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Thread: absolute maximum /minimum of the function..

  1. #1
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    absolute maximum /minimum of the function..

    The absolute maximum value of the function


    is equal to :

    a)None of these mentioned
    b)2
    c)4
    d)0
    e)-4



    I know that the steps to find absolute minimum or maximus is:

    1- find the critical number of f in {a,b}
    2-Evaluate f at each critical number in [a,b]
    3-Evaluate f at each endpoint of [a,b]
    4-The smallest of these values is the minimum and the largest of these values is the maximum.


    How can I find the absolute value in the question and the interval is not given ??
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    Whooops, I'm sorry, that was wrong....

    Someone please delete this post. Thank you
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  3. #3
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    Quote Originally Posted by change_for_better View Post
    The absolute maximum value of the function


    is equal to :

    a)None of these mentioned
    b)2
    c)4
    d)0
    e)-4



    I know that the steps to find absolute minimum or maximus is:

    1- find the critical number of f in {a,b}
    2-Evaluate f at each critical number in [a,b]
    3-Evaluate f at each endpoint of [a,b]
    4-The smallest of these values is the minimum and the largest of these values is the maximum.


    How can I find the absolute value in the question and the interval is not given ??
    $\displaystyle f(x) = 4 - \sqrt{x}$

    domain of $\displaystyle f(x)$ is $\displaystyle x \geq 0$

    remember the parent graph $\displaystyle y = \sqrt{x}$ ? you should be able to graph this function using the applicable transformations you learned in precalculus and "see" that the absolute max occurs at the left endpoint, namely $\displaystyle f(0) = 4$.

    the calculus confirms this ...

    $\displaystyle f'(x) = -\frac{1}{2\sqrt{x}}$

    $\displaystyle f'(x) < 0$ for all x in the domain of $\displaystyle f(x)$ except at $\displaystyle x = 0$, where $\displaystyle f'(x)$ is undefined ... $\displaystyle f(x)$ is decreasing throughout its domain, therefore the absolute max is at the left endpoint of the domain.
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