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 $0 < |x + 1| < \delta$
On
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 $\delta = 0.2/14 = 1/70$.
$$|(2x^3 + 5) -3| < 0.2 \iff |x^3+1| < 0.1$$ $$\iff -0.1< x^3 + 1 < 0.1$$ $$\iff -1.1< x^3 < -0.9 $$ $$\iff -\sqrt[3]{1.1}< x < -\sqrt[3]{0.9} $$ Choose $\delta = \min\{\sqrt[3]{1.1}-1,1-\sqrt[3]{0.9}\}$