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Simple Calculus Question

  1. Feb 10, 2014 #1
    Hello everyone! I'm studying out of Spivak's calculus on my own and ran into a problem I can't explain on Theorem 1 of Chapter 11 (of the third edition). It's probably a very simple problem (Spivak calls it an easy theorem), but I'm still at a roadblock.

    Spivak wants to prove that if f is a function defined on (a,b), f is differentiable at x, and x is a maximum (or minimum) point for f on (a,b), then f'(x) = 0.

    Spivak shows that, if h>0, then [f(x+h) - f(x)] / h ≤ 0. This implies that the one-sided limit, as h approaches 0 from above, of [f(x+h) - f(x)] / h ≤ 0. (Sorry, I'm not sure about how to do the limit notation on the computer).

    I wasn't sure how he made the leap that: because that function is ≤ 0, the one-sided limit is ≤ 0. I used a proof by contradiction to show that, if the one-sided limit exists, it must be ≤ 0, but this is assuming it exists. So my main issue is that I don't know how we can prove that the one-sided limit exists.

    I was thinking of using the given fact that f is differentiable at x to show that the one-sided limit must exist. However, Spivak doesn't use this argument, and also I was thinking: even if the two-sided limit at a maximum did not exist, the one-sided limits would still exist, right? (but not be equal to each other). So I was wondering if there was any way to prove that the one-sided limit exists without using the differentiability of the function at x.

    Thanks for any input
     
  2. jcsd
  3. Feb 10, 2014 #2

    olivermsun

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    Hmm. First thought — how does Spivak define a "differentiable" function?
     
  4. Feb 10, 2014 #3
    Spivak defines a function as differentiable at x if the limit, as h approaches 0, of [f(x+h) - f(x)] / h exists. In this way, I think that you could say that the one-sided limit must exist because this 2-sided limit exists at x (by virtue of the function being differentiable at x).

    But I was more wondering: if the function has a maximum at x but is not differentiable at x, then does [f(x+h) - f(x)] / h being ≤ 0 imply that the limit, as h approaches 0 from above, of [f(x+h) - f(x)] / h is ≤ 0? And if so, why?
     
  5. Feb 10, 2014 #4

    AlephZero

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    With those conditions, the limit doesn't even need to exist. For example
    f(x) = 1-x if x is rational and x >= 0,
    f(x) = 0 if x is irrational or x < 0

    Clearly f(x) has a maximum value of 1 when x = 0.

    For any h > 0, (f(0+h) - f(0))/h < 0, because f(0) = 1 and f(0+h) < 1, but the limit at x = 0 does not exist.

    On the other hand, if a function is differentiable at a point, it is also continuous at that point, which is a different ball-game.
     
  6. Feb 11, 2014 #5
    Thanks! That makes sense to me now
     
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