0

Let $H=\{ x \in \Bbb R : x+\frac3x-1 \le 0\}$. Find $\sup(H)$, $\max(H)$, $\inf(H)$, and $\min(H)$. I understand the terminology but I am new to this please just offer any first steps or any help on how to get started.

Martin Sleziak
  • 53,687
M.Matthews
  • 79

2 Answers2

1

As the expression on the left is continuous, the values of $x$ for which it is true will form one or more intervals. The transitions will either be where $x+\frac 3x-1=0$ or where it is undefined, so solve the equation, add $0$ to the set of roots because the expression is undefined there, and assess the truth of the inequality on each resulting interval. Can you then find the $\inf, \sup, \max, \min$ once you have the interval(s)?

Ross Millikan
  • 374,822
  • I found the max to be (-\sqrt{3}, -2\sqrt{3}-1) and the min to just be the positive version of the max, is that correct? If so then have I found the Sup and Inf? @RossMillikan – M.Matthews Feb 09 '17 at 15:18
  • No, I don't find any real roots of the quadratic. – Ross Millikan Feb 09 '17 at 15:21
  • i''m sorry then I cannot understand what I need to do – M.Matthews Feb 09 '17 at 17:10
  • Once you know that there are no real roots of the quadratic, the only place the sign can change is $0$. If you evaluate the expression at $x=1$ you find $x+\frac 3x-1 \gt 0$ so the positive numbers are not in $H$. Now evaluate the expression at some convenient negative number, like $x=-1$. If it is greater than zero, then $H$ is empty. If it is less than zero, $H=(-\infty,0)$. Then you need to find the $\inf, \sup, \max, \min$ of $H$ – Ross Millikan Feb 09 '17 at 17:18
  • That was extremely helpful you are a saint thank you sir – M.Matthews Feb 10 '17 at 12:14
-1

Notice that we have $$x+\frac3x \ge 2\sqrt3$$ for $x>0$. There are various ways to show this, you can have a look at several posts with proofs of similar inequality $x + \frac{1}{x} \geq 2 $.

Let us denote $f(x)=x+\frac3x-1$. For $x>0$ you have $$f(x) = x+\frac3x-1 \ge 2\sqrt3-1 >0.$$ And for $x<0$ you have $$f(x) = -\left(|x|+\frac3{|x|}\right) -1 \le -2\sqrt3 - 1 <0.$$

So you can see that $f(x)>0$ for positive values and $f(x)<0$ for negative values. (You can also try to sketch the graph to see this.)

Which basically means that $$H=\{x\in\mathbb R; f(x)\le 0 \} = \{x\in\mathbb R; x<0\}.$$ Now that you have much simpler description of $H$, it should be easier to find supermum, infimum, maximum, minimum of $H$ (if they exist).

Community
  • 1
Martin Sleziak
  • 53,687