State the domain of [f*g](x) for f(x) = 1/(x-1)^2 and g(x) = x+3.
is it all real numbers??? thanks for helping
2007-09-18 17:37:44 · 4 answers · asked by apromiseibroke 1 in Science & Mathematics ➔ Mathematics
f(g(x)) = h(x) = 1/((x+3)-1)^2
h(x) = 1/(x+2)^2
Denominator cannot equal zero. x + 2 =/= 0, or x =/= -2.
The domain is all real numbers except -2.
or u mean multifunction
2007-09-18 18:36:31 · answer #1 · answered by sanjeewa 4 · 0⤊ 1⤋
It's usually easiest to rewrite the function (until you'e had a lot of practice ☺). So..........
f(x) - 1/(x-1)² and g(x) = x+3 fives
[f*g](x) = 1/(g(x) - 1)² = 1/((x + 3) - 1)² = 1/(x + 2)²
Now....... The domain of this function is pretty obviously the set of all real numbers --except-- -2 (since, at x = -2 the function would be 1/(-2 + 2) = 1/0 which is undefined)
HTH
Doug
2007-09-18 17:46:39 · answer #2 · answered by doug_donaghue 7 · 0⤊ 1⤋
f(g(x)) = h(x) = 1/((x+3)-1)^2
h(x) = 1/(x+2)^2
Denominator cannot equal zero. x + 2 =/= 0, or x =/= -2.
The domain is all real numbers except -2.
2007-09-18 17:44:33 · answer #3 · answered by Matiego 3 · 0⤊ 1⤋
x-1 can't be 0
x can't be 1
You can't have 0 on the bottom of a fraction.
Okay, you are asking for (f*g)(x), not (fog)(x), right? This is correct for (f*g)(x). The other people have given you the answer for (fog)(x). Two TOTALLY different questions!!!!!!!!!!
2007-09-18 17:43:08 · answer #4 · answered by ccw 4 · 0⤊ 0⤋