 Use the given graph off to find a number \delta killjoy1990xb9 2022-01-07 Answered
Use the given graph off to find a number $$\displaystyle\delta$$ such that if $$\displaystyle{\left|{x}-{1}\right|}{ < }\delta$$ then $$\displaystyle{\left|{f{{\left({x}\right)}}}-{1}\right|}{ < }{0.2}$$

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$$\displaystyle{\left|{X}\right|}{ < }{a}$$ implies $$\displaystyle-{a}{ < }{X}{ < }{a}$$
Add 1 in each and simplify
$$\displaystyle{\left|{f{{\left({x}\right)}}}-{1}\right|}{ < }{0.2}$$
$$\displaystyle-{0.2}{ < }{f{{\left({x}\right)}}}-{1}{ < }{0.2}$$
$$\displaystyle{0.8}{ < }{f{{\left({x}\right)}}}{ < }{1.2}$$
The graph shows that if $$\displaystyle{0.8}{ < }{f{{\left({x}\right)}}}{ < }{1.2}$$ x must verify $$\displaystyle{0.7}{ < }{x}{ < }{1.1}$$
On the other hand we have x that must verify this unequality:
$$\displaystyle{\left|{x}-{1}\right|}{ < }\delta$$
$$\displaystyle-\delta{ < }{x}-{1}{ < }\delta$$
If we can find $$\displaystyle\delta$$ that verify $$\displaystyle-{0.3}\leq-\delta{ < }{x}-{1}{ < }\delta\leq{0.1}$$, we can make sure that $$\displaystyle{0.7}{ < }{x}{ < }{1.1}$$
so $$\displaystyle\delta$$ must verify two conditions: $$\displaystyle\delta\leq{0.1}$$ and $$\displaystyle-{0.3}\leq-\delta$$
which means that $$\displaystyle\delta\leq{0.1}$$ and $$\displaystyle{0.3}\geq\delta$$
$$\displaystyle{0.7}{ < }{x}{ < }{1.1}$$
$$\displaystyle{0.7}-{1}{ < }{x}-{1}{ < }{1.1}-{1}$$
$$\displaystyle-{0.3}{ < }{x}-{1}{ < }{0.1}$$
We can choose $$\displaystyle\delta={0.1}$$, so to be sure two conditions are satisfied, but if you want, you can choose any value.
Not exactly what you’re looking for? Toni Scott

You can find it using the graph
1. Find the intervals where $$\displaystyle{f{{\left({x}\right)}}}$$ and $$\displaystyle{x}$$ must be. In our case for $$\displaystyle{f{{\left({x}\right)}}}={\left[{0.8},{1.2}\right]}$$, for $$\displaystyle{x}={\left[{0.7},{1.1}\right]}$$
2. Find the middle of the first interval. We have $$\displaystyle{1}$$ $$\displaystyle{\left({\frac{{{0.8}+{1.2}}}{{{2}}}}={1}\right)}$$
3. Find $$\displaystyle{x}_{{0}}$$, wheree $$\displaystyle{f{{\left({x}_{{0}}\right)}}}{Z}={d}\le={1},\ {w}{e}\ {h}{a}{v}{e}\ {x}_{{0}}={1}$$
4. Calculate the distance between $$\displaystyle{x}_{{0}}$$ and the endpoints of the second interval, in our case: $$\displaystyle{\left|{x}_{{0}}-{0.7}\right|}={\left|{1}-{0.7}\right|}={0.3}$$ and $$\displaystyle{\left|{x}_{{0}}-{1.1}\right|}={\left|{1}-{1.1}\right|}={0.1}$$
5.$$\displaystyle\delta$$ is the minimum of these two values, thus it's $$\displaystyle{0.1}$$