Consider the family of equations $y=x^n$ with the integer parameter n.

Question

I graphed this family of equations with range $0<y<1$ and domain $0<x<1$. Each of the curves has has a point for which its derivative is one; for perspective, I also graphed the line with a slope of one which is tangent to these points. How would one go about finding the equation for the curve that connects every such point? (Namely, all the points for which the derivative is 1)

I could always have wolfram|alpha fit a polynomial equation by feeding it the points, but I would like to know how one would one calculate the equation, if possible.

Thanks in advance.

Answer 1

The slope of $x^n$ is, at any point, equal to $nx^{n-1}$ (this is true even for non-integer $n$). So if the slope is equal to $1$, then (for $n\neq 1$) we have $$ nx^{n-1}=1\\ x=\sqrt[n-1]{\frac1n} $$ As for the $y$-coordinate of this point, we just insert this $x$-value into the original expression for the curve: $$ y=\sqrt[n-1]{\frac1n}^n $$ So now we have a parametric expression for the curve going through all your tangent points (again, this is not valid for $n=1$) $$ \cases{x(n)=\sqrt[n-1]{\frac1n}\\ y(n)=\sqrt[n-1]{\frac1n}^n} $$ In principle, you could probably solve the first equation for $n$, and insert that into the second equation to get something of the form $y=f(x)$, but I don't know how possible it is in practice.

Answer 2

Starting from Arthur's answer, you could make things explicit since $$x=\sqrt[n-1]{\frac1n}\implies n=\frac{W(x \log (x))}{\log (x)}$$ where appears Lambert function. This makes $$\Large{y=x^{\frac{W(x \log (x))}{\log (x)}}}$$