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Math Help - few calc/alg questions

  1. #1
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    few calc/alg questions

    1) if you have 4x^(-2), and you wanted to make the power positive, would you bring the coefficient to the bottom?

    e.g. 1/(4x^2) -OR- 4/(x^2)



    2) find the derivative of (sqrt(x)-1)/(sqrt(x)+1)
    I understand the quotient rule and syntax, it just doesn't work out to the answer that I've been given... PROBABLY a simplification error... here's what I did:

    {(x+1)(1/2x-)-(x-1)(1/2x-)}/(x+1)^2

    (1/2)+(1/2x-)-(1/2)+(1/2x-)

    x-/(x+1)^2

    what I get:

    1/(x^2(x+1)^2)

    and the answer is 1/(x(x+1)^2)




    3) why is it that when you look at the cuberoot function (x^(1/3)) from left to right, you see its slope is always positive (albeit undefined at x=0), yet the derivative (1/3x^(-1/3)) shows that its slope is negative from a domain of negative infinity to 0? should it not be positive?



    thanks.
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  2. #2
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    1) The coefficient stays on top to make 4/(x^2)

    The rest I don't know
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  3. #3
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    thank you, anyone know the others?
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  4. #4
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    Hello, squidplant!

    1) if you have 4x^{-2}, and you wanted to make the power positive,
    would you bring the coefficient to the bottom? . . . No

    This is correct: \frac{4}{x^2}

    2) Find the derivative of: y \:= \:\frac{\sqrt{x}-1}{\sqrt{x}+1}

    I got: . y' \:=\:\frac{x^{-\frac{1}{2}}}{\left(x^{\frac{1}{2}}+1\right)^2} . . . Correrct!

    And that becomes: . y' \:=\:\frac{1}{x^{\frac{1}{2}}\left(x^{\frac{1}{2}} - 1\right)^2}
    You had it right . . . you moved the x^{-\frac{1}{2}} incorrectly.


    3) Why is it that when you look at the cuberoot function x^{\frac{1}{3}} from left to right,
    you see its slope is always positive (albeit undefined at x=0),
    yet the derivative \frac{1}{3}x^{-\frac{1}{3}} shows that its slope is negative
    from a domain of negative infinity to 0? .Should it not be positive?

    Your derivative is incorrect: . y' \:=\:\frac{1}{3}x^{-\frac{2}{3}} \:=\:\frac{1}{3x^{\frac{2}{3}}} \:=\:\frac{1}{3\left(\sqrt[3]{x}\right)^2}

    So the derivative is positive for all x \neq 0.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    A simple error, but do not be embarrassed.

    You are Thinking about derivatives, slopes, signs, increasing/decreasing, etc.

    Evidently you have a thorough understanding of these concepts ... impressive!

    And you thought you found a contradiction . . . good Thinking!

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  5. #5
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    Thank you very much Sorobon, it's good to know that mathematics still remains my haven of consistency and pure pragmitism with all the things occuring in my life; I owe math a cure for cancer. Take care~

    (p.s. i'm in my first year of AP Calculus at my high school, junior year. we're learning integrals in a few weeks! ^_^)
    Last edited by squidplant; November 12th 2006 at 01:24 PM. Reason: added note
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