reverse of epsilon delta criterion Sorry for any mistakes. English is not my first language and I'm having this class in german, but tried translating my question because this way I'll get help faster.
The so-called $\epsilon \delta$ criterion (or the version we learned) goes as follows: If a function $f(x)$ is continuous at $x_0$, then for any $\epsilon > 0$ there exists a $\delta > 0$ so that the following applies: $|x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < \epsilon$.
Now my question is: With a given function, what would be the formula to get $\delta$ in relation (I hope that's formulated somewhat correct) to $x_0$ and $\epsilon$?
If you need an example function: $f(x) = -11 x^2 + 13 x + 3$
 A: I'm not sure I understand your question correctly. In general there is no formula for $\delta$ because it depends on whatever function $f$ and point $x_0$ you are working with. Sometimes $\delta$ might even be independent of $\epsilon$ (for example if $f$ is constant).
To find $\delta$ it usually helps to analyze $f$ and perhaps simplify $|f(x) - f(x_0)|$ in the concrete case to try and find a $\delta$ that works. This can be done for example by "solving" the resulting inequality $|f(x)-f(x_0)| < \epsilon$ for $x-x_0$ which gives you $\delta$.
Example:
Lets pick $f(x)= x^2 +2x +5$ and $x_0 =0$. Then we have $|f(x)-f(x_0)| = |x^2 +2x|$. Note that in this case the condition $|x-x_0| < \delta$ gives $|x|<\delta$ which simplifies the problem a bit.
Solving $x^2 + 2x = \epsilon$ for $x$ gives $-1 + \sqrt{1+\epsilon}$.
If we pick $\delta := -1 + \sqrt{1+\epsilon}$ then
$$|x|<\delta \Rightarrow |f(x) - f(x_0)| = |x^2 +2x| \leq x^2 + 2|x| < \epsilon$$
where the last inequality follows if you plug in $\delta$ for x and simplify.
Now note that this "solving" does not always work and is not really formally sound as I am sort of ignoring the absolute values in some cases. However it is a good trick that helps you find a $\delta$ which you can then properly justify by showing why that $\delta$ works as I did in the last equation.
A: We want to solve $$0<|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$$ for $\delta$.
Let $f^{-1}(S)$ where $S$ is a set denote the set of all $x$ that map to a value in $S$ by $f$:
$$f^{-1}(S)=\{x:f(x)\in S\}.$$
Now let
$$S:=(f(x_0)-\epsilon,f(x_0)+\epsilon),\\ x_-:=\inf_{x\ne x_0}(f^{-1}(S)),\\x_+:=\sup_{x\ne x_0}(f^{-1}(S)).$$
If these values exist, then
$$\delta=\min(x_0-x_-,x_+-x_0)$$ is a solution.
The expanded formula is
$$\delta=\min(x_0-\inf_{x\ne x_0}(f^{-1}((f(x_0)-\epsilon,f(x_0)+\epsilon))),\sup_{x\ne x_0}(f^{-1}((f(x_0)-\epsilon,f(x_0)+\epsilon)))-x_0).$$

Note that there is a very interesting case, when $f$ is invertible, as the formula reduces to
$$\delta=\min(x_0-f^{-1}(f(x_0)-\epsilon),f^{-1}(f(x_0)+\epsilon)-x_0)$$ if $f$ is growing, or
$$\delta=\min(x_0-f^{-1}(f(x_0)+\epsilon),f^{-1}(f(x_0)-\epsilon)-x_0)$$ otherwise. It suffices to evaluate the inverse at two points.
A: You mean a general formula, in terms of $f(x), \varepsilon, $ and $x_0$ ? Interesting question, but i doubt it can be done, because for example, the function: $f(x) = x$ if $x$ is rational; $-x$ if $x$ is irrational, is continuous at $x=0$, but I doubt a formula will find us what $\delta$ is: we would need to inspect the function from a human perspective. And of course you can come up with (a lot) more complicated examples than this one, where the function is still continuous at $x_0.$
