Finding Zeros of a Specific Function

Question

Suppose that $a,b\in\langle0,1\rangle$ are two constants. Considering the following function, I want to show how many roots $f(\varepsilon)$ has in $\varepsilon$ (optimally it is only one):

$f(\varepsilon)=a^{{\varepsilon}} - \frac{1}{1+\frac{1}{\varepsilon}}a^{1+{\varepsilon}} - \frac{1}{1+{\varepsilon}}b^{1+{\varepsilon}}$

I do not know how to approach this problem (I didn't find explicit solutions for the derivatives). I would highly appreciate your help! Thank you!

Answer 1

It is equivalent to $\frac{\varepsilon}{1+\varepsilon}+\frac{1}{1+\varepsilon}(\frac{b}{a})^{1+\varepsilon}=\frac1a$.

Or

$\frac{1}{1+\varepsilon}((\frac{b}{a})^{1+\varepsilon}-1)=\frac1a-1 \gt 0$,

or

$(\frac{b}{a})^{1+\varepsilon}-1=(\frac1a-1)(1+\varepsilon)$

so if $1+\varepsilon\gt 0$ , we must have $b \gt a$; and if $1+\varepsilon\lt 0$, we also have $b \gt a$.

So it $b\le a$, there're no zeros for the equation. ($\varepsilon=-1$ is invalid for original equation)

But for $b\gt a$, since $(\frac b a)^{1+\varepsilon} -1$ is convex function, and we have found a zero $\varepsilon=-1$, there must be another zero unless -1 is duplicated zeros. wether the other zero is larger than -1 is dependent on the sign of $\ln(\frac b a)-(\frac1a-1)$