not know what the absolute min is. I use the two "edges" of the constraint and since when I solved for x I already got what the "edges" of the constraint then I determined the two "edges" to be the values to optimize with but I just do not understand what is going on. In addition, my calculator says (-1)^(2/3) does not exist but algebraically, I determined it to be = 1; who's right. me or the calculator? (My work is attached)

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    Question:
    (1 pt) Let on the interval . Find the absolute maximum and absolute minimum of on this interval.

    The absolute max occurs at ?
    The absolute min occurs at ?

    The absolute max is supposedly x = -1 and I do not know what the absolute min is. I use the two "edges" of the constraint and since when I solved for x I already got what the "edges" of the constraint then I determined the two "edges" to be the values to optimize with but I just do not understand what is going on. In addition, my calculator says (-1)^(2/3) does not exist but algebraically, I determined it to be = 1; who's right. me or the calculator? (My work is attached)

    Any help would be greatly appreciated!
    Thanks in advance!
    1. f(-1) = 5

    f(1) = 1

    2. You can show that f(0) = 0 and f(x) \geq 0 for all x \in \mathbb{R}
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  3. #3
    s3a
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    How would you know to plug in x = 0 when you did not obtain x = 0 when solving nor are the "edges" of the constraint 0?
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  4. #4
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    Smile

    hi
    x=1 the root of your first derivative is either a maximum or a minimum,try to find f'' and study its sign at that point.
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  5. #5
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    Quote Originally Posted by s3a View Post
    How would you know to plug in x = 0 when you did not obtain x = 0 when solving nor are the "edges" of the constraint 0?
    I don't know what you mean by that ...?

    3\sqrt[3]{x^2} - 2x=0~\implies~\sqrt[3]{x^2}(3-2\sqrt[3]{x})=0~\implies~x=0~\vee~x=\left(\frac32 \right)^3
    Last edited by earboth; November 22nd 2009 at 09:19 PM.
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    Quote Originally Posted by s3a View Post
    How would you know to plug in x = 0 when you did not obtain x = 0 when solving nor are the "edges" of the constraint 0?
    The max or min may occur where the derivative is 0 or where the derivative does not exist. The derivative does not exist at x= 0.
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  7. #7
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    Quote Originally Posted by Raoh View Post
    hi
    x=1 the root of your first derivative is either a maximum or a minimum,try to find f'' and study its sign at that point.
    For absolute max or min, the second derivative test is not necessary. Just calculate the values at the endpoints and critical points (where the derivative is 0 or does not exist) and see which value is highest and which lowest.
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