Question

What are the certain rules/steps in getting the domain of the operation of functions? namely sum,difference,product and quotient of functions?

Answer 1

for the sum, difference and product... the domain will be the common domain, common to both functions.

so if f + g, then the domain will be the common domain of both f and g; that is, the domain of f + g will be the same as all of the points that are common to the domain of f and the domain of g.

Answer 2

for the quotient, f/g it's still the common domain with the added bit that you have to exclude whatever makes g (the denominator) equal 0. this is because division by 0 is not allowed.

Answer 3

so just get the intersection of both functions for sum, diff and product?

Answer 4

the intersection of the domains

Answer 5

but what if the sum of the functions or diff or product. forms a rational function in the end? would I get the domain of that too?

Answer 6

well, if the sum, diff or product forms a rational function, one of the original functions had to be a rational function. there would already be restriction on the original function so that there was no division by 0.

Answer 7

f(x)=x^2-x-6 g(x)=x+2; find f(x)/g(x)
=>(x+2)(x-3)/x+2
=>x-3

Answer 8

what would the domain be?

Answer 9

oh ok got it thanks... only confused in the domain for quotient

Answer 10

so since f and g are both polynomials, their domains are all real numbers. but because you are dividing by g, you need to find if g can be 0. if it can, then that point in g's domain must be excluded from the domain of f/g.

Answer 11

so what makes g equal 0?

Answer 12

so the domain would be D={x/x not -2} ? even if the resulting function isn't a rational function?

Answer 13

even if you canceled it in the solution?

Answer 14

{\(x|x \in \mathbb{R},\,x \ne -2\)}

Answer 15

yes, even if it cancels out. the fact is that you start with f/g and x = -2 must be excluded.

Answer 16

what happens is you get a "hole" at the point x = -2.

Answer 17

ohhhhh, thanks I'm gonna survive that test now haha! thank you for youe replies and patience..... :D

Answer 18

you're welcome! thanks for sticking with it!

Answer 19

*your