Let $0<\varepsilon<1$. Consider the functions $f(x)=x-x^{\varepsilon}$, $g(x)=x^2-x^{2\varepsilon}$, and $h(x)=x^3-x^{3\varepsilon}$ in $[0,1]$.
It can be shown that f(x)-g(x), and f(x)-h(x) have each one unique (real) zero in $(0,1)$ (by using Roll's theorem).
My question is how can I show that $f(x)-g(x)$, and $f(x)-h(x)$ dont share the same zero? Numerically, the zeros of $f(x)-g(x)$, and $f(x)-h(x)$ in $(0,1)$ seem to be different.
This is an example of a more generalized equation I encountered in my research, so I am not sure about the direction I should use. Even some related references are blessed.
thanks
If $g(w)-f(w)=0, 0<w<1, 0 < \epsilon <1$,dividing by $w-w^{\epsilon} \ne 0$ we get $1=w+w^{\epsilon}$
Similarly $h(y)-f(y)=0, 0<y<1, 0 < \epsilon <1$,dividing by $y-y^{\epsilon} \ne 0$ we get $1=y^2+yy^{\epsilon}+y^{2\epsilon}$
But now if $y=w$ squaring the first relation and then subtracting the second, one gets $ww^{\epsilon}=0$ which is not possible.
Note that we can actually say more, namely that $w<y$ since otherwise $1=(w+w^{\epsilon})^2 > y^2+yy^{\epsilon}+y^{2\epsilon}=1$