# How to find infimum and supremum of an inequality?

Given $$A=\{x\in \mathbb{R}: (2x^2+x-21)(x^2+2x)<0\}$$, I want to find $$\inf(A)$$ and $$\sup(A)$$ and to if it admits minimum and/or maximum.

First of all, I've solved the inequality, which gives me that for $$]-2:-\frac{7}{2}[$$ and $$]0:3[$$ is negative, whereas the other parts of the domain are positive. I think the maximum of the function, in this case, are indeed $$-2;-\frac{7}{2};0;3$$, but I don't know how to find $$\inf(A)$$. Moreover, how do I know if it admits minimum and/or maximum?

• It's simply $inf(A)=-2$ and $sup(A)=3$. Commented Mar 5, 2019 at 13:28
• I think your solution is not right! Commented Mar 5, 2019 at 13:29

Well, you got that $$A = \left\langle -2,-\frac72\right\rangle \cup \langle 0,3\rangle$$ so $$\inf A = -2$$ and $$\sup A = 3$$. Neither is attained.
Your solution is $$A = \left[-3.5, -2 \right] \cup [0,3].$$ Here, we used to $$-\dfrac{7}{2} = -3.5$$. Let is a definition: "If $$A$$ is bounded above, then a number $$x$$ is said to be a supremum (or a least upper bound) of $$A$$ if it satisfies the two conditions:
(1) $$x$$ is an upper bound of $$A$$ and
(2) if $$y$$ is any upper bound of $$A$$, then $$x \leq y$$".
So, if $$-2$$ is not a supremum of $$A$$, because $$-2$$ is not an upper bound of $$A$$. So, the supremum is a number 3.
The same way to an infimum, i.e., $$\inf A = -\dfrac{7}{2}$$.