I3.4 If $x$ is a real number and $d$ is the maximum value of the function $y=\frac{3 x^{2}+3 x+c}{x^{2}+x+1}$, find the value of $d$.
$$
\left(x^{2}+x+1\right) y=3 x^{2}+3 x+4
$$
$(3-y) x^{2}+(3-y)+(4-y)=0 \ldots \ldots(*)$, this is a quadratic equation in $x$.
For any real value of $x, y=\frac{3 x^{2}+3 x+c}{x^{2}+x+1}$ is a well-defined function
$\therefore(*)$ must have real roots in $x$.
$$
\begin{array}{l}
\Delta=(3-y)^{2}-4(3-y)(4-y) \geq 0 \\
(3-y)(3-y-16+4 y) \geq 0 \\
(y-3)(3 y-13) \leq 0 \\
3 \leq y \leq \frac{13}{3}
\end{array}
$$
$d=$ the maximum value of $y=\frac{13}{3}$
Method 2
$$
\begin{array}{l}
y=\frac{3 x^{2}+3 x+4}{x^{2}+x+1}=3+\frac{1}{x^{2}+x+1}=3+\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}} \\
\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4} \\
y \leq 3+\frac{4}{3}=\frac{13}{3}
\end{array}
$$
$d=$ the maximum value of $y=\frac{13}{3}$