1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
  6. Nov 14, 2013 #5
    Try to clarify

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted