1. Given the quadratic function
$$
f(x)=x^{2}+b x+8(b \neq 0)
$$

has two distinct real roots $x_{1}, x_{2}$, and $x_{1}+\frac{1}{x_{2}}, x_{2}+\frac{1}{x_{1}}$ are the two zeros of a quadratic function $g(x)$ with a leading coefficient of 1. If $g(1)=f(1)$, find all possible values of $g(1)$.
1. By Vieta's formulas, we have
$$
x_{1}+x_{2}=-b, x_{1} x_{2}=8 \text {. }
$$

Let $g(x)=x^{2}+c x+d$.
Then for $g(x)$, we have
$$
\begin{array}{l}
-c=x_{1}+\frac{1}{x_{2}}+x_{2}+\frac{1}{x_{1}} \\
=-b+\frac{x_{1}+x_{2}}{x_{1} x_{2}}=-\frac{9 b}{8}, \\
d=\left(x_{1}+\frac{1}{x_{2}}\right)\left(x_{2}+\frac{1}{x_{1}}\right) \\
=x_{1} x_{2}+\frac{1}{x_{1} x_{2}}+1+1=\frac{81}{8} .
\end{array}
$$

By $g(1)=f(1)$
$$
\begin{array}{l}
\Rightarrow 1+\frac{9 b}{8}+\frac{81}{8}=1+b+8 \\
\Rightarrow b=-17 \\
\Rightarrow g(1)=f(1)=1+b+8=-8 .
\end{array}
$$