Finding the supremum of a set

Question

Consider the sets A and B defined by $$A=\{ \frac{x}{x^2+7}: x \in \Bbb R\} \ \ and \ \ b=\{ \frac{x}{x^2+7}: x \in \Bbb N\} $$ What are the values of sup A and sup B?

For the first part of the question, I did $(x-\sqrt7)^2 \ge 0$ then $2\sqrt7x \le x^2+7$, and so that $\frac{x}{x^2+7} \le \frac{x}{2x\sqrt7}$ , which means $sup A= \frac{1}{2 \sqrt7}$.

Is this part of the working correct? And also, I am not sure how I would approach the second part of the question which asks to find sup B. I think $supB= \frac{3}{16}$ but I am not sure what are the steps of working that I need to include in order to show that.

Thanks to anybody who helps.

Answer 1

What you did pretty much seems to be the way to go here.

For the first set $A$, you showed that $\frac{x}{x^2+7}\leq \frac{1}{2\sqrt{7}}$ for all real $x$, so we know that $y\in A \Rightarrow y\leq \frac{1}{2\sqrt{7}}$. Technically, that leaves us only with the knowledge that $\frac{1}{2\sqrt{7}}$ is an upper bound for the set $A$, not necessarily the smallest one.

However, this is only a technicality, since this value is actually assumed (take $x=\sqrt{7}$), so there is an $y\in A: y=\frac{1}{2\sqrt{7}}$, meaning that it is also the smallest upper bound, hence the supremum. Which also happens to be a maximum.

For the second one, you could either prove that your function is decreasing, or find an upper bound for $x$ sufficiently large - if, say, $x>10$, then $\frac{x}{x^2+7}<\frac{x}{10x+7}<\frac{1}{10}$, leaving you with only a few values for $x$ to try.

Answer 2

Your working for the case of real numbers is right. You might want to mention that $x^2+7 > 0$ and hence dividing doesn't change the sign of the inequality.

For the case of natural numbers, note that the function is decreasing for $x \in (-\infty,-\sqrt7) \cup (\sqrt7,\infty)$ and it is increasing for $x \in (-\sqrt7, \sqrt7)$. Further it is negative for $x \in (-\infty,0)$. Hence the maximum over the set of natural numbers occurs at either $x = \lfloor \sqrt7 \rfloor = 2$ or $x = \lceil \sqrt7 \rceil = 3$. Checking for $2$ and $3$, we can conclude that the maximum occurs at $3$.