Find the minimum value of the function
$$
f(x)=x^{2}-2 x-14 \sqrt{x-1}+x \sqrt{x^{2}-4 x-28 \sqrt{x-1}+61}
$$
Let $y=2 \sqrt{x-1}$, then $y^{2}=4(x-1)$. Thus,
\[
\begin{aligned}
f(x) & =\frac{1}{2}\left(2 x^{2}-4 x-14 y+2 x \sqrt{x^{2}-4 x-14 y+61}\right) \\
& =\frac{1}{2}\left(x^{2}+2 x \sqrt{x^{2}-4 x-14 y+61}+\left(x^{2}-4 x-14 y+61\right)-61\right) \\
& =\frac{1}{2}\left(\left(x+\sqrt{x^{2}-4 x-14 y+61}\right)^{2}-61\right) \\
& =\frac{1}{2}\left(\left(x+\sqrt{x^{2}-4 x-4(x-1)+y^{2}-14 y+61}\right)^{2}-61\right) \\
& =\frac{1}{2}\left(\left(x+\sqrt{\left(x^{2}-8 x+16\right)+\left(y^{2}-14 y+49\right)}\right)^{2}-61\right) \\
& =\frac{1}{2}\left(\left(x+\sqrt{(x-4)^{2}+(y-7)^{2}}\right)^{2}-61\right) .
\end{aligned}
\]

The expression $x+\sqrt{(x-4)^{2}+(y-7)^{2}}$ geometrically represents the sum of the distance from a point on the upper branch of the parabola $y^{2}=4(x-1)$ to the $y$-axis and the distance to the point $A(4,7)$. The vertex of the parabola is $(1,0)$, the focus is $F(2,0)$, and the directrix is the $y$-axis. As shown in Figure 6, draw $P D \perp y$-axis, with the foot of the perpendicular being $D$, then $x+\sqrt{(x-4)^{2}+(y-7)^{2}}=|P D|+|P A|$.

And
\[
|P D|+|P A|=|P F|+|P A|=|A F|=\sqrt{(4-2)^{2}+7^{2}}=\sqrt{53} .
\]

Solve the system of equations
\[
\left\{\begin{array}{l}
y=\frac{7}{2}(x-2), \\
y^{2}=4(x-1),
\end{array}\right.
\]

Taking the positive root, we get $x=\frac{106+4 \sqrt{53}}{49}$. That is, when $x=\frac{106+4 \sqrt{53}}{49}$, $f(x)$ attains its minimum value of -4.