Chapter 11 - Section 11.1 - Derivatives of Powers, Sums, and Constant Multiples - Exercises - Page 795: 69

$x=1$ or $x=-1$

$y=f(x)=x+\displaystyle \frac{1}{x}=x+x^{-1} \\$ $f^{\prime}(x)=1-1\displaystyle \cdot x^{-2}=1-\frac{1}{x^{2}} \\$ $f^{\prime}(x)=0$ for ... (solve for x)$\\$ $1-\displaystyle \frac{1}{x^{2}}=0 \\\\$ $\displaystyle \frac{x^{2}-1}{x^{2}}=0\\\\$ $x^{2}-1=0\ \ \ \ $...recognize a differrence of squares$\\\\$ $(x+1)(x-1)=0\\\\$ $x=1$ or $x=-1$