SOLUTION: {{{f(x)= (3x^2-6)/(x^2-4x)}}} {{{h(x)= 6x+2}}} find the domain of f(h(x)) I have no clue thanks bunches for help

Algebra ->  Rational-functions -> SOLUTION: {{{f(x)= (3x^2-6)/(x^2-4x)}}} {{{h(x)= 6x+2}}} find the domain of f(h(x)) I have no clue thanks bunches for help       Log On


   

Question 547282: f%28x%29=+%283x%5E2-6%29%2F%28x%5E2-4x%29
h%28x%29=+6x%2B2
find the domain of f(h(x))
I have no clue thanks bunches for help
Answer by mathie123(224) About Me  (Show Source):
You can put this solution on YOUR website!
To find f(h(x)), we need to plug in h(x) for any x we see in f(x). This leaves:
f%28x%29=+%283%286x%2B2%29%5E2-6%29%2F%28%286x%2B2%29%5E2-4%286x%2B2%29%29
We could expand this... though it is not needed and will take some time. Instead, we know that there is no limitations on x in the numerator. But we do know that when there is a fraction, the denominator cannot be equal to 0.
This means that
%286x%2B2%29%5E2-4%286x%2B2%29 cannot equal 0.
Well... when does this expression equal 0. In other words, lets solve:
%286x%2B2%29%5E2-4%286x%2B2%29=0
Again, we can expand this, or we could see that each term has a common factor of (6x+2) and factor this out. This leaves:
%286x%2B2%29%28%286x%2B2%29-4%29=0
%286x%2B2%29%286x-2%29=0
If two things multiply to give a product of 0, then at least one of them must be 0. This means
%286x%2B2%29=0
6x=-2
x=-1%2F3
or
%286x-2%29=0
6x=2
x=1%2F3
This means x can be anything except 1%2F3 and -1%2F3.
Hopefully this helps! Please let me know if you still do not understand:)