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Math Help - Absolute max values..

  1. #1
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    Absolute max values..

    It says find the absolute max values of

    |12x - x^3| +1 x e (-1, 3]


    I know how to find the absolute max values of normal functions but its the | | that is confusing me, every time I try my normal method, my answers are wrong,. What do the lines mean and how do I approach them?
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  2. #2
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    Quote Originally Posted by Dave J View Post
    It says find the absolute max values of

    |12x - x^3| +1 x e (-1, 3]


    I know how to find the absolute max values of normal functions but its the | | that is confusing me, every time I try my normal method, my answers are wrong,. What do the lines mean and how do I approach them?
    I can understand the |12x - x^3|, what's messing me up is the +1      x e (-1, 3]. What is this?

    I'll ignore the last half of this and just focus on the absolute value part.

    We need to find out where 12x - x^3 = 0.
    x^3 - 12x = 0

    x(x^2 - 12) = 0

    So x = 0 or x^2 - 12 = 0

    Thus the solutions are x = 0, -\sqrt{12}, \sqrt{12}.

    Now we ask, where is 12x - x^3 = 0 negative? We have 4 test intervals:
    (-\infty, -\sqrt{12}): 12x - x^3 < 0
    (-\sqrt{12}, 0): 12x - x^3 > 0
    (0, \sqrt{12}): 12x - x^3 < 0
    (\sqrt{12}, \infty ): 12x - x^3 > 0

    All of this leads to the following:
    |12x - x^3| = \left \{ \begin{array}{cc} -(12x - x^3); & x \leq -\sqrt{12} \\ 12x - x^3; & -\sqrt{12} < x \leq 0 \\ -(12x - x^3); & 0 < x \leq \sqrt{12} \\ 12x - x^3; & \sqrt{12} < x \end{array} \right.

    Now you can take your derivative on the "piece-wise defined" function, like you would any other function.

    -Dan
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    Ahh the question is

    |12x - x^3| +1

    and
    x is all real numbers from (-1, 3]

    See I know about the derivative and such, but the +1 and the two | | throw me off. Im not sure how they interact with each other.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dave J View Post
    Ahh the question is

    |12x - x^3| +1

    and
    x is all real numbers from (-1, 3]

    See I know about the derivative and such, but the +1 and the two | | throw me off. Im not sure how they interact with each other.
    Ah! The 20 W lightbulb in my head just turned on!

    You do it in the same way as I showed below. You need to find out where the argument of the absolute value function is negative. Then just take the negative of that. Simpler example:
    f(x) = |x| + 1

    f(x) = \left \{ \begin{array}{cc} -x + 1 & x \leq 0 \\ x + 1 & 0 < x \end{array} \right.

    -Dan
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  5. #5
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    Oh I see how it works, thanks
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