Given a map $\alpha: k[x] \to k[x]$ where $k$ is a field of zero characteristic defined as $\alpha(x) = x+1$. Is $\alpha$ an inner automorphism?
Since we are in a commutative ring I believe "inner" in this case means
$$\exists n\in \mathbb{N} \ \ \text{ s.t } \ \ \alpha^n(r) = r \ \forall r \in \mathbb R .$$ Ok, so given this question I think $\alpha$ is not inner. My reasoning for this is that $\alpha$ is not inner on powers of $x$ which form a basis for $k[x]$. Is this enough of an argument?
Generally, an automorphism of an algebraic structure is inner if both it and its inverse are representable by unary algebra polynomials. Here a unary polynomial of an algebra $A$ is a function $\alpha(x) = u(x,\bar{a})$ where $u(x,\bar{y})$ is a word in the language and $\bar{a}$ is a tuple of elements of $A$. Moreover, if $\alpha(x)$ is representable as $u(x,\bar{a})$ and $\alpha^{-1}(x) = v(x,\bar{b})$, then for $\alpha$ to be inner we also require that $u(v(x,\bar{b}),\bar{a}) = x = v(u(x,\bar{a}),\bar{b})$ be provable from the laws of the algebras in question. [More precisely, in the coproduct $A\sqcup \langle x\rangle$, the endomorphism that fixes $A$ and maps $x$ to $u$ is the inverse of the endomorphism that fixes $A$ and maps $x$ to $v$.]
What I said the previous paragraph is a theorem and not a definition. The definition is that an automorphism of an object $A$ of some category $\mathcal C$ is inner if it can be extended in a functorial manner to all objects of $\mathcal C$ lying under $A$. If $\mathcal C$ is the category of models of an algebraic theory, the definition can be shown to agree with what I said in the first paragraph.
In your question, $\alpha(x) = x+1$ looks like a ring polynomial whose inverse $\alpha^{-1}(x) = x-1$ is also a polynomial. But if you were truly thinking of $\alpha$ as a ring polynomial, then you would evaluate $\alpha(f(x))$ to be $f(x)+1$. I suspect you want $\alpha(f(x))$ to be $f(x+1)$. Whichever was intended, the map $f(x)\mapsto f(x)+1$ is not an endomorphism, while the map $f(x)\mapsto f(x+1)$ is not a polynomial.
It is not hard to show that the only inner automorphism of $k[t]$ is the identity function. For, if
$$\alpha(x) = a_n(t)x^n + \cdots + a_1(t)x+a_0(t) \in k[t][x]$$
is an additive automorphism of $k[t]$ and $\alpha(1)=1$, then we get that $\alpha(m) = m$ for $m = 0, 1, 2, \ldots$. Therefore the polynomial $\alpha(x)-x$ has infinitely many zeros in the integral domain $k[t]$. This implies $\alpha(x)=x$.
Of course, the map $\alpha\colon f(t)\mapsto f(t+1)$ $\underline{\textrm{is}}$ an automorphism of $k[t]$. If you are using a different definition of "inner", then this automorphism might satisfy your definition.