Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 302: 60

Answer

(a) Using the graph, we can make the following estimates: There is no absolute maximum value, but the local maximum is approximately $f(2) \approx 0.5$ The minimum value appears to be $f(0) = 0$ The exact values are as follows: The local maximum value is $f(2) = \frac{4}{e^2} \approx 0.54$ The minimum value is $f(0) = 0$ (b) The graph appears to increase most rapidly, that is it has the highest positive slope, approximately at the point $x = 0.5$ The exact value is as follows: The slope of the graph is a maximum when $x = 2 - \sqrt{2} \approx 0.586$

Work Step by Step

(a) $f(x) = x^2~e^{-x}$ Using the graph, we can make the following estimates: There is no absolute maximum value, but the local maximum is approximately $f(2) \approx 0.5$ The minimum value appears to be $f(0) = 0$ We can find the exact values: $f'(x) = 2x~e^{-x}-x^2~e^{-x}$ $f'(x) = e^{-x}(2x-x^2) = 0$ $2x-x^2 = 0$ $x(x-2) = 0$ $x = 0, 2$ $f(0) = 0^2~e^{-0} = 0$ $f(2) = 2^2~e^{-2} = \frac{4}{e^2} \approx 0.54$ The local maximum value is $f(2) = \frac{4}{e^2} \approx 0.54$ The minimum value is $f(0) = 0$ (b) The graph appears to increase most rapidly, that is it has the highest positive slope, approximately at the point $x = 0.5$ We can find $x$ when $f''(x) = 0$: $f''(x) = 2~e^{-x}-2x~e^{-x}-2x~e^{-x}+x^2~e^{-x}$ $f''(x) = x^2~e^{-x}-4x~e^{-x}+2~e^{-x}$ $f''(x) = (x^2-4x+2)~e^{-x} = 0$ $x^2-4x+2 = 0$ We can use the quadratic formula: $x = \frac{4\pm \sqrt{(-4)^2-(4)(1)(2)}}{2(1)}$ $x = \frac{4\pm \sqrt{16-8}}{2}$ $x = \frac{4\pm 2\sqrt{2}}{2}$ $x = 2 \pm \sqrt{2}$ Using the graph, we can clearly see that the negative branch is the correct solution. The slope of the graph is a maximum when $x = 2 - \sqrt{2} \approx 0.586$
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