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Thread: 2 questions i need tips with

  1. #1
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    2 questions i need tips with

    $\displaystyle
    x^\frac{1}{2}(2x^\frac{1}{2}+3x^\frac{3}{2})
    $

    How would this multiply out so it can be differentiated?

    also

    Code:
    Find the line enclosed between the curve $\displaystyle y=4-x^2$ and the line $\displaystyle y=3x$
    Iv found the two factors are x+1 and x-4 but does this mean the limit is x+1 and x-4 or x-1 and x+4?


    Thanks in advance


    edit : unfortunetly i have another tiny small problem

    $\displaystyle \frac {1}{9x^2+6x+2}$

    How would i simplify this?
    Last edited by Hydralore; Dec 10th 2007 at 02:17 PM. Reason: adding another question.
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  2. #2
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    Quote Originally Posted by Hydralore View Post
    $\displaystyle
    x^\frac{1}{2}(2x^\frac{1}{2}+3x^\frac{3}{2})
    $

    How would this multiply out so it can be differentiated?
    $\displaystyle 2x + 3x^2$
    Last edited by colby2152; Dec 11th 2007 at 07:34 AM.
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    Quote Originally Posted by colby2152 View Post
    $\displaystyle 2x + 4x^2$
    Actually $\displaystyle x^{1/2}(2x^{1/2} + 3x^{3/2}) = 2x + 3x^2$

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hydralore View Post
    Code:
    Find the line enclosed between the curve $\displaystyle y=4-x^2$ and the line $\displaystyle y=3x$
    Iv found the two factors are x+1 and x-4 but does this mean the limit is x+1 and x-4 or x-1 and x+4?
    "Find the line enclosed" between a curve and a line? Is this written correctly?

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hydralore View Post
    edit : unfortunetly i have another tiny small problem

    $\displaystyle \frac {1}{9x^2+6x+2}$

    How would i simplify this?
    In most cases you wouldn't. The quadratic doesn't factor over the rationals. If you really insist, you could write it as:
    $\displaystyle \frac {1}{9x^2+6x+2} = \frac{1}{(3x + 1 + i)(3x + 1 - i)}$
    but that's kind of overkill for most problems.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Actually $\displaystyle x^{1/2}(2x^{1/2} + 3x^{3/2}) = 2x + 3x^2$

    -Dan
    Haha, looks like I added to the coefficient as well?
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  8. #8
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    Quote Originally Posted by topsquark View Post
    "Find the line enclosed" between a curve and a line? Is this written correctly?

    -Dan

    ... no.. im just stupid.. its 'find the area enclosed'
    sorry.
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  9. #9
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    Quote Originally Posted by topsquark View Post
    In most cases you wouldn't. The quadratic doesn't factor over the rationals. If you really insist, you could write it as:
    $\displaystyle \frac {1}{9x^2+6x+2} = \frac{1}{(3x + 1 + i)(3x + 1 - i)}$
    but that's kind of overkill for most problems.

    -Dan
    the original question was
    $\displaystyle

    f(x)= \frac{1}{x^2-1} $

    $\displaystyle g(x) = 3x+1
    $
    find h(x) when h(x) = f(g(x))

    im guessing iv made a mistake?
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  10. #10
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    Quote Originally Posted by Hydralore View Post
    Find the line enclosed between the curve $\displaystyle y=4-x^2$ and the line $\displaystyle y=3x$
    Get the coordinates where these equations intersect.

    Define $\displaystyle f(x)=4-x^2$ & $\displaystyle h(x)=3x,$ now

    $\displaystyle 4-x^2=3x\implies x^2+3x-4=0\,\therefore\,(x-1)(x+4)=0.$

    Now we have to integrate a certain function on the inverval $\displaystyle [-4,1].$ The thing is, in such interval $\displaystyle f(x)\ge g(x),$ so the integral to compute will be

    $\displaystyle \int_{ - 4}^1 {\Big[ {f(x) - g(x)} \Big]\,dx} = \int_{ - 4}^1 {\left( {4 - x^2 - 3x} \right)\,dx} = \frac{{125}}
    {6}.$
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  11. #11
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    Quote Originally Posted by Hydralore View Post
    $\displaystyle

    f(x)= \frac{1}{x^2-1} $

    $\displaystyle g(x) = 3x+1
    $
    find h(x) when h(x) = f(g(x))
    $\displaystyle (f \circ g)(x) = \frac{1}
    {{(3x + 1)^2 - 1}}.$

    That's all.
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  12. #12
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    Thanks alot guys
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  13. #13
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    i have one last question i need help with ...

    $\displaystyle
    \int_{0}^p (x-1)^2 dx=3 find P.
    $
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  14. #14
    Math Engineering Student
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    Start by settin' $\displaystyle u=x-1,$ the integral becomes

    $\displaystyle \int_{ - 1}^{p - 1} {u^2 \,du} = \left. {\frac{1}
    {3}u^3 } \right|_{ - 1}^{p - 1} = \frac{{(p - 1)^3 + 1}}
    {3} = 3.$

    Now you know what to do.
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