 #1
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 Homework Statement:

Determine whether the following set is bounded (from below, above, or both). If so,
determine the infimum and/or supremum and find out whether these infima/suprema
are actually minima/maxima.
 Relevant Equations:
 ##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##
##S_3 = \left\{ \ x∈ℝ : x^2+x+1≥0 \right\}##
I am not sure if I have done this correctly. The infimum/supremum and maximum/minimum are confusing me a bit.
This is how I started:
##x^2+x+1=0##
##x^2+x+ \frac1 4\ =\frac{3} {4}\ ##
## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ = 0##
Therefore: ## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ≥ 0##
Thus ##S_3 = ℝ##
Since## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ## will never be negative, ##S_2## is not bounded above or below.
The interval of ##S_2## is ## \left ( ∞,∞ \right ) ##
Therefore: ##inf S_2 = ∞## and ##sup S_2 = ∞ ##
I do not think there will be any minima or maxima since it is an open interval.
Can someone please give me some advice or hints on this problem. I am very confused at this moment and I'm not really sure if I have done this correctly.
Thank you very much in advance.
I am not sure if I have done this correctly. The infimum/supremum and maximum/minimum are confusing me a bit.
This is how I started:
##x^2+x+1=0##
##x^2+x+ \frac1 4\ =\frac{3} {4}\ ##
## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ = 0##
Therefore: ## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ≥ 0##
Thus ##S_3 = ℝ##
Since## \left\{ x^2+\frac 1 2\ \right\} ^2 +\frac 3 4\ ## will never be negative, ##S_2## is not bounded above or below.
The interval of ##S_2## is ## \left ( ∞,∞ \right ) ##
Therefore: ##inf S_2 = ∞## and ##sup S_2 = ∞ ##
I do not think there will be any minima or maxima since it is an open interval.
Can someone please give me some advice or hints on this problem. I am very confused at this moment and I'm not really sure if I have done this correctly.
Thank you very much in advance.