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Some Pictures of this kind of function

So I'm thinking about $f(x)= a^x -x^a, \text{s.t. } (x>0)$

  1. It's easy to see that $f(a)=0 $

  2. When $x>1$, there is always a point where $x^a$ suddenly gets bigger then $a^x$, and then there's a point when $a^x$ suddenly gets bigger than $x^a$

  3. Sometimes the point where $a^x$ gets bigger than $x^a$ is $(a,0)$. However, sometimes that's the point when $a^x$ is bigger than $x^a$ (such as when $a=2$). Then, I found out that when $a$ equals $e$ and $x>1$, only $f(e)$ equals $0$, and I think that's the point where $x=a$ changes from the smaller intersection into the bigger intersection of $f(x)$ and the $x$ axis.

So, to conclude:

  • When $0<a<1$, there is only one intersection between $f(x)$ and the $x$-axis, defined as $(p, 0)$. When $0<x<p$, $f(x)>0$ and when $x>p$, $f(x)<0$.

  • When $1<a<e$, there is two intersection between $f(x)$ and the $x$-axis. Let's make the bigger one $(p,0)$ and the smaller one is $(a,0)$. I want to know how to calculate the value of this $p$. I need some help thx.

  • When $a=e$, there is again only one intersection between $f(x)$ and the $x$ axis. And $f(x)$ is also always $ >= 0$

  • When $a>e$, there is two intersection between $f(x)$ and the $x$ axis. Let's make the smaller one $(p,0)$, and the bigger one is $(a,0)$. Again, I need help calculating the value of $p$. So can u help me calculate the value of $p$? Thanks a lot!

FishDrowned
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Orange_jacky Orange Orange_jac
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1 Answers1

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I think that you are discovering the beauty of Lambert function.

If $$f(x)=a^x-x^a$$ the solution (beside the trivial $x=a$) is given by $$x=-\frac{a}{\log (a)}\, W\left(-\frac{\log (a)}{a}\right)$$ which, in the real domain, has two branches, namely $W_0(.)$ and $W_{-1}(.)$.

You must read about it : it is very much used (type Lambert in the search bar of this site : there are $4415$ entries).

Claude Leibovici
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