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The inequality which implies f(x) > 0 - Spivak's Calculus

  1. Aug 25, 2013 #1
    Hello,

    if you guys would turn to page 117 in Spivak's Calculus, there is the proof for theorem 3. At the last line he stated that this last inequality ##|f(x)-f(a)|<f(a)## implies ##f(x)>0##. How can you check this fact?

    Can we assume first that ##f(x)-f(a)<0## to eliminate the absolute value, which leads to the inequality ##f(x)>0##?

    On the second case if ##f(x)-f(a)>0##, we have ##2f(a)-f(x)>0##, how can we infer that ##f(x)>0##

    Is it sufficient to show that ##f(x)>0## with the first case? (Logically this means ##(f(x)>0)\wedge (f(x)<2f(a))##

    Thank You
     
  2. jcsd
  3. Aug 25, 2013 #2
    You have to prove the result for both cases of the sign of ##f(x) - f(a)##. If you only prove it when the quantity is negative, then you haven't proved the result. You've only proved it when it is negative.

    Fortunately, the other case isn't too hard. I can't figure out how to give a proper hint for this question so I'll just tell you the answer. If ##f(x) - f(a) \ge 0## then ##f(x) \ge f(a)##. But an assumption at the beginning of the theorem was that ##f(a) > 0##, so you're done.
     
  4. Aug 25, 2013 #3
    First attempt:

    From ##f(x)-f(a)>0## we have ##f(x)>f(a)##
    From ##f(x)-f(a)<f(a)## we have ##f(x)<2f(a)##

    so ##f(a)<f(x)<2f(a)##

    but ##f(a)<2f(a)\iff f(a)>0\implies f(x)>f(a)>0\implies f(x)>0##

    Hope this is helpful (and correct xD)

    [Edit]: I don't have the book
     
  5. Aug 25, 2013 #4
    Ok thanks you all, I understand it now!
     
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