Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Review - Exercises - Page 359: 3

Answer

The absolute maximum is $f(2) = \frac{2}{5}$ The absolute minimum is $f(-\frac{1}{3}) = -\frac{9}{2}$ There is no local maximum. $f(-\frac{1}{3}) = -\frac{9}{2}$ is a local minimum.

Work Step by Step

$f(x) = \frac{3x-4}{x^2+1}$ Note that this function is continuous on the interval $[-2,2]$ and differentiable on the interval $(-2,2)$. We can find the points where $f'(x) = 0$: $f'(x) = \frac{3(x^2+1)-2x(3x-4)}{(x^2+1)^2}=0$ $3x^2+3-6x^2+8x=0$ $-3x^2+8x+3=0$ $3x^2-8x-3=0$ $(3x+1)(x-3)=0$ $x = -\frac{1}{3}, 3$ We can verify the value of the function when $f'(x) = 0$ and at the endpoints of the interval $[-2,2]$: $f(-2) = \frac{3(-2)-4}{(-2)^2+1} = -2$ $f(-\frac{1}{3}) = \frac{3(-\frac{1}{3})-4}{(-\frac{1}{3})^2+1} = -\frac{9}{2}$ $f(2) = \frac{3(2)-4}{(2)^2+1} = \frac{2}{5}$ The absolute maximum is $f(2) = \frac{2}{5}$ The absolute minimum is $f(-\frac{1}{3}) = -\frac{9}{2}$ There is no local maximum. $f(-\frac{1}{3}) = -\frac{9}{2}$ is a local minimum.
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