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Math Help - Applications of Derivatives

  1. #1
    Member purplec16's Avatar
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    Applications of Derivatives

    Ok I have to identify the extram i.e.the maximum and minimum points of this function and...i'm kinda confused f(x)=x(\sqrt{4-{x}^{2 } } )

    f'(x)=\frac{4-2{x}^{2 } }{\sqrt{4-{x}^{2 } } }

    Thats what i got for my derivative so i just factor out numerator and denominator right?

    To get:

    4-2{x}^{ 2}= 0

    \sqrt{4-{x}^{2 } } = 0

    as my zeroes
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  2. #2
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    ok ... solve for x in each case to determine critical values.

    4-2x^2 = 2(2-x^2) = 2(\sqrt{2}-x)(\sqrt{2}+x) = 0

    \sqrt{4-x^2} = \sqrt{(2-x)(2+x)} = 0

    Now what? ... btw, pay attention to the domain of the original function.
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  3. #3
    Member purplec16's Avatar
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    for the numerator i got \pm 2and the denominator \pm \sqrt{2}

    but how u solved the numerator i thought if you can't factor of x's you would just say -2{x}^{2 } = -4

    and then solve from there
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  4. #4
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    Quote Originally Posted by purplec16 View Post
    for the numerator i got \pm 2and the denominator \pm \sqrt{2}

    but how u solved the numerator i thought if you can't factor of x's you would just say -2{x}^{2 } = -4

    ... but you can factor 4 - 2x^2 ,just as I presented it.
    you've got it backwards ... roots of the numerator are x = \pm \sqrt{2} , denominator ... x = \pm 2

    one more time ... domain of f(x) is important in this problem.
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  5. #5
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    Remember! Division by zero is not defined.

    skeeter got the roots excluding 2 and by separating the contents using the formula for the difference of two squares,
    that is 4-2x^2 = 2(2-x^2) = 2(\sqrt{2} - x)(\sqrt{2} + x), just as he had written, because  (a^2 - b^2) =(a-b)(a+b) .

    Now, considering that division by zero isn't defined, what has to be zero for the fraction to equal zero?
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