Let $k = F_p$, and let $k(x)$ be the rational function field in one variable over $k$. Define $φ : k(x) \to k(x)$ by $φ(x) = x+1$. Show that $φ$ has finite order in $Gal(k(x)/k)$. Determine this order, find a $u$ so that $k(u)$ is the fixed field of $φ$, determine the minimal polynomial over $k(u)$ of $x$, and find all the roots of this minimal polynomial.
Now order of $\phi$ is $p$ but what about the rest? I can see that if I consider $u=x^p -x $ then $\phi (x^p -x)=(x+1)^p-(x+1)= x^p -x$ So, $\phi$ fixes this $k(u)$. Now, is this the fixed field? If it is then why? and what about the rest?
I know from experience that this can get incredibly confusing. You have correctly identified the fixed field of $\varphi:x\mapsto x+1$ as $k(x^p-x)$, and called your generating element $u$. If you determine that $\bigl[k(x)\colon k(u)\bigr]=p$, you will have identified the complete fixed field as $k(u)$ because the degrees are right.
So all you need to do is find the $k(u)$-polynomial $\text{Irr}\bigl(x,k(u)[X]\bigr)$: an irreducible polynomial in $X$ with coefficients in $k(u)$ having $x$ as a root. In fact, the polynomial I’m going to show you has coefficients in the UFD $k[u]$.
The polynomial is $X^p-X-u$. Notice that when you plug in $x$ for $X$, you get zero. You see at a glance what the other roots are, and it’s not hard to see why the polynomial is irreducible.