Find $\sup \{x: x^2 + x< 1\}$

Question

Find $\sup \{x: x^2 + x< 1\}$

If I factor, I get $x(x+1) < 1$. Should I use first derivative and find critical points?

Answer 1

Since$^1$ $$x^{2}+x<1\Leftrightarrow \frac{-1-\sqrt{5}}{2}<x<\frac{-1+\sqrt{5}}{2},$$ we have

$$\sup \{x: x^2 + x< 1\}=\sup\, ]\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}[\ =\frac{-1+\sqrt{5}}{2}.$$

The last equality follows from the definition of the supremum of a set of real numbers.

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$^1$ In general, if $f(x)=ax^2+bx+c$ and the discriminant $\Delta =b^2-4ac>0$, then $f(x)$ has two real roots $x_1,x_2$ and $f(\alpha)=a(\alpha -x_1)(\alpha -x_2)$. If $a>0$, then $f(\alpha)<0$ for $\alpha\in ]x_1,x_2[$. If $a<0$, then $f(\alpha)>0$ for $\alpha\in ]x_1,x_2[$.

Answer 2

Hint: the region where $x^2+x<1$ and the region where $x^2+x>1$ are separated by points where $x^2+x=1$. What are those points?

Answer 3

Hint: Complete the square. Just write $x^2+x = (x+\frac{1}{2})^2 - \frac{1}{4}$. So you are trying to find $$\sup \{x: (x+\frac{1}{2})^2 < \frac{5}{4}\}$$