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Spivak Derivate Question

  1. Nov 11, 2006 #1
    Suppose
    f(a) = g(a) = h(a)
    and f(x) <= g(x) <= (x) for all x

    Prove g(x) is differentible and that
    f'(a) = g'(a) = h'(a).

    So.. I need to prove that the following limit exists:

    lim h -->0 (g(x+h) - g(x)) / h

    but how can i use the fact that f(x) <= g(x) <= (x) for all x?

    Thanks
     
  2. jcsd
  3. Nov 11, 2006 #2

    StatusX

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    Homework Helper

    Obviously you're gonna have to tell us something about f and h.
     
  4. Nov 11, 2006 #3
    f(a) = g(a) = h(a)
    and f(x) <= g(x) <= h(x) for all x

    and

    f'(a) = h'(a)
     
    Last edited: Nov 11, 2006
  5. Nov 11, 2006 #4

    HallsofIvy

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    Staff Emeritus
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    Is it possible that f and h are given as differentiable? Otherwise, just take f(x)= g(x)= h(x) to be any non-differentiable functions and the hypotheses are satisfied while the conclusion is not true!
     
  6. Nov 11, 2006 #5
    you're right

    *I seem to have typed the question wrong many times*
    sorry.

    So, given:

    f(a) = g(a) = h(a)
    and
    f(x) <= g(x) <= h(x) for all x
    and
    f'(a) = h'(a)

    Prove g is differentiable at a, and that g'(a) = g'(a) = h'(a).
     
  7. Nov 11, 2006 #6

    StatusX

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    Do you know the squeeze theorem?
     
  8. Nov 11, 2006 #7
    oh!
    sol instead of just including g(x) in the inequality.. i include more:

    f(x) <= g(x) <= h(x)
    f(a+h) <= g(a+h) <= h(a+h)
    f(a+h) - f(a) <= g(a+h) - f(a) <= h(a+h) - f(a)
    f(a+h) - f(a) <= g(a+h) - f(a) <= h(a+h) - f(a)

    f'(a) <= g'(a) <= h'(a)

    but f'(a) = h'(a)

    so (how exactly do i phrase this?)

    f'(a) = g'(a) = h'(a)

    ooh. that was a good hint/spark =) thanks!
     
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