Answer
a) $x^3+5x^2-1$; all real numbers
b) $x^3-x^2+1$; all real numbers
c) $3x^5+6x^4-x^3-2x^2$; all real numbers
d) $\frac{x^3+2x^2}{3x^2-1}$; $(-\infty,-\sqrt{\frac{1}{3}})\cup(-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}})\cup(\sqrt{\frac{1}{3}},\infty)$
Work Step by Step
a)
$f+g=(x^3+2x^2)+(3x^2-1)$
$=x^3+5x^2-1$
You can put any number into x and get a valid output, so the domain is all real numbers.
b)
$f-g=(x^3+2x^2)-(3x^2-1)$
$=x^3-x^2+1$
You can put any number into x and get a valid output, so the domain is all real numbers.
c)
$fg=(x^3+2x^2)(3x^2-1)$
Since we are multiplying two binomials, we'll FOIL.
$fg=3x^5-x^3+6x^4-2x^2$
You can put any number into x and get a valid output, so the domain is all real numbers.
d) $f/g=\frac{x^3+2x^2}{3x^2-1}$
Since we don't want the denominator to equal $0$, we must find all the $x$ values that give us a denominator of $0$.
$3x^2-1=0$
$x=-\sqrt{\frac{1}{3}},\thinspace\sqrt{\frac{1}{3}}$
So the domain is all real numbers except for $x=-\sqrt{\frac{1}{3}}$ and $x=\sqrt{\frac{1}{3}}$, or $(-\infty,-\sqrt{\frac{1}{3}})\cup(-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}})\cup(\sqrt{\frac{1}{3}},\infty)$
