$f$ and $g$ are functions of real variables, strictly increasing and strictly decreasing respectively on $\Bbb R$ (both surjections), I'm asked to prove that there exists at most one solution to the equation $f(x)=g(x)$.
My attempt: Suppose $f(x)=g(x)=m,$ for some $x\in \Bbb R \implies \exists a,b \in \Bbb R | a\lt x\lt b$. Then : $$f(a)\lt f(x)\lt f(b)$$ and $$-g(a)\lt -g(x)\lt -g(b).$$ Hence $$f(a)-g(a)\lt f(x)-g(x)\lt f(b)-g(b).$$ So if we can prove that $f(a)-g(a)\lt 0$ and $f(b)-g(b)\gt 0$. Hence for $f(a)-g(a)=p\lt 0$ and $f(b)-g(b)=q\gt 0$, we will have $p\lt 0\lt q \equiv \top$, but I couldn't prove it. Any help would be appreciated.