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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Expert Answer

Philip Williams
Answered 2022-01-08 Author has 3715 answers
\(\displaystyle{\left|{X}\right|}{ < }{a}\) 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.
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Toni Scott
Answered 2022-01-09 Author has 3431 answers

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 \(\displaystyle{0.1}\)

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