Example 1 Let $f(x)=x^{n}, x \in D, n \in \mathbf{N}^{*}$, determine whether $f(x)=x^{n}$ is a solution to the functional inequality
$$
f(x)+f(1-x)>1
$$

If it is, find the domain $D$; if not, explain the reason.
If $f(x)$ is a solution to inequality (1), it is equivalent to $x^{n}+(1-x)^{n}>1$ having a solution.
When $n=1$, equation (2) becomes $1>1$, which is a contradiction.
When $n>1$, let $t=x-\frac{1}{2}$, then equation (2) is equivalent to $\left(t+\frac{1}{2}\right)^{n}+\left(\frac{1}{2}-t\right)^{n}>1$. Since $\left(\frac{1}{2}+t\right)^{n}+\left(\frac{1}{2}-t\right)^{n}=2\left(\frac{1}{2}\right)^{n}+C_{n}^{2}\left(\frac{1}{2}\right)^{n-2} t^{2}+C_{n}^{4}\left(\frac{1}{2}\right)^{n-4} t^{4}+\cdots$. Let $g(u)=2\left(\frac{1}{2}\right)^{n}+C_{n}^{2}\left(\frac{1}{2}\right)^{n-2} u+C_{n}^{4}\left(\frac{1}{2}\right)^{n-4} u^{2}+\cdots$.
Then $g(u)$ is a monotonically increasing function on $[0,+\infty)$.
Equation (3) $\Leftrightarrow g\left(t^{2}\right)>g\left(\frac{1}{4}\right) \Leftrightarrow t^{2}>\frac{1}{4} \Leftrightarrow\left(x-\frac{1}{2}\right)^{2}>\frac{1}{4} \Leftrightarrow x>1$ or $x<0$.
When $n>1$, $f(x)$ can be a solution to inequality (1), with its domain being $D=(-\infty, 0) \cup(1,+\infty)$, or a subset $D^{\prime}$ (satisfying $x \in D^{\prime}$, necessarily $(1-x) \in D^{\prime}$).