Answer
$\begin{array}{llllll}
& & domain & & & \\
a. & 2x^{2}+x-1, & x\in \mathbb{R} & \text{ }& e. & 20\\
b. & -2x^{2}+x-1, & x\in \mathbb{R} & & f. & -29\\
c. & 2x^{3}-2x^{2}, & x\in \mathbb{R} & & g. & 8\\
d. & \dfrac{x-1}{2x^{2}}, & x\neq 0 & & h. & 0
\end{array}$
Work Step by Step
$f(x)=x-1 g(x)=2x^{2}$
$\mathrm{a}.\ (f+g)(x)=f(x)+g(x)=$
$=x-1+2x^{2}$
$=2x^{2}+x-1$
Domain: $\{x|x \in \mathbb{R}\}$.
$\mathrm{b}.\ (f-g)(x)=f(x)-g(x)$
$=(x-1)-(2x^{2})$
$=x-1-2x^{2}$
$=-2x^{2}+x-1$
Domain: $\{x|x \in \mathbb{R}\}$.
$\mathrm{c}.\ (f\cdot g)(x)=f(x)\cdot g(x)$
$=(x-1)(2x^{2})$
$=2x^{3}-2x^{2}$
Domain: $\{x|x \in \mathbb{R}\}$.
$\mathrm{d}.\ (\displaystyle \frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{x-1}{2x^{2}}$
Domain: $\{x|x\neq 0\}$.
$\mathrm{e}.\ (f+g)(3)= $... substitute x=3 in the result of a.
$=2(3)^{2}+3-1$
$=2(9)+3-1$
$=18+3-1$
$=20$
$\mathrm{f}.\ (f-g)(4)= $... substitute x=4 in the result of b.
$=-2(4)^{2}+4-1$
$=-2(16)+4-1$
$=-32+4-1=-29$
$\mathrm{g}.\ (f\cdot g)(2)=$ ... substitute x=2 in the result of c.
$=2(2)^{3}-2(2)^{2}$
$=2(8)-2(4)$
$=16-8$
$=8$
$\mathrm{h}.\ (\displaystyle \frac{f}{g})(1)= $... substitute x=1 in the result of d.
$=\displaystyle \frac{1-1}{2(1)^{2}}$
$=\displaystyle \frac{0}{2(1)}$
$=0$