Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 223: 27

Answer

$$\eqalign{ & {\text{max}} = - 2 - 2\sqrt 2 {\text{ at }}x = - 1 - \sqrt 2 . \cr & {\text{There is no minimum value}}{\text{.}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{{x^2} + 1}}{{x + 1}};{\text{ }}\left( { - 5,1} \right) \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{{\left( {x + 1} \right)\left( {2x} \right) - \left( {{x^2} + 1} \right)\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{2{x^2} + 2x - {x^2} - 1}}{{{{\left( {x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{{{x^2} + 2x - 1}}{{{{\left( {x + 1} \right)}^2}}} \cr & f'\left( x \right) = 0 \cr & \frac{{{x^2} + 2x - 1}}{{{{\left( {x + 1} \right)}^2}}} = 0 \cr & {x^2} + 2x - 1 = 0 \cr & {\text{By the quadratic formula}} \cr & x = \frac{{ - 2 \pm \sqrt {{{\left( 2 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{2} \cr & x = \frac{{ - 2 \pm \sqrt 8 }}{2} = \frac{{ - 2 \pm 2\sqrt 2 }}{2} \cr & x = - 1 \pm \sqrt 2 \cr & {\text{Both values are on the interval }}\left( { - 5,1} \right) \cr & {\text{Evaluate }}f{\text{ at all the critical points and at the end points }}a{\text{ and }}b \cr & f\left( { - 5} \right) = - \frac{{13}}{2} = - 6.5 \cr & f\left( { - 1 - \sqrt 2 } \right) = - 2 - 2\sqrt 2 \approx - 4.828 \cr & f\left( { - 1 + \sqrt 2 } \right) = - 2 + 2\sqrt 2 \approx 0.82842 \cr & f\left( 1 \right) = 1 \cr & {\text{The largest value of }}f{\text{ in the interval }}\left( { - 5,1} \right){\text{ is }} - 2 + 2\sqrt 2 ,{\text{ so}} \cr & {\text{there is a max}} = - 2 - 2\sqrt 2 {\text{ at }}x = - 1 - \sqrt 2 . \cr & {\text{There is no minimum value}}{\text{.}} \cr} $$
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