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\[x^2+4x+4\neq0 \] \[(x+2)^2\neq0\] so the domain interval is:
\[x^2+4x+4\neq0 \] \[(x+2)^2\neq0\] so the domain interval is: \[x \in(-\infty,-2)\cup(-2,+\infty)\]
Check your nos nena..
to check what?
I am not understanding...i was told to find the domain of the function using the factored form F(x)=[-2(x-2)(x+2)]/[3(x+2)(x+2)]....how do I work this out
Sorry my bad
Once factored you must use fact the Denominator cannot be zero for any real no
it's ok...:D kcbrosell what part you don't understand?
all of it I am failing this class big time...I dont understand all the factoring or how to put it into and equation and so on
every time you want to find the domain of the function which is given in fraction form denominator must not be equal to zero, so when you factor x^2+4x+4 you get the result I wrote.....
I am not understanding how you get to the result u wrote
well it's a basic binomial formula which you must know if you want to do some serious math in the future....
I got to this point x^2-4=(x-2)(x+2) & x^2+4x+4=(x^2+2)
you got wrong x^2+2x+4...it is (x+2)^2
o crud...so after I get that worked out do I have to put them into another equation
if terms like (x+2) cancel, there would normally be a "hole" in the graph...but note that there is still a (x+2) in the denominator, so you must set it equal to zero and solve....this is the value that won't work....so the domain is all of the other numbers....the numbers less than -2 and the numbers greater than -2.
so then my answer would be (-oo, -2) (-2, oo)
so it all has to do with your neg and pos and how they factor out
think about what would make the denominator equal to zero...you are correct you often times must factor first to be able to tell.
okay i thinki understand it a little bit better now thank you for your help