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?
Best Answer
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.