Question 198289
# 1

{{{f+g=f(x)+g(x)=2/(x+3)+x/(x+3)=(2+x)/(x+3)}}}



So {{{f+g=(2+x)/(x+3)}}} where the domain is: x can be any real number except {{{x<>-3}}}



----------------------------------------------


# 2

{{{f-g=f(x)-g(x)=2/(x+3)-x/(x+3)=(2-x)/(x+3)}}}



So {{{f-g=(2-x)/(x+3)}}} where the domain is: x can be any real number except {{{x<>-3}}}




----------------------------------------------


# 3

{{{fg=f(x)g(x) = (2/(x+3))(x/(x+3))=(2x)/(x+3)}}}



So {{{fg=(2x)/(x+3)}}} where the domain is: x can be any real number except {{{x<>-3}}}



----------------------------------------------


# 4

{{{f/g=(f(x))/(g(x))=(2/(x+3))/(x/(x+3))=(2/(x+3))((x+3)/x)=(2(x+3))/(x(x+3))=(2*cross((x+3)))/(x*cross((x+3)))=2/x}}}



So {{{f/g=2/x}}} where the domain is: x can be any real number except {{{x<>-3}}} or {{{x<>0}}}



Note: all of the domain restrictions are to avoid division by zero.