Chapter 10 - Exponents and Radicals - 10.1 Radical Expressions and Functions - 10.1 Exercise Set - Page 635: 110

$(f+g)(x)=\dfrac{3x^2-x+1}{x}$

Since $(f+g)(x)=f(x)+g(x),$ with $f(x)=3x-1$ and $g(x)=\dfrac{1}{x},$ then \begin{array}{l}\require{cancel} (f+g)(x)=f(x)+g(x) \\\\ (f+g)(x)=3x-1+\dfrac{1}{x} \\\\ (f+g)(x)=(3x-1)\cdot\dfrac{x}{x}+\dfrac{1}{x} \\\\ (f+g)(x)=\dfrac{3x^2-x}{x}+\dfrac{1}{x} \\\\ (f+g)(x)=\dfrac{3x^2-x+1}{x} .\end{array}