IS. 3 Given that $c$ is the minimum value of $f(x)=\frac{x^{2}-2 x+\frac{1}{b}}{2 x^{2}+2 x+1}$. Find the value of $c$.
Let $y=f(x)=\frac{x^{2}-2 x-3}{2 x^{2}+2 x+1}$
$2 y x^{2}+2 y x+y=x^{2}-2 x-3$
$$
(2 y-1) x^{2}+2(y+1) x+(y+3)=0
$$

For any values of $x$, the above quadratic equation has real solution.
$$
\begin{array}{l}
\therefore \Delta \geq 0 \\
(y+1)^{2}-(2 y-1)(y+3) \geq 0 \\
y^{2}+2 y+1-\left(2 y^{2}+5 y-3\right) \geq 0 \\
-y^{2}-3 y+4 \geq 0 \\
y^{2}+3 y-4 \leq 0 \\
(y+4)(y-1) \leq 0 \\
-4 \leq y \leq 1 \\
c=\text { the minimum of } y=-4
\end{array}
$$