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Homework Help: Estimating values

  1. Nov 28, 2004 #1
    f(x) is 1.8 when x=1
    f(x) is 0.5 when x=2.8
    f(x) is 1.5 when x=1.2

    what is the best estimate for the value of (f^-1)'1?

    I now that f'(1) is -1.6 because of linear apporximation:
    (2.8-1.2)/(.5-1.5)=1.6/-1=-1.6

    so then (f^-1)'1 = 1/1.6?
     
  2. jcsd
  3. Nov 28, 2004 #2
    How do you know that f(x) is linear?
     
  4. Nov 28, 2004 #3
    sorry, that method is "local approximation", not linear approximation
     
  5. Nov 29, 2004 #4

    HallsofIvy

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    Your confusing f(x) with f-1(x) so you have your formula "upside down".
    I would interpret this as saying that f-1(0.5)= 2.8, f-1(1.0)= 1.8, and f-1(1.5)= 1.2. That's clearly a decreasing function
    You are given three equally spaced points and a good estimate for the derivative of a function, in that situation, is (f(a+h)- f(a-h))/(2h). In terms of the inverse function, we have a= 1, h= 0.5 so that
    f-1(1- 0.5)= f-1(0.5)= 2.8 and f-1(1+ 0.5)= f-1(1.5)= 1.2.
    The estimate for the derivative of f-1at x=1 (not f' (1))
    is (1.2- 2.8)/(2(0.5))= -1.6. That is the derivative of f-1(x) at x= 1, not of f(x).
     
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