Chapter 3: Derivatives - Section 3.9 - Linearization and Differentials - Exercises 3.9 - Page 174: 3

$ L(x)=2$

The linearization of $ f(x)$ at $ x=a $ is given as: $ L(x)=f(a)+f'(a)(x-a)$ Knowing that $ f(x)=x+\frac{1}{x}$, $ f'(x)=1-\frac{1}{x^{2}}$ and $ a=1$, we have $ L(x)=(1+\frac{1}{1})+(1-\frac{1}{1^{2}})(x-1)=2$