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

1. Nov 14, 2013

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. Nov 14, 2013

### pasmith

You're not told that $f(7) = 2$, you're told that $f(7) > 2$.

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.

3. Nov 14, 2013

### PeroK

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:

$Let \ \epsilon = f(7) - 2$

$Then \ \exists \ \delta > 0 \ s.t. |x-7| < \delta \Rightarrow \ |f(x) - f(7)| < \epsilon$

$\Rightarrow f(x) > f(7) - \epsilon \ = \ 2$

Note that (in case you're wondering):

$|f(x) - f(7)| < \epsilon \ \Rightarrow \ f(x) \in (f(7)-\epsilon, f(7)+\epsilon)$

$\Rightarrow \ f(7)-\epsilon < f(x) < f(7)+\epsilon$

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.

4. Nov 14, 2013

Thanks

5. Nov 14, 2013