Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.1 Derivative of f(x)=bx and the Number e - Exercises - Page 327: 59



Work Step by Step

Given $$f(x)=x^{2} e^{-x} $$ Since $f(a) = a^2e^{-a} $, then \begin{align*} f^{\prime}(x)&=-x^{2} e^{-x}+2 x e^{-x}\\ &=e^{-x}\left(2 x-x^{2}\right)\\ f'(a) &=(2a - a^2)e^{-a} \end{align*} Then the tangent line to $f$ at $x = a$ is given by \begin{align*} \frac{y-y_1}{x-x_1}&= f'(a) \\ \frac{y- a^2e^{-a} }{x-a}&= (2a - a^2)e^{-a}\\ y&=\left(2 a-a^{2}\right) e^{-a}(x-a)+a^{2} e^{-a} \end{align*} The line passes through $(0,0) $ when \begin{align*} \left(2 a-a^{2}\right) e^{-a}(-a)+a^{2} e^{-a}&=0\\ e^{-a}\left(a^{2}-2 a^{2}+a^{3}\right)&=0\\ a^{2} e^{-a}(a-1)&=0 \end{align*} Then $a=0$ or $ a=1 $ and since $a>0 $, then $a=1$
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