Answer
$f+g,f-g,fg$: all real numbers; $\frac{f}{g}$: all real numbers except $1$
Work Step by Step
For all but $\frac{f}{g}$, the domain includes all real numbers. To find the domain of $\frac{f}{g}$, set the denominator equal to $0$.
$f+g=(6x^2-x-1)+(x-1)$
$f+g=6x^2-2$
$(-∞,∞)$
$f-g=(6x^2-x-1)-(x-1)$
$f-g=6x^2-2x$
$(-∞,∞)$
$fg=(6x^2-x-1)(x-1)$
$fg=6x^3-7x^2+1$
$(-∞,∞)$
$\frac{f}{g}=\frac{6x^2-x-1}{x-1}$
$x\ne1$