How does the real root of a higher order equations change with the order? We know $a^{n+1} - (1+d) a + d = 0$ has a root $a^*$ in the interval $(0, 1)$, where $n = 1,2, \cdots $ and $n > d > 0$, then how does the root $a^*$ change if we increase $n$?

Let $f(x) = x^{n+1}-(1+d)x + d, x \in (0, 1)$, $f^{\prime}(x)=(n+1)x^n -(1+d)$ and $f^{\prime \prime}(x) = (n+1)n x^{n-1} > 0$. Since $f^{\prime}(0) < 0$ while $f^{\prime}(1) > 0$, we know $f(x)$ decreases and then increases. $f(0) = d > 0$ and $f(1) = 0$. So we have a real root $a^*(n)$ which depends on $n$ in the interval $(0, 1)$. But I have no idea how to study its changes when we change $n$.
 A: $a^{n+1} - (1+d) a + d = 0 \implies n = \dfrac{\ln\left((1+d)a - d\right)}{\ln(a)} - 1=g(a)\,$, which gives $\,n\,$ as a function $\,g\,$ of $\,a \in \left(d/(d+1),1\right)\,$. The inverse of this function $\,a=g^{-1}(n)\,$ gives the root $\,a\,$ as a function of $\,n\,$. Unsurprisingly, however, the inverse $\,g^{-1}\,$ does not have a closed form in the general case.
The qualitative behavior of the root $\,a\,$ in terms of $\,n\,$ becomes more apparent if writing the equation as $\,a^{n+1}=(1+d)a - d\,$. The LHS is a power function $\,y=x^{n+1}\,$ which becomes "flatter" in $\,[0,1]\,$ as $\,n\,$ increases, while the RHS is a linear function $\,y=(1+d)x-d\,$ independent of $\,n\,$. Moreover, in the limit case $\,n=d\,$ the two curves are tangent at point $\,(1,1)\,$. For $\,n \gt d\,$ the tangency point splits into two simple intersections $\;-\;$ one of them always at $\,(1,1)\,$ and the other one moving down the line $\,y=(1+d)x-d\,$ towards the limit point $\,\left(\dfrac{d}{d+1},0\right)\,$ when $\, n \to \infty\,$. Below is a graph showing the behavior for $\,d=1\,$ and $\,n=1,2,3,4\,$.

