I am unsure how to solve the following
Using the graph $f(x)=x^2/3$ find the largest possible $\delta$ such that if
$$0< |x-3|< \delta$$ then $$0< \left|\frac{x^2}{3}-3\right|<1.$$
I am not sure what to do.
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Such that if what? – Git Gud Jan 26 '13 at 23:44
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Such that if what? – Sigur Jan 26 '13 at 23:44
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Surely there is more to the question. What is the rest of the sentence at line 2? – Shaun Ault Jan 26 '13 at 23:44
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1I think there is something missing in your question. – Tomás Jan 26 '13 at 23:44
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@GitGud, you were 5 seconds faster than me... lol – Sigur Jan 26 '13 at 23:45
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Lynch the infidel! – Git Gud Jan 26 '13 at 23:45
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I dont know why but what I wrote did not show up.. – Fernando Martinez Jan 26 '13 at 23:47
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A related problem. – Mhenni Benghorbal Jan 27 '13 at 00:01
1 Answers
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First, you can ignore the $0 \lt $ part as unless $x=3$ the contents of the absolute value sign will be non-zero, so just concentrate on keeping it less than $1$. For the absolute value to be exactly $1$, you must have $\frac {x^2}3=2$ or $\frac {x^2}3=4$. So solve both of those and take the smaller difference from $3$ as $\delta$
Ross Millikan
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I solved both of them and got square root(6) and square root(12) does this mean I subtract square root 12 - square root(6) and the subtract this from 3 to get delta. – Fernando Martinez Jan 27 '13 at 19:16
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@FernandoMartinez: you are right that you need $x \in (\sqrt 6,\sqrt{12})$ for $|\frac {x^2}3-3|\lt 1$. But that interval is not symmetric around $3$, while having $|x-3| \lt \delta$ produces a symmetric interval. So you need to choose the "worst case" direction. That is why I said to take the smaller difference from $3$. – Ross Millikan Jan 28 '13 at 00:35