Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 122: 48

Answer

$1$

Work Step by Step

f(x) = $x^{2}e^{-x}$ f'(x) = $-x^{2}e^{-x}+2xe^{-x}$ f(a) = $a^{2}e^{-a}$ f'(a) = $-a^{2}e^{-a}+2ae^{-a}$ y = f'(a)(x-a)+f(a) y = $(-a^{2}e^{-a}+2ae^{-a})(x-a)+a^{2}e^{-a}$ origin (0, 0) replace x = 0 and y =0 0 = $(-a^{2}e^{-a}+2ae^{-a})(0-a)+a^{2}e^{-a}$ 0 = $(a^{3}-2a^{2}+a^{2})e^{-a}$ 0 = $(a^{3}-a^{2})e^{-a}$ 0 = $(a-1)a^{2}e^{-a}$ a = 0, 1 a > 0 so a = 1
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