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If f : R −→ R is continuous and f (7) > 2, then ∃δ > 0 such that

  1. Nov 14, 2013 #1
    If f : R −→ R is continuous and f (7) > 2, then ∃δ > 0 such that ....

    If f : R −→ R is continuous and f (7) > 2, then ∃δ > 0 such that f (x) > 2 ∀x ∈ Vδ (7).

    I know the definition of continuous at a point. However, the question does not specific any particular point. Will it still work?

    Could anyone help get me started?

    What I got so far
    For all epsilon > 0 there exist δ > 0 such that whenever x in R and |x-7|< δ, it follow that |f(x) - f(7)| < |f(x) - 2|< sth that I am not sure

    Am I going the right direction?
     
  2. jcsd
  3. Nov 14, 2013 #2

    pasmith

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    You're not told that [itex]f(7) = 2[/itex], you're told that [itex]f(7) > 2[/itex].

    The definition of continuity says that something is true for all strictly positive numbers, so you may want to see what follows if you take a particular strictly positive number.
     
  4. Nov 14, 2013 #3

    PeroK

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    First, you need to understand why this is true. Then you turn that understanding into the e-d formal proof.

    So, f(7) > 2. If f is continuous, then it must be > 2 for a (perhaps small) distance either side of 2. Note that:

    f may be > 2 for all x.
    f may only be greater than 2 close to 7 (x = 7 may be a local maximum of 2.1, say).

    But, one thing f can't do is jump down to less than 2 arbitrily close to 7.

    If you draw a graph of f, it must remain > 2 for a finite distance round 7.

    You can use e-d proofs as follows:

    [itex]Let \ \epsilon = f(7) - 2[/itex]

    [itex]Then \ \exists \ \delta > 0 \ s.t. |x-7| < \delta \Rightarrow \ |f(x) - f(7)| < \epsilon[/itex]

    [itex]\Rightarrow f(x) > f(7) - \epsilon \ = \ 2[/itex]

    Note that (in case you're wondering):

    [itex]|f(x) - f(7)| < \epsilon \ \Rightarrow \ f(x) \in (f(7)-\epsilon, f(7)+\epsilon)[/itex]

    [itex]\Rightarrow \ f(7)-\epsilon < f(x) < f(7)+\epsilon[/itex]

    And we only needed the first inequality in this case.

    The trick in my view with these problems is not to think about e-d until you've understood the geometry of the situation.
     
  5. Nov 14, 2013 #4
    Thanks
     
  6. Nov 14, 2013 #5
    Try to clarify

    Good
     
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