Infimum and supremum for $\{ x \in \Bbb R\mid x^2-2x-1 < 0 \} $

Question

I've been asked to obtain the infimum, supremum, minimum and maximum for:

$\{ x \in \Bbb R \mid x^2-2x-1 < 0 \} $ and $\{x^2-2x-1\mid x \in \Bbb R\} $

So for the first, I used the quadratic formula and came up with: $$\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot1\cdot-1}}{2\cdot1}=\frac{2\pm\sqrt{8}}{2} \Rightarrow 1-\sqrt 2<x<1+\sqrt 2$$ So based on this I suggested there is no minimum or maximum and supremum and infimum is $ 1+\sqrt 2$ and $ 1-\sqrt 2$ respectively.

For the second, I was given the answer as $[2, \infty)$, but I do not understand how this was obtained. Any suggestions?

Answer 1

Note that $x^2-2x-1=(x-1)^2-2$, so that the value of the expression is actually in the range $[-2,\infty)$. Moreover, given any $y\in[-2,\infty)$, we can set $x:=\sqrt{y+2}+1$ to show that it is in the set.

Answer 2

You can use high-school results on quadratic polynomials: as the leading coefficient of $x^2-2x-1$ is positive, this quadratic has no supremum, and it has a minimum attained at $$x=-\frac b{2a} \text{ standard notation) }=1.$$ Now $\;(1)^2-2\cdot 1 - 1=-2$. This is made obvious by completing the square: $$x^2-2x-1=x^2-2x+1-2=(x-1)^2-2\ge -2.$$