 # I have a continious function f that is strictly increasing. And a continious function g rzfansubs87 2022-06-29 Answered
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 $f\left(x\right)=g\left(x\right)$ 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?
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This is not necessarily true, take $f\left(x\right)={e}^{x}$ and $g\left(x\right)=-{e}^{x}$ then $f\left(x\right)=g\left(x\right)$ has no solution, if $f\left(x\right)=g\left(x\right)$ has a solution then it's unique!

Say $f\left(x\right)=g\left(x\right)$ has a solution

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

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

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

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