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Math Help - 2 questions i need tips with

  1. #1
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    2 questions i need tips with

     <br />
x^\frac{1}{2}(2x^\frac{1}{2}+3x^\frac{3}{2})<br />

    How would this multiply out so it can be differentiated?

    also

    Code:
    Find the line enclosed between the curve 
    
    
    
    
    y=4-x^2 and the line 
    
    
    
    
    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

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

    How would i simplify this?
    Last edited by Hydralore; December 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
     <br />
x^\frac{1}{2}(2x^\frac{1}{2}+3x^\frac{3}{2})<br />

    How would this multiply out so it can be differentiated?
    2x + 3x^2
    Last edited by colby2152; December 11th 2007 at 07:34 AM.
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  3. #3
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  4. #4
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    Quote Originally Posted by colby2152 View Post
    2x + 4x^2
    Actually 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 
    
    
    
    
    y=4-x^2 and the line 
    
    
    
    
    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

     \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:
     \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
    GAMMA Mathematics
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    Quote Originally Posted by topsquark View Post
    Actually 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:
     \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
     <br /> <br />
f(x)= \frac{1}{x^2-1}

     g(x) = 3x+1<br />
    find h(x) when h(x) = f(g(x))

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

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

    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 [-4,1]. The thing is, in such interval f(x)\ge g(x), so the integral to compute will be

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

     g(x) = 3x+1<br />
    find h(x) when h(x) = f(g(x))
    (f \circ g)(x) = \frac{1}<br />
{{(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 ...

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

    \int_{ - 1}^{p - 1} {u^2 \,du} = \left. {\frac{1}<br />
{3}u^3 } \right|_{ - 1}^{p - 1} = \frac{{(p - 1)^3 + 1}}<br />
{3} = 3.

    Now you know what to do.
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