Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises: 54

Answer

$f(1)=1$ is the absolute maximum value. $f(0)=0$ is the absolute minimum value.

Work Step by Step

The Closed Interval Method 1) Function $f$ must be continuous on the closed interval $[a, b]$ 2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints. 3) Compare the values. - The largest is the absolute maximum value. - The smallest is the absolute minimum value. $$f(x)=\frac{x}{x^2-x+1},\hspace{1.5cm}[0, 3]$$ First, $f(x)$ is continuous on the interval $[0,3]$, so the Closed Interval Method can be applied here. 1) Find critical points of $f$ $$f'(x)=\frac{1(x^2-x+1)-x(2x-1)}{(x^2-x+1)^2}$$ $$f'(x)=\frac{x^2-x+1-2x^2+x}{(x^2-x+1)^2}$$ $$f'(x)=\frac{-x^2+1}{(x^2-x+1)^2}$$ $f'(x)$ does not exist when $$x^2-x+1=0.$$ However, as $x^2-x+1\ne0$ for all real x, there are no values of $x$ so that $f'(x)$ does not exist. Also, $$f'(x)=0$$ $$-x^2+1=0$$ $$x^2=1$$ $$x=\pm1$$ $x=\pm1$ all belong to the domain of $f$, so they are all critical points. However, only $x=1$ lies in interval $(0, 3)$ given, so only $x=1$ will move to the next sections. 2) Calculate the values of $f$ at critical points and endpoints. - Critical points: $x=1$ $$f(1)=\frac{1}{1^2-1+1}=\frac{1}{1}=1$$ - End points: $x=0$ and $x=3$ $$f(0)=\frac{0}{0^2-0+1}=0$$ $$f(3)=\frac{3}{3^2-3+1}=\frac{3}{7}$$ 3) Comparison Among the results: - $f(1)=1$ is the largest, so $f(1)=1$ is the absolute maximum value. - $f(0)=0$ is the smallest, so $f(0)=0$ is the absolute minimum value.
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