Answer
$(f+g)(x)=3x^2+4x+1\space\space\{x\in \Re \}$
$(f-g)(x)=3x^2-2x+1\space\space\{x\in \Re \}$
$(f\cdot g)(x)=9x^3+3x^2+3x\space\space\{x\in \Re \}$
$(\dfrac{f}{g})(x)=\dfrac{3x^2+x+1}{3x}\space\space\{x|x\ne 0\}$
Work Step by Step
$(f+g)(x)=3x^2+x+1+3x=3x^2+4x+1\space\space\{x\in \Re \}$
$(f-g)(x)=3x^2+x+1-3x=3x^2-2x+1\space\space\{x\in \Re \}$
$(f\cdot g)(x)=(3x^2+x+1)(3x)=9x^3+3x^2+3x\space\space\{x\in \Re \}$
$(\dfrac{f}{g})(x)=\dfrac{3x^2+x+1}{3x}\space\space\{x|x\ne 0\}$
