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

• Apr 3rd 2011, 04:57 AM
alice8675309
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+$\displaystyle \delta$)?
• Apr 3rd 2011, 05:45 AM
Plato
Quote:

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 $\displaystyle \varepsilon = \frac{{f(a)}}{2} > 0$.
• Apr 3rd 2011, 05:53 AM
alice8675309
Oops sorry it is supposed to be intersection. But how to I go about starting it?
• Apr 3rd 2011, 05:59 AM
Plato
Quote:

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.
$\displaystyle \left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| <\dfrac{{f(a)}}{2}.$
• Apr 3rd 2011, 06:44 AM
alice8675309
Quote:

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.
• Apr 3rd 2011, 07:13 AM
TheEmptySet
Quote:

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