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: 23

Answer

$${\text{no minimum; maximum value }}f\left( 1 \right) = 1$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 4{x^3} - 3{x^4};\,\,\left( { - \infty , + \infty } \right) \cr & {\text{The function is a polynomial }} \cr & {\text{Since }}f(x){\text{ has even }}\,{\text{degree and the leading coefficient is negative}} \cr & f(x) \to - \infty {\text{ as }}x \to \pm \infty .{\text{ Thus}}{\text{, there is an absolute maximum but}} \cr & {\text{not absolute minimum}}{\text{. From }}\,{\text{theorem}}\,{\text{ 3}}{\text{.4}}{\text{.3}}{\text{, }}\,{\text{the absolute maximum}} \cr & {\text{must occur at a critical point }}f. \cr & {\text{Since }}f{\text{ is differentiable everywhere}}{\text{, we can find all critical points }} \cr & {\text{by solving the equation }}f'(x) = 0.{\text{ This equation}}\,{\text{is}} \cr & f'\left( x \right) = 12{x^2} - 12{x^3} \cr & f'\left( x \right) = 0 \cr & 12{x^2} - 12{x^3} = 0 \cr & 12{x^2}\left( {1 - x} \right) = 0 \cr & x = 0{\text{ and }}x = 1 \cr & {\text{Evaluating }}f{\text{ at the critical points}} \cr & f\left( 0 \right) = 4{\left( 0 \right)^3} - 3{\left( 0 \right)^4} \cr & f\left( 0 \right) = 0 \cr & f\left( 1 \right) = 4{\left( 1 \right)^3} - 3{\left( 1 \right)^4} \cr & f\left( 1 \right) = 1 \cr & \cr & {\text{Therefore}}{\text{, }}f{\text{ has a maximum value }}f\left( 1 \right) = 1 \cr} $$
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