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Differentiate the square of the distance

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data


    Find the point on the graph of f(x)=√x that is closest to the value (4,0).

    (differentiate the square of the distance from a point (x,√x) on the graph of f to the point (4,0).)

    2. Relevant equations


    f(x)=√x
    3. The attempt at a solution
    Square of distance from (4,0)
    D(x)= ((x-4)^2+x)
    For minimum, D'(x)=0
    2(x-4)+1=0
    x=7/2
     
    Last edited by a moderator: Dec 6, 2011
  2. jcsd
  3. Dec 7, 2011 #2

    Mark44

    Staff: Mentor

    D(x) doesn't represent the distance - it's the square of the distance. As it turns out, though, mininimizing the distance is equivalent to minimizing the square of the distance.
    This is the correct value of x, but you haven't answered the question, which is to find the point on the graph of f(x)=√x that is closest to the [STRIKE]value[/STRIKE] point (4,0).
     
  4. Dec 7, 2011 #3
    I don't get it?
     
  5. Dec 7, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Mark44 told you you have the right x value. Now what point is that on the graph?
     
  6. Dec 7, 2011 #5
    Is it (7/2,(√7/2))?
     
  7. Dec 7, 2011 #6

    Mark44

    Staff: Mentor

    It depends on what you mean by (√7/2).

    This is [itex]\sqrt{7}/2[/itex].
     
  8. Dec 7, 2011 #7
    Oh I mean √7/√2
     
  9. Dec 7, 2011 #8

    Mark44

    Staff: Mentor

    Which is the same as √(7/2). So, yes, (7/2, √(7/2)) is the right point on the graph of f(x) = √x.
     
  10. Dec 7, 2011 #9
    Yay thanks
     
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