Answer
(a)$2x^2 + x$; $Domain$ = $all$ $Real$ $numbers$
(b) $x + 2$; $Domain$ = $all$ $Real$ $numbers$
(c) $x^4 + x^3 - x - 1$; $Domain$ = $all$ $Real$ $numbers$
(d) $\frac{x^2 + x + 1}{x^2 - 1}$; $Domain$ = $all$ $Real$ $numbers$ $EXCEPT$ $values$ $\frac{+}{} 1$.
Work Step by Step
$$f(x) = x^2 + x + 1$$ $$g(x) = x^2 - 1$$
(a) $(f + g)(x)$ $$(x^2 + x + 1) + (x^2 - 1)$$ $$2x^2 + x$$ where the $Domain$ is $all$ $Real$ $Numbers$ because the function is a polynomial.
(b) $(f - g)(x)$ $$(x^2 + x + 1) - (x^2 - 1)$$ $$x^2 - x^2 + x + 1 + 1$$ $$x + 2$$ where the $Domain$ is $all$ $Real$ $Numbers$ because the function is a polynomial.
(c) $(fg)(x)$ $$(x^2 + x + 1)(x^2 - 1)$$ $$x^4 - x^2 + x^3 - x + x^2 - 1$$ $$x^4 + x^3 - x - 1$$ where the $Domain$ is $all$ $Real$ $Numbers$ because the function is a polynomial.
(d) $(\frac{f}{g})(x)$ $$\frac{x^2 + x + 1}{x^2 - 1}$$ Since function $f(x)$ cannot be factored, the function $(\frac{f}{g})(x)$ is considered to be simplified. The $Domain$ is, therefore, limited to values where the denominator is not equal to zero, that is: $$x^2 - 1 \neq 0$$ $$(x-1)(x+1) \neq 0$$ $$x \neq \frac{+}{}1$$ The $Domain$ is then $all$ $Real$ $numbers$ $EXCEPT$ $values$ $\frac{+}{} 1$.
