Answer
$(f+g)(x)=2x+3\space\space\{x\in \Re \}$
$(f-g)(x)=-4x+1\space\space\{x\in \Re \}$
$(f\cdot g)(x)=-3x^2+5x+2\space\space\{x\in \Re \}$
$(\dfrac{f}{g})(x)=\dfrac{2-x}{3x+1}\space\space\{x|x\ne-\frac{1}{3}\}$
Work Step by Step
$(f+g)(x)=2-x+3x+1=2x+3\space\space\{x\in \Re \}$
$(f-g)(x)=2-x-(3x+1)=-4x+1\space\space\{x\in \Re \}$
$(f\cdot g)(x)=(2-x)(3x+1)=6x+2-3x^2-x=-3x^2+5x+2\space\space\{x\in \Re \}$
$(\dfrac{f}{g})(x)=\dfrac{2-x}{3x+1}\space\space\{x|x\ne-\frac{1}{3}\}$
