$\delta-\epsilon$ Question on Ordered Field $\mathbb{R}$

Question

I got came across this question with the $\delta-\epsilon$ definition of a limit, but I do not know how to use it to solve the context of this problem:

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Problem: Let $f:\mathbb{R}\to\mathbb{R}$ be defined as follows: $f(x)=3-2x$, for all $x\in\mathbb{R}$. Let $\epsilon\in(0,\infty)$. Find (in terms of $\epsilon$) the largest real numbers $\delta$ possible such that $$(\forall x\in\mathbb{R}) \ (|x-1|<\delta\implies|f(x)-f(1)|<\epsilon).$$

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I had only gotten up to the point where you plug in $f(x)=3-2x$ and $f(1)$ into the centered inequality greater than $\epsilon$. However I have no idea on how to go from there. Hints, approaches, and spoilers are welcome!

Answer 1

Note $$|f(x) - f(1)| = |(3 - 2x) - (3 - 2)| = |2 - 2x| = 2|x - 1|.$$ So $|f(x) - f(1)| < \varepsilon$ if $|x - 1| < \frac{\varepsilon}{2}$. If you choose $\delta > \frac{\varepsilon}{2}$, then set $x = 1 + \frac{\varepsilon}{2}$. We have $|x - 1| = \frac{\varepsilon}{2} < \delta$, but $|f(x) - f(1)| = 2|x - 1| = \varepsilon$. Hence, the largest $\delta$ is $\frac{\varepsilon}{2}$.

Answer 2

Approach the question geometrically. Let $\epsilon>0$ and consider those $x$ values in the domain for which $f(x)=1\pm\epsilon$. I leave it to you to prove that $\delta\leq\epsilon/2$.

Answer 3

The function $f$ considered as a map $$f:\quad{\mathbb R}\to{\mathbb R},\qquad x\mapsto y=f(x)$$ multiplies all distances by $2$. Therefore, if your $y$-tolerance is $\epsilon$, you should make sure that the allowed $x$-deviation is $\leq{\epsilon\over2}$.