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: 24

Answer

$${\text{no maximum; min}} = - 3{\text{ at }}x = - 1$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^4} + 4x;\,\,\left( { - \infty , + \infty } \right) \cr & {\text{The function is a polynomial }} \cr & {\text{Since }}f(x){\text{ has even }}\,{\text{degree and the leading coefficient is positive}} \cr & f(x) \to + \infty {\text{ as }}x \to \pm \infty .{\text{ Thus}}{\text{, there is an absolute maximum but}} \cr & {\text{not absolute minimum}}{\text{. From }}\,{\text{theorem}}\,{\text{ 3}}{\text{.4}}{\text{.3}}{\text{, }}\,{\text{the absolute maximum}} \cr & {\text{must occur at a critical point }}f. \cr & {\text{Since }}f{\text{ is differentiable everywhere}}{\text{, we can find all critical points }} \cr & {\text{by solving the equation }}f'(x) = 0.{\text{ This equation}}\,{\text{is}} \cr & f'\left( x \right) = 4{x^3} + 4 \cr & f'\left( x \right) = 0 \cr & 4{x^3} + 4 = 0 \cr & 4\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) = 0 \cr & x = - 1 \cr & {\text{Evaluating }}f{\text{ at the critical points}} \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^4} + 4\left( { - 1} \right) \cr & f\left( { - 1} \right) = 1 - 4 \cr & f\left( { - 1} \right) = - 3 \cr & \cr & {\text{Therefore}}{\text{, }}f{\text{ has an absolute minimum of}} - 3{\text{ at }}x = - 1 \cr} $$
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