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Proving local minimum

  1. Nov 24, 2004 #1
    Suppose that c1 < c2 and that f takes on local maxima at c1 and c2. Prove that if f is continuous on [c1, c2], then there is at least one c in (c1, c2) at which f takes on a local minimum.

    This question seems common sense, but does anyone know how to actually prove this?
     
  2. jcsd
  3. Nov 24, 2004 #2

    Tide

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    Hint: Sketch the graph of f ' (x). Since f ' ' (c1) < 0 and f ' ' (c2) < 0 the graph must decrease as you move away from x = c1 and toward x = c2. Likewise, the graph must increase as you move away from x = c2 and toward x = c1. Therefore, f ' (x) must pass from a positive value to a negative value somewhere in (c1, c2).
     
  4. Nov 24, 2004 #3

    Hurkyl

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    Don't forget that f might not be differentiable... though looking at derivatives might jump start your understanding of the problem.


    Trap: do you know anything, in general, about the minima of continuous functions on closed intervals?
     
  5. Nov 24, 2004 #4
    thanks everyone for the responds.

    I do not know about the minima of continuous functions on closed intervals, can u provide an answer to this? thanks :smile:
     
  6. Nov 24, 2004 #5

    Hurkyl

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    Any continuous function on a compact set (such as a closed interval [a, b]) has a global minimum and maximum.

    So, for your problem, it would suffice to prove that a and b can't be the (only) global minimum.


    I don't know if you've had this theorem yet, though.
     
  7. Nov 24, 2004 #6
    m....I haven't learnt about global min/max theorems, but thanks anyways for your help and explainations.
     
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