Shortcuts to solving inequalities

Question

While working on inequalities in an assignment, I happen to patterns and shortcuts for solving a handful of inequalities, taking the example $x^2-\left| 3x+2 \right|\le 8x$, it's solutions is $I=\left[ \frac{1}{2}(11-\sqrt{129}),\frac{1}{2}(11+\sqrt{129}) \right]$. First thing I notice is that the endpoints are equal to the zeroes of $x^2-\left| 3x+2 \right|= 8x.$ Other examples like $\frac{1}{3}x^{3}\lt 2x+1 $ also hold true, I hypothesize that there must be an implication of some sort where

If $f(x)=ax+b,$ where $a,b \in \mathbb{R^{+}}$ and $g(x) \neq f(x)+c$, where $c\in\mathbb{R}$, then the endpoints of $I$, the solution of the inequalities $f(x)\le g(x),f(x)\lt g(x),f(x)\ge g(x),f(x)\gt g(x)$ are the roots of $f(x)-g(x)$ greater than $0$.

Second thing is the relation between the solutions of $f(x)\le g(x)$ and $f(x)\gt g(x)$. Using the same examples , the solution of $x^2-\left| 3x+2 \right|\gt 8x$ is $I'=\left]-\infty ,\frac{1}{2}(11-\sqrt{129}) \right[\cup \left] \frac{1}{2}(11+\sqrt{129}) ,+\infty \right[$ . Which is exactly $I^{C}$, the complementary of $I$ to $\mathbb{R}$.By using other examples as evidence led me to believe that the following is true.

Let $I$ be the solution of $f(x)\le g(x),$ then the solution of $f(x)\gt g(x)$ is $I^{C}$.

My issue with this is that I am assuming all of these based on intuition, I would like help on how to prove the 2 statements if they are true or false. Help is appreciated! :)

Answer 1

The first is almost true. Let me state a true, more general version.

Claim. Let $f,g:\mathbb R\to \mathbb R$ be continuous functions. If $I=\{x\mid f(x)>g(x)\}$, then $I$ is a countable union of disjoint open intervals whose endpoints are roots of $f-g$. Moreover, if $f-g$ has $k<\infty$ roots then $I$ is a union of at most $k/2+1$ intervals.

Note. By adding $-g$ to everything, we may assume that $g=0$. Also, note that the endpoints of $I$ do not necessarily contain all roots of $f-g$. E.g. $f=x^2$ and $g=0$. The intuition using IVT is that as $f$ "travels from $>0$ to $\leq 0$", it must pass through the point $0$. The same statement is true for $<$.

Proof. First note that $I=f^{-1}(0,\infty)$ is a preimage of an open set, and thus open. Every open set in $\mathbb R$ is a countable union of disjoint open intervals. Let $x$ be an endpoint of an interval in $I$. Without loss of generality, assume $x$ is an upper endpoint. There is some $y<x$ such that $f(y)>0$ and $(y,x)\subseteq I$. Suppose now that $f(x)<0$. By the Intermediate Value Theorem, there is some $z\in (y,x)$ with $f(z)=0$, contradicting our assumption that $(y,x)\subseteq I$. This means $f(x)\geq 0$. But $f(x)\notin I$, so $f(x)=0$. Thus the endpoints of intervals in $I$ are roots of $f$. For the 'moreover' part, note that every interval has two endpoints, with the exception of $-\infty$ and $\infty$.

Your second Claim is spot-on!

Claim. Let $f,g:\mathbb R\to \mathbb R$ be arbitrary functions. If $I=\{x\mid f(x)>g(x)\}$ and $J=\{x\mid f(x)\leq g(x)\}$, then $I^C=J$.

Proof Hint. You must prove that if $x\in I$, then $x\notin J$, and vice versa.

Note. Using the claims together (assuming $f,g$ continuous), we see that $J$ must be a finite union of closed intervals.