Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - Chapter Review Exercises - Page 222: 34


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 $
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