Let f(x) =

Given

f (x)  = x2 and g (x) = 2x + 1

Clearly, D (f) =  R and D (g) = R

$\therefore D(f \pm g)=D(f) \cap D(g)=R \cap R=R$

$D(f g)=D(f) \cap D(g)=R \cap R=R$

$D\left(\frac{f}{g}\right)=D(f) \cap D(g)-\{x: g(x)=0\}=R \cap R-\left\{-\frac{1}{2}\right\}=R-\left\{-\frac{1}{2}\right\}$

Thus,

$(f+g)(x): R \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=x^{2}+2 x+1=(x+1)^{2} .$

$(f-g)(x): R \rightarrow R$ is given by $(f-g)(x)=f(x)-g(x)=x^{2}-2 x-1$

$(f g)(x): R \rightarrow R$ is given by $(f g)(x)=f(x) \cdot g(x)=x^{2}(2 x+1)=2 x^{3}+x^{2} .$

$\left(\frac{f}{g}\right): R-\left\{-\frac{1}{2}\right\} \rightarrow R$ is given by $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^{2}}{2 x+1}$