Thread: prove that there exists a delta>0 such that f(x)>0...

1. prove that there exists a delta>0 such that f(x)>0...

Let f: D->R be continuous at a point a$\displaystyle \in$D and assume f(a)>0. How would I prove that there exists a $\displaystyle \delta$>0 such that f(x)>0 for all x$\displaystyle \in$D$\displaystyle \cup$(a-$\displaystyle \delta$, a+$\displaystyle \delta$)?

2. Originally Posted by alice8675309 Let f: D->R be continuous at a point a$\displaystyle \in$D and assume f(a)>0. How would I prove that there exists a $\displaystyle \delta$>0 such that f(x)>0 for all x$\displaystyle \in$D$\displaystyle \cup$(a-$\displaystyle \delta$, a+$\displaystyle \delta$)?
You have a mistake. It must be intersection not union.
As in $\displaystyle x\in D\cap (a-\delta, a+\delta)$
BTW: Look a the correct LaTeX: $$x\in D\cap (a-\delta, a+\delta)$$.

In the definition of continuity use $\displaystyle \varepsilon = \frac{{f(a)}}{2} > 0$.

3. Oops sorry it is supposed to be intersection. But how to I go about starting it?

4. Originally Posted by alice8675309 Oops sorry it is supposed to be intersection. But how to I go about starting it?
I have already told you.
$\displaystyle \left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| <\dfrac{{f(a)}}{2}.$

5. Originally Posted by Plato I have already told you.
$\displaystyle \left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| <\dfrac{{f(a)}}{2}.$
Right. My biggest problem is going from what we have to what I need, which is basically writing the proof. Following the format of my notes and other proofs, I've been trying to put together a proof. I don't know but is this right:

Assume that f is continuous at a and that f(a)>0. Now we choose epsilon=f(a)/2>0. By the definition of continuous, there exists a delta>0 such that for any x, |x-a|<delta then |f(x)-f(a)|<f(a)/2.

I'm not sure about this but its what I managed to get out.

6. Originally Posted by alice8675309 Right. My biggest problem is going from what we have to what I need, which is basically writing the proof. Following the format of my notes and other proofs, I've been trying to put together a proof. I don't know but is this right:

Assume that f is continuous at a and that f(a)>0. Now we choose epsilon=f(a)/2>0. By the definition of continuous, there exists a delta>0 such that for any x, |x-a|<delta then |f(x)-f(a)|<f(a)/2.

I'm not sure about this but its what I managed to get out.
You need to keep going with the algebga. Plato gave you what you need.

Hint: $\displaystyle |y-a|<b \iff a-b < y < a+b$

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