## Calculus (3rd Edition)

The maximum is $R(1)= 1/3$; the minimum is $R(0)=0$
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 $R(0)=0$