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Math Help - absolute max/ absolute min

  1. #1
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    absolute max/ absolute min

    Consider the function f(x) = x4 − 72 x2 + 5, −5 ≤ x ≤ 13. This function has an absolute minimum value equal to __
    and an absolute maximum value equal to ____

    what i did is this:


    f'(x) = 4(x^3)-144x
    0=4(x^3)- 144x
    0= 4x [ x^2 - 36 ]
    36=x^2
    x = 6

    is this right so far?

    now to check the endpoints ....

    i have
    -5
    6
    and
    13

    that's all i have to check right?
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  2. #2
    Grand Panjandrum
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    Re: absolute max/ absolute min

    Quote Originally Posted by habibixox View Post
    Consider the function f(x) = x4 − 72 x2 + 5, −5 ≤ x ≤ 13. This function has an absolute minimum value equal to __
    and an absolute maximum value equal to ____

    what i did is this:


    f'(x) = 4(x^3)-144x
    0=4(x^3)- 144x
    0= 4x [ x^2 - 36 ]
    36=x^2
    x = 6

    is this right so far?

    now to check the endpoints ....

    i have
    -5
    6
    and
    13

    that's all i have to check right?
    You will find it easier for others to follow if you tell us in words what you are trying to do rather than just dump a load of manipulation on the page.

    The global maximum (minimum) will be either a calculus type maximum (minimum) or it will occur at a boundary point. You have the derivative correct:

    f'(x)=4x^3-144x=4x(x^2-36)

    Now a calculus type critical point occurs where the derivative is zero so we look at the roots of:

    f'(x)=4x(x^2-36)=0

    which are x=0, and x=\pm 6, but we drop the x=-6 as it is outside the interval of interest.

    So now you need to examine the value of the function at x=-5, x=0, x=6 and x=13.

    (By the way the question is asking for the maximum (minimum) of the function not the value of x where it attains the maximum (minimum) )

    CB
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