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Math Help - Precalculus help

  1. #1
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    Precalculus help

    I think i know this, but im not exactly sure....

    1. simplify:

    (x^(-2)-y^(-2))(y+x)^(-1)

    2. If f(x) = (4/x), find and simplify (f(x+h)-f(x))/(h)

    3. express the area A of an equilateral triangle as a function of the length S of a side of the triangle.

    4. solve for the real number x: ln x - ln(x-1) = 1

    5. solve for x in the interval [0, 2(pi)]: sin2x-sinx = 0

    thanks!!
    Last edited by skabani; September 3rd 2007 at 09:45 PM. Reason: #4
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by skabani View Post
    I think i know this, but im not exactly sure...

    simplify:

    (x^(-2)-y^(-2))(y+x)^(-1)
    \frac {x^{-2} - y^{-2}}{y + x} = \frac {\frac {1}{x^2} - \frac {1}{y^2}}{y + x}

    ............... = \frac {\frac {y^2 - x^2}{x^2y^2}}{y + x}

    ............... = \frac {(y + x)(y - x)}{x^2 y^2} \cdot \frac {1}{y + x}

    ............... = \frac {y - x}{x^2 y^2}

    doesn't look much simpler to me, but whatever.


    EDIT: wow, where'd all those other questions come from. anyway, i think i'm going to leave now. you'll have to wait on someone else to help
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by skabani View Post
    2. If f(x) = (4/x), find and simplify (f(x+h)-f(x))/(h)
    Aside from a bit of work, this one is straightforward:
    f(x) = \frac{4}{x}

    f(x + h) = \frac{4}{x + h}

    So
    \frac{f(x + h) - f(x)}{h} = \frac{ \frac{4}{x + h} - \frac{4}{x} }{h}

    = \frac{ \frac{4x}{x(x + h)} - \frac{4(x + h)}{x(x + h)} }{h}

    = \frac{ \frac{4x - 4(x + h)}{x(x + h)} } {h}

    = \frac{4x - 4x - 4h}{hx(x + h)}

    = \frac{-4h}{hx(x + h)}

    = -\frac{4}{x(x + h)}

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by skabani View Post
    4. solve for the real number x: ln x - ln(x-1) + 1
    You can't solve this. It isn't an equation. You could simplify it though...

    ln(x) - ln(x - 1) + 1 = ln(x) - ln(x - 1) + ln(e) = ln \left ( \frac{ex}{x - 1} \right )

    Quote Originally Posted by skabani View Post
    5. solve for x in the interval [0, 2(pi)]: sin2x-sinx = 0
    sin(2x) = 2sin(x)cos(x)

    so...
    sin(2x) - sin(x) = 0

    2sin(x)cos(x) - sin(x) = 0

    sin(x) ( 2cos(x) - 1 ) = 0

    So either
    sin(x) = 0 \implies x = 0, ~ \pi
    or
    2cos(x) - 1 = 0 \implies cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, ~ \frac{5\pi}{3}

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    You can't solve this. It isn't an equation. You could simplify it though...

    ln(x) - ln(x - 1) + 1 = ln(x) - ln(x - 1) + ln(e) = ln \left ( \frac{ex}{x - 1} \right )



    sin(2x) = 2sin(x)cos(x)

    so...
    sin(2x) - sin(x) = 0

    2sin(x)cos(x) - sin(x) = 0

    sin(x) ( 2cos(x) - 1 ) = 0

    So either
    sin(x) = 0 \implies x = 0, ~ \pi
    or
    2cos(x) - 1 = 0 \implies cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, ~ \frac{5\pi}{3}

    -Dan

    thanks..and number 4, was not + 1, it was =1...sorry abt that..
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by skabani View Post
    4. solve for the real number x: ln x - ln(x-1) = 1
    Same idea as before:
    ln(x) - ln(x - 1) = 1

    ln \left ( \frac{x}{x - 1} \right ) = 1

    \frac{x}{x - 1} = e

    x = e(x - 1)

    x = ex - e

    x - ex = -e

    x(1 - e) = -e

    x = \frac{-e}{1 - e} = \frac{e}{e - 1}

    -Dan
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    \frac {x^{-2} - y^{-2}}{y + x} = \frac {\frac {1}{x^2} - \frac {1}{y^2}}{y + x}

    ............... = \frac {\frac {y^2 - x^2}{x^2y^2}}{y + x}

    ............... = \frac {(y + x)(y - x)}{x^2 y^2} \cdot \frac {1}{y + x}

    ............... = \frac {y - x}{x^2 y^2}
    Another solution

    \begin{aligned}\frac{x^{-2}-y^{-2}}{y+x}&=\frac{x^2y^2\left(x^{-2}-y^{-2}\right)}{x^2y^2(y+x)}\\&=\frac{y^2-x^2}{x^2y^2(y+x)}\\&=\frac{y-x}{x^2y^2}\end{aligned}
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  8. #8
    Math Engineering Student
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    Quote Originally Posted by topsquark View Post
    Same idea as before:
    ln(x) - ln(x - 1) = 1

    ln \left ( \frac{x}{x - 1} \right ) = 1

    \frac{x}{x - 1} = e

    x = e(x - 1)

    x = ex - e

    x - ex = -e

    x(1 - e) = -e

    x = \frac{-e}{1 - e} = \frac{e}{e - 1}
    Other solution

    \ln x-\ln(x-1)=1\implies\ln\frac{x-1}x=-1\implies\frac1e=1-\frac1x\,\therefore\,x=\frac e{e-1}
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