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 ( 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?
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Answers (1)

Johnathan Morse
Answered 2022-06-30 Author has 18 answers
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 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!

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