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So there is this question

Given $f(x) = 2x^3 + 5$. Find the $\delta$ such that if $0 < |x+1| < \delta$ then $|(2x^3 + 5) -3| < 0.2$.

So I recognised this is an epsilon delta problem. But I haven't done this in a while so I don't remember how to do it.

This is what I tried so far

$\forall \epsilon > 0$, $\exists \delta > 0$, such that $|(2x^3 + 5) -3| < 0.2$ for all $x$ with $0 < |x + 1| < \delta$

GNUSupporter 8964民主女神 地下教會
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Itsnhantransitive
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  • $x+1$ is a factor of $2x^3 + 2$ find and upper bound for the other factor. $\frac {2x^3 + 2}{x+1} < M$ then $\delta = \frac {0.2}{M}$ will be your answer. – Doug M Feb 06 '18 at 21:00
  • You have for all $\epsilon$ with no $\epsilon$ in the inequality. – Calvin Khor Feb 06 '18 at 21:00
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    Solve $0<|x+1|<\delta$. Its solutions are the points in $(-1-\delta,-1)\cup(-1,-1+\delta)$. Solve $0.2>|(2x^3+5)-3|$. This is the same as $0.1>x^3+1>-0.1$. Or $-1+0.1>x^3>-1-0.1$. Or $\sqrt[3]{-1+0.1}>x>\sqrt[3]{-1-0.1}$. This is the interval $(\sqrt[3]{-1-0.1}, \sqrt[3]{-1+0.1})$ Now find $\delta$ such that the first set lies inside this interval. This can be achieved if $-1-\delta>\sqrt[3]{-1-0.1}$ and $-1+\delta<\sqrt[3]{-1+0.1}$. So, $0<\delta <\min(-1-\sqrt[3]{-1-0.1}, \sqrt[3]{-1+0.1} + 1)$. – orole Feb 06 '18 at 21:01
  • My answer here might be helpful in understanding the general pattern for these problems. – Ivo Terek Feb 06 '18 at 21:07

3 Answers3

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Hint: Factorise!

$$2x^3 + 5 -3 = 2x^3 + 2 = (x+1) \cdot \ ...$$

Zestylemonzi
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$$|(2x^3 + 5) -3| < 0.2 \iff |x^3+1| < 0.1$$ $$\iff -0.1< x^3 + 1 < 0.1$$ $$\iff -1.1< x^3 < -0.9 $$ $$\iff -\sqrt[3]{1.1}< x < -\sqrt[3]{0.9} $$ Choose $\delta = \min\{\sqrt[3]{1.1}-1,1-\sqrt[3]{0.9}\}$

GNUSupporter 8964民主女神 地下教會
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To solve this in general, note that $$\begin{align}|2x^3+5-3| &= |2x^3+2| = 2|x^3+1|=2|x+1||x^2-x+1| \\ &\leq 2|x+1|(|x|^2+|x|+1).\end{align}$$Note that $|x+1|<\delta$ implies $|x| < 1+\delta$, so that assuming $\delta < 1$ gives $$|2x^3+5-3| \leq 2\delta(4+2+1) = 14\delta.$$Meaning that $\delta = \min\{ \epsilon/14, 1\} > 0$ will always work. For $\epsilon = 0.2$, you can take $\delta = 0.2/14 = 1/70$.

Ivo Terek
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