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Math Help - prove that there exists a delta>0 such that f(x)>0...

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

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

    In the definition of continuity use \varepsilon  = \frac{{f(a)}}{2} > 0.
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    Oops sorry it is supposed to be intersection. But how to I go about starting it?
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    Quote Originally Posted by alice8675309 View Post
    Oops sorry it is supposed to be intersection. But how to I go about starting it?
    I have already told you.
    \left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| <\dfrac{{f(a)}}{2}.
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    Quote Originally Posted by Plato View Post
    I have already told you.
    \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.
    Last edited by alice8675309; April 3rd 2011 at 07:01 AM.
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    Quote Originally Posted by alice8675309 View Post
    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: |y-a|<b \iff a-b < y < a+b
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