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: 49


$ x=1$, local minima at $ x=1$

Work Step by Step

To find the critical point, we put $ f'(x)=0$, so we have $$ f(x)=\frac{e^{x}}{x}\Longrightarrow f'(x)=\frac{xe^x-e^x}{x^2}=0,$$ then $ xe^{x}=e^x $ and hence $ x=1$. So the critical point is $ x=1$. Moreover, we have $$ f''(x)=\frac{x^2(xe^x)-2x(xe^x-e^x)}{x^4} =\frac{x^3-2x(x-1)}{x^4}e^x $$ then $$ f''(1)=e>0 $$ then $ f(x) $ has a local minima at $ x=1$.
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