For $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$. Neither is convergent. $f(0) < g(0)$ but $f'(x)>g'(x)$ for all $x>x_0$. Is it true that there always exists an $a$ where $a>x_0$ such that $f(a) = g(a)$?
This is for one of my Physics homework that involves proving that a transcendental equation has exactly one solution. Graphically I can show that a solution seems to always exist, but I am having trouble proving it algebraically. Since the equation is transcendental, there is no analytical solution so I thought maybe a general theorem like the one I stated could be helpful. Unfortunately, because I'm bad at Maths, I don't know if that statement is actually true.
The equation is: $$e^{-ka/2} + e^{ka/2} = \frac{\hbar^2k}{m\alpha}e^{ka/2}$$ And I need to prove that there is always exactly one solution for $k$ for every $a$, $m$, and $\alpha$.
No, this is not true in general. Consider
$$f(x)= 1-e^{-x}$$ and $$g(x) = 2.$$
Then $f(0) < g(0)$ and $$f'(x) = e^{-x} > 0 = g'(x),$$
but you can see there is no intersection.
However, if you have instead that for some $a,b$ with $a<b$
$$f(a) < g(a)$$ and
$$f(b) > g(b)$$
you will have a $c$, such that $a<c<b$ where
$$f(c)=g(c)$$ as long as $f$ and $g$ are continuous. This follows from applying the intermediate value theorem to $h(x)=f(x)-g(x)$.