Intermediate value theorem on infinite interval R 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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Intermediate value theorem on infinite interval       R  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&#039;t have the interval to apply it on. I don&#039;t know when f crosses g and therefore can&#039;t take any interval. Or is there some sort of axiom applied here that I am missing?

neobuzdanio · Accepted Answer

Step 1Note that if   f  (  x  )  =  g  (  x  ) has a solution       x    0   then it can have no other solution       x    1    &amp;gt;      x    0   because   f  (      x    1    )  &amp;gt;  f  (      x    0    )  =  g  (      x    0    )  &amp;gt;  g  (      x    1    ). Similarly, it cannot have a solution       x    2    &amp;lt;      x    0  . So, maximum number of possible solution is 1Step 2But 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.

Karsyn Beltran · Accepted Answer

Step 1This 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&#039;s unique!Say   f  (  x  )  =  g  (  x  ) has a solutionConsider 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&#039;s strictly increasing.Step 2(As if   h  (  x  )  =  h  (  y  ) for   x  ≠  y then either   x  &amp;gt;  y or   y  &amp;gt;  x so either   f  (  x  )  &amp;gt;  f  (  y  ) or   f  (  y  )  &amp;gt;  f  (  x  ), anyway we get a contradiction! )Hence if   f  (  x  )  =  g  (  x  ) has a solution then it&#039;s unique!

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?

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