Let $S=\{x>0:3x^2-8x-3\le0\}$. Find the supremum and infimum.
I want to solve this question, although I have no idea how to do it, I think I should guess the $\sup S$ and $\inf S$ first and try to prove it. I have solve the equation $3x^2-8x-3=0$ and get $x=3$. Then I found that if we have an equation $3x^2-8x-3=p$, for some $p\le0$, then we have $$x=\frac{8\pm \sqrt{64+12(3+p)}}{6}$$ and $$\frac{8\pm \sqrt{64+12(3+p)}}{6}\le3 \iff p\le0$$ Hence, I think I have proven that $3$ is a upper bound of $S$.
I know if I want to show that $3=\sup S$, I need to show that for any $\varepsilon>0$, there exists a number $t$ in $S$ such that $t>3-\varepsilon$, but I don't know how to prove that. And I have no idea how to find a lower bound of $S$ and $\inf S.$
Best Answer
Rewrite the inequality as follows:
$$3x^2-8x-3 \le 0 \land x>0 $$
$$\iff(3x+1)(x-3) \le 0 \land x > 0$$
$$\iff -\frac13 \le x \le 3 \land x > 0$$
$$\iff 0 < x \le 3$$
and it should be clear that the supermum is indeed $3$ and the infimum is $0$.