# Find the supremum and infimum

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.

• Hint: What does the graph of $x^2-x-1$ look like? Jun 25, 2016 at 10:22
• If you don't show your thought process, nobody can read your mind to figure out why you got the wrong answer. Jun 25, 2016 at 10:23
• 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? Jun 25, 2016 at 10:23

$$x^2-x-1 = (x-\frac{1+\sqrt{5}}{2})(x-\frac{1-\sqrt{5}}{2}) < 0$$

=> $$x \in (\frac{1-\sqrt5}{2},\frac{1+\sqrt5}{2})$$

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.

HINT

Solving the inequation $x^2 -x -1 \lt 0$ we get $x \in (\frac {1-\sqrt5}2,\frac {1+\sqrt5}2)$