# Use the given graph off to find a number \delta

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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Philip Williams
$$\displaystyle{\left|{X}\right|}{ < }{a}$$ 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 youâ€™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 $$\displaystyle{0.1}$$