Answer
The maximum is $R(1)= 1/3$; the minimum is $R(0)=0 $
Work Step by Step
Given $$R(t)=\frac{t}{t^{2}+t+1}, \quad[0,3]$$
Since
$$ R'(t) =\frac{\frac{d}{dt}\left(t\right)\left(t^2+t+1\right)-\frac{d}{dt}\left(t^2+t+1\right)t}{\left(t^2+t+1\right)^2}= \frac{-t^2+1}{\left(t^2+t+1\right)^2} $$
We find the critical points
\begin{align*}
g'(\theta)&=0\\
\frac{-t^2+1}{\left(t^2+t+1\right)^2}&=0\\
1-t^2&=0
\end{align*}
Then $R(t) $ has critical points at $t=-1, \ t=1$, but we eliminate $-1$ because $-1\notin [0,3]$.
We check the values at the critical points and end points:
\begin{align*}
R(0)&= 0 \\
R(1)&= 1/3 \\
R(3) &= 3/13
\end{align*}
Then the maximum is $R(1)= 1/3$, and the minimum is $R(0)=0 $
