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Local linearization

  1. Apr 7, 2005 #1
    the question is here

    I got a different answer than what the book says...

    so I need to find the formula of the graph.

    [tex]H'(3)=f(3)=2[/tex]
    [tex]m=\frac{\Delta y}{\Delta x}[/tex]

    [tex]2=\frac{\Delta y}{x- \int^3_0 f(t)dt}[/tex]
    [tex]y=2(x+2)[/tex]
    [tex]y=2x+4[/tex]

    the book's answer is 2x-8

    where did I go wrong?
     
  2. jcsd
  3. Apr 7, 2005 #2

    Hurkyl

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    Through what point did you want your line to go through? It looks like you used (-2, 0)...
     
  4. Apr 7, 2005 #3
    you mean I should do:
    [tex]2=\frac{y-2}{x-3}[/tex]
    [tex]y=2x-4[/tex]
    ??

    what I did was...

    [tex]H'(x)=\frac{y-\int_{-2}^yf(t)dt}{x-\int_0^xf(t)dt}[/tex]
     
  5. Apr 7, 2005 #4

    Hurkyl

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    Why do you want your line to go through the point (3, 2)?

    What you need to do is stop guessing and think it through. Working through a simpler problem might help.

    What is the local linearization of the function f(x) = x2 near x = -1? First tell me what that means geometrically, then work out the answer algebraically.
     
  6. Apr 7, 2005 #5
    f'(x)= (y2-y1) / (x2-x1)
    -2= (y2-1) / (x+1)
    y=-2x-1
     
  7. Apr 7, 2005 #6
    thank you, I used your example to get the right answer
     
  8. Apr 7, 2005 #7

    Hurkyl

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    I notice you didn't try a geometric explanation. :tongue2:

    Anyways, that's exactly right. Now, why did you pick the point (x1, y1) = (-1, 1)? Apply the same reasoning to your problem.
     
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