Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 82

Answer

$x= 1$ is a local minimum

Work Step by Step

Given $$ g(x)=x-\ln x$$ Since \begin{align*} g^{\prime}(x)&= 1-\frac{1}{x} \end{align*} Then $g(x)$ has critical points when \begin{align*} g'(x)&=0\\ \frac{x-1}{x}&=0 \end{align*} Then $x=1$ is a critical point. Now, we use the second derivative to check $x=1 $ \begin{align*} g''(x) &=\frac{1}{x^2} \end{align*} Hence $$g''(1)=1>0 $$ Since the second derivative at the critical point is positive, we have a minimum. Then $x= 1$ is a local minimum.
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