Intermediate value theorem on infinite interval $\mathbb{R}$

Question

I have a continious function $f$ that is strictly increasing. And a continious function $g$ that is strictly decreasing. How to I rigorously prove that $f(x)=g(x)$ has a unique solution?

Intuitively, I understand that if I take limits to infinity, then $f$ grows really large and $g$ grows very small so the difference is less than zero. If I take the limits to negative infinity then the opposite happens. Using the intermediate value theorem, there must be an intersection, and since they are strictly increasing/decreasing, only one intersection happens.

My question is how do I apply the intermediate value theorem here? I don't have the interval to apply it on. I don't know when $f$ crosses $g$ and therefore can't take any interval. Or is there some sort of axiom applied here that I am missing?

Answer 1

This is not necessarily true, take $f(x)=e^x$ and $g(x)=-e^x$ then $f(x)=g(x)$ has no solution, if $f(x)=g(x)$ has a solution then it's unique!

Say $f(x)=g(x)$ has a solution

Consider the function $h(x)=f(x)-g(x)$, $h$ is continuous, and clearly $h$ is strictly incresing, so $h(x)=0$ cannot have more than $1$ real roots as it would violate that it's strictly increasing.

(As if $h(x)=h(y)$ for $x\neq y$ then either $x>y$ or $y>x$ so either $f(x)>f(y)$ or $f(y)>f(x)$, anyway we get a contradiction! )

Hence if $f(x)=g(x)$ has a solution then it's unique!

Answer 2

Note that if $f(x)=g(x)$ has a solution $x_0$ then it can have no other solution $x_1 >x_0$ because $f(x_1) > f(x_0) = g(x_0) > g(x_1)$ .
Similarly, it cannot have a solution $x_2 <x_0$ .
So, maximum number of possible solution is $1$

But however it is wrong to conclude that they will always have a solution, because we have counterexamples : $f(x)=e^x$ and $g(x)=-e^x$ contradicts your claim.