Supremum and infimum of set with absolute value

Question

I am looking for the supremum and infimum of $A = \{x \in \mathbb{R}: x^2-5 \lvert x \rvert + 4 <0\}$. I consider two separate cases ($x \ge 0$ and $x < 0$) and find that $x \in (-4; -1) \cup (1;4)$, so $\text{sup}A = 4$ and $\text{inf}A = -4$. This seems almost too easy. Is it?

Answer 1

You're right. Write $A= \{x \in \mathbb{R} | f(x) < 0\}$ with $f(x) = x^2 -5|x| + 4$.

For any $x<0$ $x\in A \iff f(x) = x² + 5x + 4 < 0$.

For any $x\geq 0$ $x\in A \iff f(x) = x² - 5x + 4 < 0$.

Basic calculus now gives, that $A = (-4,-1) \cup (1,4)$. Indeed it is just that simple as you said.

If you want to make this more plausible to you, calculate the derivative of $f$ and see that $f^\prime(x) <0$ for $x < -4$ and $f^\prime(x) > 0$ for $x > 4$. Together with having a root at $x = \pm 4$ you can conclude, that $f(x) > 0$ for $|x| > 4$.