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Math Help - Point on the graph closet to set coords.

  1. #1
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    Point on the graph closet to set coords.

    Which point on the graph

    y=4x^3+2

    closest to 2,0

    I just need a general idea of what to do and Ill do the work
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  2. #2
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    Quote Originally Posted by sk8erboyla2004 View Post
    Which point on the graph

    y=4x^3+2

    closest to 2,0

    I just need a general idea of what to do and Ill do the work
    The general idea is illustrated in these threads:

    http://www.mathhelpforum.com/math-he...e-problem.html

    http://www.mathhelpforum.com/math-he...-x-origin.html
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  3. #3
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    Ok If I use the distance forumla I get


    D=\sqrt{(x-2)^2+(4x^3+2)^2}

    If I simplify the inside which I assume I can I get

    D=\sqrt{16x^6+17x^2+8}

    ?

    Which when I do the derivative of

    \frac{96x^5+34x}{2\sqrt{16x^6+17x^2+8}}

    then I solve the numerator for 0? which I dont know how I would do that
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  4. #4
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    Quote Originally Posted by sk8erboyla2004 View Post
    Ok If I use the distance forumla I get


    D=\sqrt{(x-2)^2+(4x^3+2)^2}

    If I simplify the inside which I assume I can I get

    D=\sqrt{16x^6+17x^2+8}

    ?

    Which when I do the derivative of

    \frac{96x^5+34x}{2\sqrt{16x^6+17x^2+8}}

    then I solve the numerator for 0? which I dont know how I would do that
    96x^5+34x=0 \iff x(96x^4+34)=0

    So either x=0 \mbox{ or } 96x^4+34=0

    so x=0. we reject the other because it only has complex roots.
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  5. #5
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    So all steps taking the derivative where correct?

    If so much appreciated either way!

    So if that is correct the point closet to graph is


    (2,0) ?
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by sk8erboyla2004 View Post
    So all steps taking the derivative where correct?

    If so much appreciated either way!

    So if that is correct the point closet to graph is


    (2,0) ?
    Yes.

    The numerator was correct. Just as a note in this type of problem you can just minimize the square of the distance .
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  7. #7
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    Ok If I have this


    2x-3y=12 closet to (7,8)

    I solve for y

    I get y=\frac{12-2x}{3}

    so would my distance formula look like this?


    \sqrt({x-7})^2+({\frac{12-2x}{3}-8})^2


    ?
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  8. #8
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    Quote Originally Posted by sk8erboyla2004 View Post
    Ok If I have this


    2x-3y=12 closet to (7,8)

    I solve for y

    I get y=\frac{12-2x}{3}

    so would my distance formula look like this?


    \sqrt({x-7})^2+({\frac{12-2x}{3}-8})^2


    ?
    Almost- you have the square root only around the (x-7)^2. It should be around the entire expression. You do that by putting the entire expression in { }. But as TheEmptySet said earlier, to minimize \sqrt{f(x)}, it is sufficient to minimize f(x). So find the x and y that minimize (x- 7)^2+ \left(\frac{12- 2x}{3}- 8\right)^2.
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  9. #9
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    So differentiate for x and and find y
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