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: 21

Answer

$${\text{no maximum; min}} = - 9/4{\text{ at }}x = 1/2$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^2} - x - 2;\,\,\left( { - \infty , + \infty } \right) \cr & {\text{The function is a polynomial }} \cr & {\text{Since }}f(x){\text{ has even degree and the leading coefficient is positive }} \cr & f(x) \to + \infty {\text{ as }}x \to \pm \infty .{\text{ Thus}}{\text{, there is an absolute minimum but}} \cr & {\text{not absolute maximum}}{\text{. From theorem 3}}{\text{.4}}{\text{.3}}{\text{, the absolute minimum}} \cr & {\text{must occur at a critical point }}f. \cr & {\text{Since }}f{\text{ is differentiable everywhere}}{\text{, we can find all critical points }} \cr & {\text{by solving the equation }}f'(x) = 0.{\text{ This equation}}\,{\text{is}} \cr & f'\left( x \right) = 2x - 1 \cr & {\text{Let }}f'\left( x \right) = 0 \cr & 2x - 1 = 0 \cr & x = \frac{1}{2} \cr & {\text{Evaluating }}f{\text{ at the critical point}} \cr & f\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^2} - \left( {\frac{1}{2}} \right) - 2 \cr & f\left( {\frac{1}{2}} \right) = - \frac{9}{4} \cr & {\text{Therefore}}{\text{, }}f{\text{ has an absolute minimum of}} - \frac{9}{4}{\text{ at }}x = \frac{1}{2} \cr} $$
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