G2.3 Let $f(x)=\frac{x+a}{x^{2}+\frac{1}{2}}$, where $x$ is a real number and the maximum value of $f(x)$ is $\frac{1}{2}$ and the minimum value of $f(x)$ is -1 . If $t=f(0)$, find the value of $t$.
Let $y=\frac{x+a}{x^{2}+\frac{1}{2}}=\frac{2 x+2 a}{2 x^{2}+1} \Rightarrow 2 y x^{2}+y=2 x+2 a \Rightarrow(2 y) x^{2}-2 x+(y-2 a)=0$
For real values of $x, \Delta=(-2)^{2}-4(2 y)(y-2 a) \geq 0$
$$
1-\left(2 y^{2}-4 a y\right) \geq 0 \Rightarrow 2 y^{2}-4 a y-1 \leq 0
$$
$\qquad$
Given that $-1 \leq y \leq \frac{1}{2} \Rightarrow(y+1)(2 y-1) \leq 0 \Rightarrow 2 y^{2}+y-1 \leq 0 \ldots \ldots(* *)$
(*) is equivalent to (**) $\therefore a=-\frac{1}{4}$
$$
\begin{array}{l}
f(x)=\frac{x-\frac{1}{4}}{x^{2}+\frac{1}{2}} \\
t=f(0)=\frac{-\frac{1}{4}}{\frac{1}{2}} \\
=-\frac{1}{2}
\end{array}
$$