Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 37

Answer

The maximum is $f(0)=f(3)=0$ and the minimum is $f(1) =-1$

Work Step by Step

Given $$y=x-\frac{4 x}{x+1}, \quad[0,3]$$ Since \begin{align*} \frac{dy}{dx}&=1-\frac{4}{\left(x+1\right)^2}\\ &=\frac{x^2+2x-3}{\left(x+1\right)^2} \end{align*} Find the critical points \begin{align*} f'(x)&=0\\ \frac{x^2+2x-3}{\left(x+1\right)^2}&=0\\ x^2+2x-3&=0\\ (x-1)(x+3)&=0 \end{align*} Then $x=1,\ \ x= -3$. Because $-3\notin [0,3]$, then $f(x)$ has a critical point at $x=1 $, since \begin{align*} f(0)&= 0\\ f(3)&=0 \\ f(1)&= -1 \end{align*} Then the maximum is $f(0)=f(3)=0$ and the minimum is $f(1) =-1$
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