Results 1 to 8 of 8

Thread: Precalculus help

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    55

    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; Sep 3rd 2007 at 09:45 PM. Reason: #4
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    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)
    $\displaystyle \frac {x^{-2} - y^{-2}}{y + x} = \frac {\frac {1}{x^2} - \frac {1}{y^2}}{y + x}$

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

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

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

    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    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:
    $\displaystyle f(x) = \frac{4}{x}$

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

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

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

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

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

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

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    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...

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

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

    so...
    $\displaystyle sin(2x) - sin(x) = 0$

    $\displaystyle 2sin(x)cos(x) - sin(x) = 0$

    $\displaystyle sin(x) ( 2cos(x) - 1 ) = 0$

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2007
    Posts
    55
    Quote Originally Posted by topsquark View Post
    You can't solve this. It isn't an equation. You could simplify it though...

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



    $\displaystyle sin(2x) = 2sin(x)cos(x)$

    so...
    $\displaystyle sin(2x) - sin(x) = 0$

    $\displaystyle 2sin(x)cos(x) - sin(x) = 0$

    $\displaystyle sin(x) ( 2cos(x) - 1 ) = 0$

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

    -Dan

    thanks..and number 4, was not + 1, it was =1...sorry abt that..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by skabani View Post
    4. solve for the real number x: ln x - ln(x-1) = 1
    Same idea as before:
    $\displaystyle ln(x) - ln(x - 1) = 1$

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

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

    $\displaystyle x = e(x - 1)$

    $\displaystyle x = ex - e$

    $\displaystyle x - ex = -e$

    $\displaystyle x(1 - e) = -e$

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by Jhevon View Post
    $\displaystyle \frac {x^{-2} - y^{-2}}{y + x} = \frac {\frac {1}{x^2} - \frac {1}{y^2}}{y + x}$

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

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

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

    $\displaystyle \begin{aligned}\frac{x^{-2}-y^{-2}}{y+x}&=\frac{x^2y^2\left(x^{-2}-y^{-2}\right)}{x^2y^2(y+x)}\\&=\frac{y^2-x^2}{x^2y^2(y+x)}\\&=\frac{y-x}{x^2y^2}\end{aligned}$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by topsquark View Post
    Same idea as before:
    $\displaystyle ln(x) - ln(x - 1) = 1$

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

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

    $\displaystyle x = e(x - 1)$

    $\displaystyle x = ex - e$

    $\displaystyle x - ex = -e$

    $\displaystyle x(1 - e) = -e$

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

    $\displaystyle \ln x-\ln(x-1)=1\implies\ln\frac{x-1}x=-1\implies\frac1e=1-\frac1x\,\therefore\,x=\frac e{e-1}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 21st 2009, 12:22 PM
  2. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2009, 01:20 PM
  3. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2009, 01:50 AM
  4. Precalculus
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2009, 01:01 AM
  5. Need PreCalculus Help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Feb 1st 2006, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum