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

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

Originally Posted by alice8675309

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

You have a mistake. It must be intersection not union.
As in
BTW: Look a the correct LaTeX: [tex]x\in D\cap (a-\delta, a+\delta)[/tex].

In the definition of continuity use .

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

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.

Originally Posted by Plato

I have already told you.

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.

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: