I have a set $E = \{x: x^2-x-1 < 0 \}$ for which I need to find the infimum and supremum (and minimum and maximum if exists). I'm not sure how to do it but after some calculation I cam up with $Inf(E) = - \infty \ Sup(E) = \frac{1}{2}$, and there is no minimum nor maximum.
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$\begingroup$ Hint: What does the graph of $x^2-x-1$ look like? $\endgroup$– celtschkJun 25, 2016 at 10:22
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$\begingroup$ If you don't show your thought process, nobody can read your mind to figure out why you got the wrong answer. $\endgroup$– user21820Jun 25, 2016 at 10:23
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$\begingroup$ Before worrying about $\sup$ and $\inf$, you should try to figure out exactly what $E$ is. What do you think $E$ is, and why do you think that is? $\endgroup$– ArthurJun 25, 2016 at 10:23
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2 Answers
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$$x^2-x-1 = (x-\frac{1+\sqrt{5}}{2})(x-\frac{1-\sqrt{5}}{2}) < 0$$
=> $$x \in (\frac{1-\sqrt5}{2},\frac{1+\sqrt5}{2})$$
Therefore E = $(\frac{1-\sqrt5}{2},\frac{1+\sqrt5}{2})$
It's clear that $ \sup E = \frac{1+\sqrt5}{2}$, $\inf E = \frac{1-\sqrt5}{2}$
EDIT: What may obfuscate you is that the supremum is not the x value where f(x) gets maximum. It is the x value in the set E which gives the set a upper boundary.
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HINT
Solving the inequation $x^2 -x -1 \lt 0$ we get $x \in (\frac {1-\sqrt5}2,\frac {1+\sqrt5}2)$