Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 174: 46

$\frac{4-x}{4}$

Linearization of $ f(x)$ at $ x=a $ is $ L(x)=f(a)+f'(a)(x-a)$ Knowing that $ f(x)=\frac{1}{x}$, $ f'(x)=-\frac{1}{x^{2}}$ and $ a=2$, we have $ L(x)=\frac{1}{2}-\frac{1}{2^{2}}(x-2)=\frac{1}{2}-\frac{x}{4}+\frac{1}{2}=1-\frac{x}{4}=\frac{4-x}{4}$